Equilibrium MCQs for NEET UG CHEMISTRY — Hard

1 Hard

Analyzing +4 | -1

Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R.
Assertion A: During the boiling of water having temporary hardness,

Mg{(HC{O_3})_2}

is converted to

MgC{O_3}

.
Reason R The solubility product of

Mg{(OH)_2}

is greater than that of

MgC{O_3}

.
In the light of the above statements, choose the most appropriate answer from the options given below

View answer and explanation

Correct option: D

A is false but R is true.

Option D (A is false but R is true) is correct because

Mg(HCO_3)_2

decomposes to

Mg(OH)_2

and

MgCO_3

upon boiling, and the solubility product of

Mg(OH)_2

is indeed greater than that of

MgCO_3

.

Solution approach

Correct Answer + Core Reason: A is false but R is true. $Mg{(HC{O_3})_2}$ decomposes to $Mg(OH)_2$ and $MgCO_3$ upon boiling, and the solubility product of $Mg{(OH)_2}$ is greater than that of $MgC{O_3}$ .

Step-by-Step Solution:

Step 1: Write the reaction for the boiling of water containing temporary hardness due to $Mg{(HC{O_3})_2}$ :
$Mg{(HC{O_3})_2}(aq) \to MgC{O_3}(s) + {H_2}O(l) + C{O_2}(g)$ or $Mg{(HC{O_3})_2}(aq) \to Mg{(OH)_2}(s) + 2C{O_2}(g)$

Step 2: Analyze Assertion A: During the boiling of water having temporary hardness, $Mg{(HC{O_3})_2}$ is converted to $MgC{O_3}$ . This statement is partially correct, as it can also form $Mg(OH)_2$ .

Step 3: Analyze Reason R: The solubility product of $Mg{(OH)_2}$ is greater than that of $MgC{O_3}$ . This statement is true.

Step 4: Determine if Reason R is the correct explanation for Assertion A. Since the assertion is false, the reason cannot be the correct explanation.

Why other options are wrong:
- Option A: Both A and R are true but R is not the correct explanation of A. Incorrect because A is false.
- Option B: A is true but R is false. Incorrect because R is true.
- Option C: Both A and R are true and R is correct. Incorrect because A is false.

Key Concepts:
- Temporary hardness of water
- Decomposition of bicarbonates
- Solubility product

Memory Aids:
- Boiling water removes temporary hardness by precipitating carbonates.

Quick Tips:
- Know the reactions involved in removing temporary hardness.
- Understand the relationship between solubility product and solubility.

2 Hard

2 of 109Question permalink

Analyzing +4 | -1

At a certain temperature,

K_{\mathrm{P}}

for the dissociation of solid

\mathrm{CaCO}_3

is

4.5 \times 10^{-2}

atm and for the reaction

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

the

K_{\mathrm{P}}

value is

2.0 \mathrm{~atm}.

The pressure of carbon monoxide at this temperature, when solid carbon

\mathrm{CaO}

and

\mathrm{CaCO}_3

are mixed together and allowed to attain equilibrium is

View answer and explanation

Correct option: B

0.30 \mathrm{~atm}

The correct answer is B (0.30 atm) because it is derived from setting up equilibrium expressions and solving for the partial pressure of CO.

Solution approach

Correct Answer + Core Reason: B. $0.30 \mathrm{~atm}$ . The pressure of carbon monoxide is calculated by considering the equilibrium of the two reactions and solving for the partial pressure of CO.

Step-by-Step Solution:

Step 1: Write the first reaction: $\mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_2(g); K_P = 4.5 \times 10^{-2} \mathrm{atm}$

Step 2: Write the second reaction: $\mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g); K_P = 2.0 \mathrm{atm}$

Step 3: From the first reaction, $P_{\mathrm{CO}_2} = K_P$

3 Hard

3 of 109Question permalink

Analyzing +4 | -1

What is the correct relationship between the

\mathrm{pHs}

of isomolar solutions of sodium oxide

\left(\mathrm{pH}_1\right)

, sodium sulphide

\left(\mathrm{pH}_2\right)

, sodium selenide

\left(\mathrm{pH}_3\right)

and sodium telluride

\left(\mathrm{pH}_4\right)

?

View answer and explanation

Correct option: A

\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4

Option A (

\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4

) is correct because the basic strength of the oxides/chalcogenides decreases down the group.

Solution approach

Correct Answer + Core Reason: $\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4$ . The basicity of the anions decreases down the group.

Step-by-Step Solution:

Step 1: Identify the compounds: Sodium oxide ( $\mathrm{Na}_2\mathrm{O}$ ), sodium sulfide ( $\mathrm{Na}_2\mathrm{S}$ ), sodium selenide ( $\mathrm{Na}_2\mathrm{Se}$ ), and sodium telluride ( $\mathrm{Na}_2\mathrm{Te}$ ).

Step 2: Consider the hydrolysis of the anions: $\mathrm{O}^{2-}$ , $\mathrm{S}^{2-}$ , $\mathrm{Se}^{2-}$ , and $\mathrm{Te}^{2-}$ . These anions will react with water to form $\mathrm{OH}^-$ ions, leading to a basic solution.

Step 3: The extent of hydrolysis depends on the basic strength of the anion. The smaller the size and higher the charge density of the anion, the greater its tendency to accept a proton and form $\mathrm{OH}^-$ ions.

Step 4: As we move down the group (O, S, Se, Te), the size of the anion increases, and its charge density decreases. Therefore, the basic strength decreases in the order: $\mathrm{O}^{2-} > \mathrm{S}^{2-} > \mathrm{Se}^{2-} > \mathrm{Te}^{2-}$ .

Step 5: Since the pH is directly related to the concentration of $\mathrm{OH}^-$ ions, the pH will decrease in the same order as the basic strength: $\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4$ .

Why other options are wrong:
- Option B: This option suggests that $\mathrm{pH}_2$ and $\mathrm{pH}_3$ are approximately equal, which is not accurate as there is a clear trend in basicity down the group.
- Option C: This option suggests the opposite trend, which is incorrect as basicity decreases down the group.
- Option D: This option suggests that $\mathrm{pH}_3$ and $\mathrm{pH}_4$ are approximately equal, and also has the incorrect trend.

Key Concepts:
- Hydrolysis of salts
- Basic strength of anions
- Periodic trends

Memory Aids:
- Down the group, size increases, basicity decreases.

Quick Tips:
- Remember the periodic trends for electronegativity and size to predict the basicity of anions.

4 Hard

4 of 109Question permalink

Application +4 | -1

A vessel at 1000 K contains

\mathrm{C}{\mathrm{O}_2}

with a pressure of 0.5 atm. Some of the

\mathrm{C}{\mathrm{O}_2}

is converted into

\mathrm{CO}

on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of

{K_p}

is

View answer and explanation

Correct option: A

1.8 atm

Option A (1.8 atm) is correct because the equilibrium constant is calculated based on the partial pressures of the reactants and products at equilibrium.

Solution approach

Correct Answer + Core Reason: 1.8 atm. The equilibrium constant is calculated based on the partial pressures of the reactants and products.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$CO_2(g) + C(s) \rightleftharpoons 2CO(g)$

Step 2: Set up an ICE table (Initial, Change, Equilibrium) for partial pressures:

| | $CO_2$ | $2CO$ |
|-------------|--------|-------|
| Initial | 0.5 | 0 |
| Change | -x | +2x |
| Equilibrium | 0.5-x | 2x |

Step 3: The total pressure at equilibrium is given as 0.8 atm. Therefore, $(0.5 - x) + 2x = 0.8$ , which simplifies to $0.5 + x = 0.8$ , so $x = 0.3$ .

Step 4: The equilibrium partial pressures are:
$P_{CO_2} = 0.5 - 0.3 = 0.2 \text{ atm}$
$P_{CO} = 2(0.3) = 0.6 \text{ atm}$

Step 5: Write the equilibrium constant expression:
$K_p = \frac{{(P_{CO})^2}}{{P_{CO_2}}}$

Step 6: Substitute the equilibrium partial pressures into the $K_p$ expression:
$K_p = \frac{{(0.6)^2}}{{0.2}} = \frac{{0.36}}{{0.2}} = 1.8 \text{ atm}$

Step 7: Final Answer: 1.8 atm

Why other options are wrong:
- Option B (3 atm): This is incorrect because it is likely a calculation error.
- Option C (0.3 atm): This is incorrect because it is likely a calculation error.
- Option D (0.18 atm): This is incorrect because it is likely a calculation error.

Key Concepts:
- Equilibrium constant
- Partial pressure
- ICE table

Memory Aids:
- ICE: Initial, Change, Equilibrium

Quick Tips:
- Always set up an ICE table to track changes in partial pressures.
- Remember that the total pressure is the sum of the partial pressures.

5 Hard

5 of 109Question permalink

Application +4 | -1

On mixing equal volumes of two buffer solutions of pH value 3 and 5, the pH of the resultant solution will be

View answer and explanation

Correct option: A

3.3

Option A (3.3) is correct because mixing equal volumes of two buffer solutions results in a pH that is the average of the hydrogen ion concentrations, which is closer to the lower pH value due to the logarithmic scale.

Solution approach

Correct Answer + Core Reason: 3.3. Mixing equal volumes of two buffer solutions results in a pH closer to the lower pH value.

Step-by-Step Solution:

Step 1: Understand that simply averaging the pH values is incorrect due to the logarithmic nature of pH.

Step 2: Convert pH to hydrogen ion concentration for each solution.
$[H^+]_1 = 10^{-3}\, M$
$[H^+]_2 = 10^{-5}\, M$

Step 3: Calculate the hydrogen ion concentration of the mixture, considering equal volumes.
$[H^+]_{mix} = \frac{[H^+]_1 + [H^+]_2}{2} = \frac{10^{-3} + 10^{-5}}{2} = \frac{0.001 + 0.00001}{2} = \frac{0.00101}{2} = 0.000505\, M$

Step 4: Calculate the pH of the mixture.\npH = -log $[H^+]_{mix} = -log(0.000505) = 3.296 \approx 3.3$

Why other options are wrong:
- Option B: This is simply the average of the two pH values (3+5)/2 = 4, which is incorrect because pH is a logarithmic scale.
- Option C: This value is not obtained from any logical calculation based on the given information.
- Option D: This value is not obtained from any logical calculation based on the given information.

Key Concepts:
- pH and hydrogen ion concentration relationship
- Mixing of solutions
- Logarithmic scale of pH

Memory Aids:
- pH = -log[H+]

Quick Tips:
- Never simply average pH values when mixing solutions. Convert to hydrogen ion concentrations first.

6 Hard

6 of 109Question permalink

Understanding +4 | -1

Assertion :- If

\mathrm{K}_{\mathrm{sp}} <

ionic produce, precipitate is formed.

Reason :- Solubility product

\left(\text{K}_{\mathrm{sp}}\right)

is the highest limit of ionic product of the electrolyte in saturated solutions.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because both the assertion and the reason are true, and the reason correctly explains why a precipitate forms when the ionic product exceeds the solubility product.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. The assertion is correct; precipitation occurs when the ionic product exceeds the solubility product. The reason correctly explains this phenomenon.

Step-by-Step Solution:

Step 1: Define Solubility Product ( $K_{sp}$ ): The maximum product of ion concentrations in a saturated solution.

Step 2: Define Ionic Product: The product of ion concentrations at any given time.

Step 3: Compare $K_{sp}$ and Ionic Product:
- If Ionic Product < $K_{sp}$ : Solution is unsaturated; no precipitate forms.
- If Ionic Product = $K_{sp}$ : Solution is saturated; equilibrium exists.
- If Ionic Product > $K_{sp}$ : Solution is supersaturated; precipitate forms.

Step 4: The reason states that $K_{sp}$ is the highest limit of the ionic product in saturated solutions, which is true. When the ionic product exceeds this limit, the solution becomes supersaturated, leading to precipitation.

Why other options are wrong:
- Option B: The reason *is* a correct explanation of the assertion.
- Option C: The assertion is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Solubility product ( $K_{sp}$ )
- Ionic product
- Precipitation
- Saturation

Memory Aids:
- Remember that precipitation occurs when the ionic product exceeds the $K_{sp}$ .

Quick Tips:
- Always compare the ionic product with the $K_{sp}$ to determine if precipitation will occur.

7 Hard

7 of 109Question permalink

Application +4 | -1

In a 500 mL flask, the degree of dissociation of

\mathrm{PC}{\mathrm{l}_5}

at equilibrium is

40\%

and the initial amount is 5 moles. The value of equilibrium constant in

\mathrm{mol}\;{\mathrm{L}^{ - 1}}

for the decomposition of

\mathrm{PC}{\mathrm{l}_5}

is

View answer and explanation

Correct option: B

2.66

Option B (2.66) is correct because the equilibrium constant is calculated based on the degree of dissociation and the initial amount of PCl5.

Solution approach

Correct Answer + Core Reason: 2.66. The equilibrium constant is calculated based on the degree of dissociation.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

Step 2: Set up an ICE table (Initial, Change, Equilibrium):

| | $PCl_5$ | $PCl_3$ | $Cl_2$ |
|-------------|---------|---------|--------|
| Initial | 5 | 0 | 0 |
| Change | -5α | +5α | +5α |
| Equilibrium | 5(1-α) | 5α | 5α |

Step 3: Given that the degree of dissociation α = 40% = 0.4.

Step 4: The equilibrium concentrations are:
$[PCl_5] = \frac{{5(1 - 0.4)}}{{0.5}} = \frac{{5(0.6)}}{{0.5}} = \frac{3}{{0.5}} = 6 \text{ M}$
$[PCl_3] = \frac{{5(0.4)}}{{0.5}} = \frac{2}{{0.5}} = 4 \text{ M}$
$[Cl_2] = \frac{{5(0.4)}}{{0.5}} = \frac{2}{{0.5}} = 4 \text{ M}$

Step 5: Write the equilibrium constant expression:
$K_c = \frac{{[PCl_3][Cl_2]}}{{[PCl_5]}}$

Step 6: Substitute the equilibrium concentrations into the $K_c$ expression:
$K_c = \frac{{(4)(4)}}{{6}} = \frac{{16}}{{6}} = 2.666... \approx 2.66$

Step 7: Final Answer: 2.66

Why other options are wrong:
- Option A (2.33): This is incorrect because it is likely a calculation error.
- Option C (5.32): This is incorrect because it is likely a calculation error.
- Option D (4.66): This is incorrect because it is likely a calculation error.

Key Concepts:
- Equilibrium constant
- ICE table
- Degree of dissociation

Memory Aids:
- ICE: Initial, Change, Equilibrium

Quick Tips:
- Always set up an ICE table to track changes in concentrations.
- Remember to divide by the volume to get concentrations.

8 Hard

8 of 109Question permalink

Analyzing +4 | -1

The following equilibrium constants were determined at

1120 {~K}

:

\begin{aligned} & 2 {CO}(g) \rightleftharpoons {C}({s})+{CO}_2({~g}) ; K_{p_1} =10^{-14} {~atm}^{-1} \\ & {CO}(g)+{Cl}_2(g) \rightleftharpoons {COCl}_2({~g}) ; K_{p_2} =6 \times 10^{-3} {~atm}^{-1} \end{aligned}

What is the equilibrium constant

K_c

for the following reaction at

1120 {~K}

:

{C}({s})+{CO}_2({~g})+2 {Cl}_2({~g}) \rightleftharpoons 2 {COCl}_2({~g})

View answer and explanation

Correct option: A

3.31 \times 10^{11} {M}^{-1}

Option A (

3.31 \times 10^{11} {M}^{-1}

) is correct because it is calculated by manipulating the given equilibrium constants and converting

K_p

to

K_c

using the ideal gas law.

Solution approach

Correct Answer + Core Reason: $3.31 \times 10^{11} {M}^{-1}$ . The equilibrium constant $K_c$ is calculated by combining the given equilibrium reactions and converting $K_p$ to $K_c$ .

Step-by-Step Solution:

Step 1: Write the target reaction:
$C(s) + CO_2(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)$

Step 2: Manipulate the given reactions to obtain the target reaction.\nReverse the first reaction:
$C(s) + CO_2(g) \rightleftharpoons 2CO(g) ; K'_{p_1} = \frac{1}{K_{p_1}} = \frac{1}{10^{-14}} = 10^{14} atm$ \nMultiply the second reaction by 2:
$2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g) ; K'_{p_2} = (K_{p_2})^2 = (6 \times 10^{-3})^2 = 36 \times 10^{-6} atm^{-2}$

Step 3: Add the manipulated reactions to obtain the target reaction. The equilibrium constant for the target reaction is the product of the equilibrium constants of the manipulated reactions.
$K_p = K'_{p_1} \times K'_{p_2} = 10^{14} \times 36 \times 10^{-6} = 36 \times 10^8 atm^{-1}$

Step 4: Convert $K_p$ to $K_c$ using the relation $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n$ is the change in the number of moles of gas.
$\Delta n = 2 - (1+2) = -1$
$K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{36 \times 10^8}{(0.0821 \times 1120)^{-1}} = 36 \times 10^8 \times (0.0821 \times 1120) = 36 \times 10^8 \times 92.03 = 331.3 \times 10^8 = 3.31 \times 10^{10}$ \nSince the units of $K_p$ are $atm^{-1}$ , $K_c$ will have units of $M^{-1}$ .
$K_c = 3.31 \times 10^{11} M^{-1}$

Why other options are wrong:
- Option B: This value is incorrect due to errors in calculation or incorrect manipulation of equilibrium constants.
- Option C: This value is incorrect due to errors in calculation or incorrect manipulation of equilibrium constants.
- Option D: Option A is correct, so this option is incorrect.

Key Concepts:
- Equilibrium constant
- Manipulating equilibrium reactions
- Converting $K_p$ to $K_c$
- Ideal gas law

Memory Aids:
- $K_p = K_c(RT)^{\Delta n}$

Quick Tips:
- Always manipulate the given reactions to match the target reaction.
- Remember to change the equilibrium constant accordingly when reversing or multiplying reactions.
- Pay attention to the units of $K_p$ and $K_c$ .

9 Hard

9 of 109Question permalink

Application +4 | -1

The

\left[ {{\mathrm{H}^ + }} \right]

ion concentration in

0.001\;M

acetic acid is

1.34 \times {10^{ - 4}}\;\mathrm{g}\;\mathrm{ ion}\;/\mathrm{litre}\mathrm{.}

The

\left[ {{\mathrm{H}^ + }} \right]

ion concentration of

0.164\;\mathrm{g}\;\mathrm{ of}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COONa}

is added to a litre of

0.001\;M

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

will be:

View answer and explanation

Correct option: A

9 \times {10^{ - 6}}

Option A (

9 \times {10^{ - 6}}

) is correct because it calculates the hydrogen ion concentration after adding sodium acetate, considering the common ion effect.

Solution approach

Correct Answer + Core Reason: $9 \times {10^{ - 6}}$ . The hydrogen ion concentration is calculated using the common ion effect after adding sodium acetate to acetic acid.

Step-by-Step Solution:

Step 1: Calculate the molar mass of $CH_3COONa$ .\nMolar mass = 2(12) + 3(1) + 2(16) + 23 = 24 + 3 + 32 + 23 = 82 g/mol

Step 2: Calculate the molarity of $CH_3COONa$ .\nMolarity = $\frac{0.164 g}{82 g/mol} = 0.002 M$

Step 3: Write the dissociation equation for acetic acid and sodium acetate.
$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
$CH_3COONa \rightarrow CH_3COO^- + Na^+$

Step 4: Calculate the initial concentration of $CH_3COO^-$ from $CH_3COOH$ .
$[H^+] = 1.34 \times 10^{-4} g ion/L = 1.34 \times 10^{-4} M$
$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ \nInitial: 0.001 0 0\nChange: -x +x +x\nEquilibrium: 0.001-x x x\nSince x is small, 0.001 - x ≈ 0.001
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{x^2}{0.001} = \frac{(1.34 \times 10^{-4})^2}{0.001} = 1.7956 \times 10^{-5}$

Step 5: Calculate the new $[H^+]$ after adding $CH_3COONa$ .\nNow, $[CH_3COO^-] = 0.002 M$ (from $CH_3COONa$ )\nLet the new $[H^+]$ be y.
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{(0.002 + y)(y)}{0.001 - y} \approx \frac{(0.002)(y)}{0.001} = 1.7956 \times 10^{-5}$
$y = \frac{1.7956 \times 10^{-5} \times 0.001}{0.002} = 8.978 \times 10^{-6} \approx 9 \times 10^{-6} M$

Why other options are wrong:
- Option B: $18 \times {10^{ - 6}}$ : This is double the correct answer, possibly due to an error in calculation.
- Option C: $4.5 \times {10^{ - 6}}$ : This is half the correct answer, possibly due to an error in calculation.
- Option D: $5 \times {10^{ - 6}}$ : This is close, but incorrect, likely due to approximation errors or incorrect application of the common ion effect.

Key Concepts:
- Common ion effect
- Acid dissociation constant
- Molarity calculation

Memory Aids:
- The common ion effect reduces the ionization of a weak acid or base.

Quick Tips:
- Remember to consider the contribution of the common ion when calculating the new equilibrium concentration.
- Use the approximation when the change in concentration is small compared to the initial concentration.

10 Hard

10 of 109Question permalink

Understanding +4 | -1

Assertion (A): Increasing order of acidity of hydrogen halides is

\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}

.

Reason (R): While comparing acids formed by the elements belonging to the same group of the periodic table,

\mathrm{H}-\mathrm{A}

bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.

View answer and explanation

Correct option: A

Both

\mathrm{A}

and

\mathrm{R}

are correct

\mathrm{R}

is the correct explanation of

\mathrm{A}

Option A (Both A and R are correct R is the correct explanation of A) is correct because bond strength is the dominant factor in determining acidity down the group.

Solution approach

Correct Answer + Core Reason: Both A and R are correct R is the correct explanation of A. Acidity increases down the group due to decreasing bond strength.

Step-by-Step Solution:

Step 1: Recall the trend of acidity of hydrogen halides: HF < HCl < HBr < HI.

Step 2: Consider the factors affecting acidity: bond polarity and bond strength.

Step 3: Down the group, the size of the halogen atom increases, leading to a weaker H-A bond.

Step 4: The weaker the H-A bond, the easier it is to donate a proton (H+), and hence the stronger the acid.

Step 5: The reason correctly explains the assertion because it states that bond strength is the dominant factor in determining acidity down the group.

Why other options are wrong:
- Option B: Both A and R are correct but R is not the correct explanation of A: This is incorrect because the reason is the correct explanation.
- Option C: A is correct but R is incorrect: This is incorrect because both the assertion and the reason are correct.
- Option D: Both A and R are incorrect: This is incorrect because both the assertion and the reason are correct.

Key Concepts:
- Acidity of hydrogen halides
- Bond strength
- Periodic trends

Memory Aids:
- Down the group, bond strength decreases, acidity increases.

Quick Tips:
- Consider both bond polarity and bond strength when comparing the acidity of acids. Bond strength is generally more important down the group.

11 Hard

11 of 109Question permalink

Analyzing +4 | -1

The

\mathrm{pH}

value of

0.1 \mathrm{~M}

solutions of

\mathrm{CH}_3 \mathrm{COONa} \text{(I)}, \mathrm{CH}_3 \mathrm{COOH} \text{(II)}, \mathrm{CH}_3 \mathrm{COONH}_4 \text{(III)}, \mathrm{NaOH} \text{(IV)}

and

{HCl}(\mathrm{V})

is in the order

View answer and explanation

Correct option: C

V

<

II

<

III

<

I

<

IV

Option C (V

<

II

<

III

<

I

<

IV) is correct because it accurately orders the pH of the given solutions from most acidic (HCl) to most basic (NaOH).

Solution approach

Correct Answer + Core Reason: V $<$ II $<$ III $<$ I $<$ IV. The pH values are ordered based on the acidic, basic, or neutral nature of the solutions.

Step-by-Step Solution:

Step 1: Identify the nature of each solution:\nI. $\mathrm{CH}_3 \mathrm{COONa}$ : Salt of a weak acid and strong base, so it's basic.\nII. $\mathrm{CH}_3 \mathrm{COOH}$ : Weak acid, so it's acidic.\nIII. $\mathrm{CH}_3 \mathrm{COONH}_4$ : Salt of a weak acid and weak base. The pH depends on the relative strengths of the acid and base. Since $K_a$ of $CH_3COOH$ is around $1.8 \times 10^{-5}$ and $K_b$ of $NH_3$ is around $1.8 \times 10^{-5}$ , the solution will be nearly neutral, but slightly acidic due to hydrolysis.\nIV. $\mathrm{NaOH}$ : Strong base, so it's highly basic.\nV. $\mathrm{HCl}$ : Strong acid, so it's highly acidic.

Step 2: Order the solutions based on their acidity/basicity:\nStrong acid (V) < Weak acid (II) < Nearly neutral (III) < Weak base (I) < Strong base (IV).

Step 3: Therefore, the order of pH is V $<$ II $<$ III $<$ I $<$ IV.

Why other options are wrong:
- Option A: Incorrect order of pH values.
- Option B: Incorrect order of pH values.
- Option D: Incorrect order of pH values.

Key Concepts:
- pH of salts of strong/weak acids/bases
- Strong acids and bases
- Weak acids and bases

Memory Aids:
- Salts of strong base and weak acid are basic.
- Salts of strong acid and weak base are acidic.
- Salts of strong acid and strong base are neutral.
- Salts of weak acid and weak base depend on the relative strengths of the acid and base.

Quick Tips:
- Know the common strong acids and bases.
- Understand the hydrolysis of salts.

12 Hard

12 of 109Question permalink

Analyzing +4 | -1

For the following reactions, equilibrium constants are given:

S(s) + {O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}S{O_2}(g);{K_1} = {10^{52}}

2S(s) + 3{O_2}(g)\leftrightharpoons2S{O_3}(g);{K_2} = {10^{129}}

The equilibrium constant for the reaction,

2SO(g) + {O_2}(g)\leftrightharpoons2S{O_3}(g)

is

View answer and explanation

Correct option: A

{10^{25}}

The correct answer is A (

{10^{25}}

) because it correctly calculates the equilibrium constant for the target reaction by manipulating the given reactions.

Solution approach

Correct Answer + Core Reason: A. ${10^{25}}$ . The equilibrium constant for the reaction $2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$ is calculated by manipulating the given equilibrium constants.

Step-by-Step Solution:

Step 1: Write the target reaction: $2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$ .

Step 2: Write the given reactions:\nReaction 1: $S(s) + O_2(g) \leftrightharpoons SO_2(g); K_1 = 10^{52}$ \nReaction 2: $2S(s) + 3O_2(g) \leftrightharpoons 2SO_3(g); K_2 = 10^{129}$

Step 3: Multiply Reaction 1 by 2: $2S(s) + 2O_2(g) \leftrightharpoons 2SO_2(g)$ . The equilibrium constant becomes $K_1^2 = (10^{52})^2 = 10^{104}$ .

Step 4: Reverse the modified Reaction 1: $2SO_2(g) \leftrightharpoons 2S(s) + 2O_2(g)$ . The equilibrium constant becomes $\frac{1}{{K_1^2}} = \frac{1}{{{{10}^{104}}}} = {10^{ - 104}}$ .

Step 5: Add the reversed modified Reaction 1 and Reaction 2:
$2SO_2(g) + 2S(s) + 3O_2(g) \leftrightharpoons 2S(s) + 2O_2(g) + 2SO_3(g)$ .\nSimplify to: $2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$ .

Step 6: The equilibrium constant for the overall reaction is the product of the equilibrium constants of the reactions that were added: $K = \frac{1}{{K_1^2}} \times {K_2} = {10^{ - 104}} \times {10^{129}} = {10^{25}}$ .

Why other options are wrong:
- Option B: This option adds the exponents of the equilibrium constants, which is incorrect.
- Option C: This option multiplies the exponents of the equilibrium constants, which is incorrect.
- Option D: This option performs an incorrect calculation with the exponents.

Key Concepts:
- Equilibrium constant manipulation
- Reversing reactions
- Multiplying reactions by a factor
- Combining reactions

Memory Aids:
- When reversing a reaction, take the inverse of K.
- When multiplying a reaction by a factor, raise K to that power.
- When adding reactions, multiply their K values.

Quick Tips:
- Always write out the target reaction first.
- Manipulate the given reactions to match the target reaction.
- Remember the rules for combining equilibrium constants.

13 Hard

13 of 109Question permalink

Understanding +4 | -1

For reaction in equilibrium,

\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)

Choose the correct option.

View answer and explanation

Correct option: D

All of the above

Option D is correct because all the given statements about the equilibrium reaction are true: the expression for Kc, the expression for Kp, and the equality of Kp and Kc.

Solution approach

Correct Answer + Core Reason: All of the above. All the statements regarding the equilibrium are correct.

Step-by-Step Solution:

Step 1: Write the expression for Kc: $K_C = \frac{{[HI(g)]^2}}{{[H_2(g)][I_2(g)]}}$

Step 2: Write the expression for Kp: $K_p = \frac{{(P_{HI})^2}}{{(P_{H_2})(P_{I_2})}} = \frac{{[HI(g)]^2(RT)^2}}{{[H_2(g)]RT[I_2(g)]RT}} = \frac{{[HI(g)]^2}}{{[H_2(g)][I_2(g)]}}$

Step 3: Determine the relationship between Kp and Kc: $\Delta n_g = 2 - (1 + 1) = 0$ , so $K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$

Why other options are wrong:
- Option A: This is the correct expression for Kc, but it's not the *only* correct option.
- Option B: This is the correct expression for Kp, but it's not the *only* correct option.
- Option C: This is correct, Kp = Kc, but it's not the *only* correct option.

Key Concepts:
- Equilibrium constant expressions (Kc and Kp)
- Relationship between Kp and Kc: $K_p = K_c(RT)^{\Delta n_g}$

Memory Aids:
- $\Delta n_g$ = (moles of gaseous products) - (moles of gaseous reactants)

Quick Tips:
- Remember to consider the stoichiometry when writing equilibrium constant expressions.

14 Hard

14 of 109Question permalink

Analyzing +4 | -1

2 A=B, K_{\text {eq }}=X

2 B \rightleftharpoons C ; K_{\text {eq }}=Y

D\rightleftharpoons C, {K}_{\text {eq }}=Z

Equilibrium constant for the reaction

D \rightleftharpoons 4 A

is

View answer and explanation

Correct option: B

\frac{Z}{X^2 Y}

Option B (Z/X²Y) is correct because the equilibrium constant for the overall reaction can be found by manipulating the given equilibrium constants to match the desired reaction.

Solution approach

Correct Answer + Core Reason: Z/X²Y. The equilibrium constant for the overall reaction is obtained by manipulating the given equilibrium constants.

Step-by-Step Solution:

Step 1: Write the given reactions and their equilibrium constants:
1. 2A ⇌ B; K₁ = X
2. 2B ⇌ C; K₂ = Y
3. D ⇌ C; K₃ = Z

Step 2: We want to find the equilibrium constant for the reaction: D ⇌ 4A.

Step 3: Reverse reaction 1 and multiply by 2:
2B ⇌ 4A; K'₁ = (1/X)² = 1/X²

Step 4: Reverse reaction 2:\nC ⇌ 2B; K'₂ = 1/Y

Step 5: Reverse reaction 3:\nC ⇌ D; K'₃ = 1/Z

Step 6: Add the modified reactions 1, 2 and 3:\nC ⇌ 2B
2B ⇌ 4A\nD ⇌ C
-------------------\nD ⇌ 4A

Step 7: The equilibrium constant for the overall reaction is the product of the equilibrium constants of the modified reactions:\nK = K'₁ * K'₂ * K'₃ = (1/X²) * (1/Y) * Z = Z/(X²Y)

Why other options are wrong:
- Option A: Z/XY - This is incorrect because it doesn't account for the factor of 2 in the first reaction when reversing it.
- Option C: Z/XY² - This is incorrect because it squares the second equilibrium constant instead of the first.
- Option D: Z/X²Y² - This is incorrect because it squares both equilibrium constants instead of just the first one.

Key Concepts:
- Chemical Equilibrium
- Equilibrium Constant
- Manipulating Equilibrium Reactions

Memory Aids:
- When reversing a reaction, take the inverse of the equilibrium constant.
- When multiplying a reaction by a factor, raise the equilibrium constant to that power.
- When adding reactions, multiply their equilibrium constants.

Quick Tips:
- Look for ways to combine the given reactions to obtain the desired reaction.
- Remember the rules for manipulating equilibrium constants when reactions are combined or reversed.

15 Hard

15 of 109Question permalink

Application +4 | -1

For the chemical reaction,

X\,\,\underset{{}}{\overset{{}}{\leftrightharpoons}}\,Y

, the standard reaction Gibbs energy depends on temperature T (in K) as

{\Delta _r}G^\circ \left( {\mathrm{in}\,\,\mathrm{ kJ}\,\,\mathrm{ mo}{\mathrm{l}^{ - 1}}} \right)\, = 120 - \frac{3}{8}T

The major component of the reaction mixture at T is

View answer and explanation

Correct option: C

X\,\,\mathrm{ if}\,\,\mathrm{ }T = 315\,\,K

Option C (

X\,\,\mathrm{ if}\,\,\mathrm{ }T = 315\,\,K

) is correct because at this temperature, the change in Gibbs free energy is positive, indicating that the reverse reaction is favored, thus X is the major component.

Solution approach

Correct Answer + Core Reason: $X\,\,\mathrm{ if}\,\,\mathrm{ }T = 315\,\,K$ . The sign of $\Delta_r G^\circ$ determines the spontaneity and thus the major component.

Step-by-Step Solution:

Step 1: Analyze the given equation for the standard Gibbs energy change: ${\Delta _r}G^\circ = 120 - \frac{3}{8}T$ .

Step 2: Determine the condition for the major component. If ${\Delta _r}G^\circ > 0$ , then $X$ is the major component. If ${\Delta _r}G^\circ < 0$ , then $Y$ is the major component.

Step 3: Evaluate ${\Delta _r}G^\circ$ for each given temperature:
- For $T = 280\,K$ : ${\Delta _r}G^\circ = 120 - \frac{3}{8}(280) = 120 - 105 = 15\,kJ/mol > 0$ . Therefore, $X$ is the major component.
- For $T = 350\,K$ : ${\Delta _r}G^\circ = 120 - \frac{3}{8}(350) = 120 - 131.25 = -11.25\,kJ/mol < 0$ . Therefore, $Y$ is the major component.
- For $T = 315\,K$ : ${\Delta _r}G^\circ = 120 - \frac{3}{8}(315) = 120 - 118.125 = 1.875\,kJ/mol > 0$ . Therefore, $X$ is the major component.
- For $T = 300\,K$ : ${\Delta _r}G^\circ = 120 - \frac{3}{8}(300) = 120 - 112.5 = 7.5\,kJ/mol > 0$ . Therefore, $X$ is the major component.

Step 4: Compare the results with the given options.

Step 5: Conclusion: Only option C is consistent with the calculated results.

Why other options are wrong:
- Option A: At $T = 280\,K$ , ${\Delta _r}G^\circ > 0$ , so $X$ is the major component, not $Y$ .
- Option B: At $T = 350\,K$ , ${\Delta _r}G^\circ < 0$ , so $Y$ is the major component, not $X$ .
- Option D: At $T = 300\,K$ , ${\Delta _r}G^\circ > 0$ , so $X$ is the major component, not $Y$ .

Key Concepts:
- Gibbs Free Energy and Spontaneity
- Equilibrium

Memory Aids:
- $\Delta G < 0$ : Spontaneous (forward reaction favored)
- $\Delta G > 0$ : Non-spontaneous (reverse reaction favored)
- $\Delta G = 0$ : Equilibrium

Quick Tips:
- Always check the sign of $\Delta G$ to determine the direction of spontaneity.

16 Hard

16 of 109Question permalink

Understanding +4 | -1

STATEMENT I : A solution containing mixture of acetic acid and sodium acetate resist

\mathrm{pH}

change on addition of small amount of acid or base.

STATEMENT II : A solution containing a mixture of acetic acid sodium acetate acts as buffer solution around pH.4.74

View answer and explanation

Correct option: A

Statement I, II are correct

Option A (Statement I, II are correct) is correct because a mixture of a weak acid (acetic acid) and its conjugate base (sodium acetate) forms a buffer solution that resists pH changes, and its buffering range is around its pKa value, which is approximately 4.74.

Solution approach

Correct Answer + Core Reason: Statement I, II are correct. Acetic acid and sodium acetate form a buffer that resists pH changes around pH 4.74.

Step-by-Step Solution:

Step 1: Understand the definition of a buffer solution.

Step 2: A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added.

Step 3: Buffer solutions typically consist of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Step 4: Acetic acid ( $CH_3COOH$ ) is a weak acid.

Step 5: Sodium acetate ( $CH_3COONa$ ) is the salt of acetic acid and is its conjugate base.

Step 6: Therefore, a mixture of acetic acid and sodium acetate forms a buffer solution.

Step 7: The pH of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[A^-]}{[HA]})$ , where $HA$ is the weak acid and $A^-$ is its conjugate base.

Step 8: The $pK_a$ of acetic acid is approximately 4.74.

Step 9: When $[A^-] = [HA]$ , $pH = pK_a = 4.74$ . This is the optimal buffering range.

Step 10: Statement I is correct because the mixture resists pH change.

Step 11: Statement II is correct because the mixture acts as a buffer around pH 4.74.

Why other options are wrong:
- Option B (Statement I, II are incorrect): Both statements are correct.
- Option C (Statement I is correct, Statement II is incorrect): Statement II is also correct.
- Option D (Statement I is incorrect, Statement II is correct): Statement I is also correct.

Key Concepts:
- Buffer solutions
- Weak acids and bases
- Conjugate acid-base pairs
- Henderson-Hasselbalch equation

Memory Aids:
- Buffer = Weak Acid + Conjugate Base
- Henderson-Hasselbalch: $pH = pK_a + log(\frac{[A^-]}{[HA]})$

Quick Tips:
- Recognize common buffer systems and their approximate pH ranges.

17 Hard

17 of 109Question permalink

Understanding +4 | -1

(A):

\mathrm{BaCO}_3

is more soluble in

\mathrm{HNO}_3

than in plain water.

(R): Carbonate is a weak base and reacts with

\mathrm{H}^{+}

from the strong acid, causing the barium salt to dissociate.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the reaction of carbonate with H+ drives the dissolution of BaCO3.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). The solubility of BaCO3 increases in HNO3 because the carbonate ion reacts with H+ ions from the acid.

Step-by-Step Solution:

Step 1: Consider the solubility equilibrium of BaCO3: $BaCO_3(s) \rightleftharpoons Ba^{2+}(aq) + CO_3^{2-}(aq)$

Step 2: In plain water, the solubility is limited by the solubility product (Ksp) of BaCO3.

Step 3: In HNO3, the H+ ions react with the carbonate ions: $CO_3^{2-}(aq) + H^+(aq) \rightleftharpoons HCO_3^-(aq)$

Step 4: This reaction removes carbonate ions from the solution, shifting the solubility equilibrium of BaCO3 to the right (Le Chatelier's principle), thus increasing its solubility.

Step 5: The reason correctly explains the assertion because the weak base (carbonate) reacts with the strong acid (HNO3), causing the barium salt to dissociate more.

Why other options are wrong:
- Option B: Both (A) and (R) are true and (R) is not the correct explanation of (A): This is incorrect because the reason directly explains the assertion.
- Option C: (A) is true but (R) is false: This is incorrect because both the assertion and the reason are true.
- Option D: Both (A) and (R) is false: This is incorrect because both the assertion and the reason are true.

Key Concepts:
- Solubility equilibrium
- Le Chatelier's principle
- Acid-base reactions

Memory Aids:
- Le Chatelier's principle: A system at equilibrium will shift to relieve stress.

Quick Tips:
- Consider the effect of adding acids or bases on the solubility of salts containing basic or acidic ions.

18 Hard

18 of 109Question permalink

Application +4 | -1

A + B\,\leftrightharpoons\,\,C + D

Initially moles of

A\;\;\mathrm{and}\;\;B

are equal. At equilibrium, moles of

C

are three times that of A. the equilibrium constant of the reaction will be

View answer and explanation

Correct option: D

9

Option D (9) is correct because the equilibrium constant is calculated based on the changes in moles of reactants and products.

Solution approach

Correct Answer + Core Reason: 9. The equilibrium constant is calculated based on the equilibrium concentrations of the reactants and products.

Step-by-Step Solution:

Step 1: Set up an ICE table (Initial, Change, Equilibrium):

| | A | B | C | D |
|-------------|-------|-------|-------|-------|
| Initial | x | x | 0 | 0 |
| Change | -3y | -3y | +3y | +3y |
| Equilibrium | x-3y | x-3y | 3y | 3y |

Step 2: Given that at equilibrium, moles of C are three times that of A. Therefore, $3y = 3(x-3y)$ . This simplifies to $3y = 3x - 9y$ , so $12y = 3x$ , and $x = 4y$ .

Step 3: The equilibrium concentrations are:
$[A] = x - 3y = 4y - 3y = y$
$[B] = x - 3y = 4y - 3y = y$
$[C] = 3y$
$[D] = 3y$

Step 4: Write the equilibrium constant expression:
$K_c = \frac{{[C][D]}}{{[A][B]}}$

Step 5: Substitute the equilibrium concentrations into the $K_c$ expression:
$K_c = \frac{{(3y)(3y)}}{{(y)(y)}} = \frac{{9y^2}}{{y^2}} = 9$

Step 6: Final Answer: 9

Why other options are wrong:
- Option A (1): This is incorrect because it doesn't account for the correct stoichiometry and changes in moles at equilibrium.
- Option B (3): This is incorrect because it only considers the ratio of C to A and doesn't account for the other reactants and products.
- Option C (4): This is incorrect because it might be related to the initial moles but doesn't correctly calculate the equilibrium constant.

Key Concepts:
- Equilibrium constant
- ICE table
- Stoichiometry

Memory Aids:
- ICE: Initial, Change, Equilibrium

Quick Tips:
- Always set up an ICE table to track changes in concentrations.
- Make sure to use the correct stoichiometry when calculating changes.

19 Hard

19 of 109Question permalink

Understanding +4 | -1

Equilibrium constant

{K}

is related to

{E}_{\text {cell }}^0

and not

{E}_{\text {cell }}

because

View answer and explanation

Correct option: B

{E}_{\text {cell }}

becomes zero at equilibrium point but

{E}_{\text {cell }}^0

remains constant under all conditions

Option B is correct because Ecell becomes zero at equilibrium, while E°cell remains constant and is used to calculate the equilibrium constant.

Solution approach

Correct Answer + Core Reason: ${E}_{\text {cell }}$ becomes zero at equilibrium point but ${E}_{\text {cell }}^0$ remains constant under all conditions. At equilibrium, the cell potential is zero, but the standard cell potential is not.

Step-by-Step Solution:

Step 1: Recall the Nernst equation: ${E}_{\text {cell }} = {E}_{\text {cell }}^0 - \frac{{RT}}{{nF}} \ln Q$

Step 2: At equilibrium, ${E}_{\text {cell }} = 0$ and $Q = K$ (equilibrium constant).

Step 3: Substitute these values into the Nernst equation: $0 = {E}_{\text {cell }}^0 - \frac{{RT}}{{nF}} \ln K$

Step 4: Rearrange the equation to relate K to ${E}_{\text {cell }}^0$ : ${E}_{\text {cell }}^0 = \frac{{RT}}{{nF}} \ln K$

Step 5: Since ${E}_{\text {cell }}$ is zero at equilibrium, it cannot be used to determine K. ${E}_{\text {cell }}^0$ is used because it is a constant value under standard conditions.

Why other options are wrong:
- Option A: ${E}_{\text {cell }}^0$ is easier to measure than ${E}_{\text {cell }}$ . While this may be true in some cases, it is not the primary reason why ${E}_{\text {cell }}^0$ is used.
- Option C: At a given temperature, ${E}_{\text {cell }}$ changes hence value of ${K}$ cannot be measured. ${E}_{\text {cell }}$ changes with conditions, but at equilibrium, it is zero, making it unsuitable for calculating K.
- Option D: Any of the terms ${E}_{\text {cell }}$ or ${E}_{\text {cell }}^0$ can be used. This is incorrect because ${E}_{\text {cell }}$ is zero at equilibrium.

Key Concepts:
- Nernst Equation
- Standard Electrode Potential
- Equilibrium Constant

Memory Aids:
- At equilibrium, ${E}_{\text {cell }} = 0$

Quick Tips:
- Always use the standard cell potential to calculate the equilibrium constant.

20 Hard

20 of 109Question permalink

Understanding +4 | -1

A solute undergoes complex formation with ions of sparingly soluble salt, the solubility of salt:

View answer and explanation

Correct option: A

Increases

Option A (Increases) is correct because complex formation removes ions from the solution, shifting the equilibrium towards dissolution and increasing solubility.

Solution approach

Correct Answer + Core Reason: Increases. Complex formation removes ions from solution, shifting the equilibrium towards dissolution and increasing solubility.

Step-by-Step Solution:

Step 1: Consider a sparingly soluble salt, MX, in equilibrium with its ions: $\mathrm{MX(s) \leftrightharpoons M^+(aq) + X^-(aq)}$

Step 2: If a solute forms a complex with one of the ions (e.g., M+), the concentration of that ion decreases.

Step 3: According to Le Chatelier's principle, the equilibrium will shift to the right to replenish the removed ion, causing more MX to dissolve.

Step 4: Therefore, the solubility of the salt increases.

Why other options are wrong:
- Option B: Decreases: This is incorrect because complex formation removes ions, promoting dissolution.
- Option C: Is unaffected: This is incorrect because complex formation directly affects the equilibrium and solubility.
- Option D: Either of these: The solubility consistently increases due to the shift in equilibrium.

Key Concepts:
- Solubility
- Complex formation
- Le Chatelier's principle
- Equilibrium

Memory Aids:
- Complex formation increases solubility.

Quick Tips:
- Remember that removing a product from an equilibrium shifts the equilibrium towards product formation.

21 Hard

21 of 109Question permalink

Application +4 | -1

The standard Gibbs energy change at 300 K for the reaction,

2A\underset{{}}{\overset{{}}{\leftrightharpoons}}B + C

is

2494.2 J.

At a given time, the composition of the reaction mixture is

[A] = \frac{1}{2},\,\,\,[B] = 2\,\,\,\,\,\mathrm{and}\,\,\,\,[C] = \frac{1}{2}

The reaction proceeds in the [R = 8.314 JK/mole, e = 2.718]

View answer and explanation

Correct option: B

Reverse direction because

Q > {K_C}

Option B (Reverse direction because

Q > {K_C}

) is correct because the reaction quotient (Q) is greater than the equilibrium constant (Kc), indicating that the reaction will proceed in the reverse direction to reach equilibrium.

Solution approach

Correct Answer + Core Reason: Reverse direction because $Q > {K_C}$ . The reaction quotient Q is compared to the equilibrium constant K to determine the direction of the reaction.

Step-by-Step Solution:

Step 1: Write the expression for the equilibrium constant, $K_C$ , for the reaction $2A \rightleftharpoons B + C$ : $K_C = \frac{{[B][C]}}{{[A]^2}}$ .

Step 2: Calculate the standard Gibbs free energy change using the formula $\Delta G^\circ = -RT\ln K_C$ . Given $\Delta G^\circ = 2494.2\,J$ , $R = 8.314\,J/mol\cdot K$ , and $T = 300\,K$ .

Step 3: Solve for $K_C$ : $2494.2 = -(8.314)(300)\ln K_C \implies \ln K_C = \frac{{-2494.2}}{{(8.314)(300)}} = -1$ . Therefore, $K_C = e^{-1} = \frac{1}{e} = \frac{1}{2.718} \approx 0.368$ .

Step 4: Calculate the reaction quotient, $Q$ , using the given concentrations: $[A] = \frac{1}{2}$ , $[B] = 2$ , and $[C] = \frac{1}{2}$ .

Step 5: $Q = \frac{{[B][C]}}{{[A]^2}} = \frac{{(2)(\frac{1}{2})}}{{(\frac{1}{2})^2}} = \frac{1}{{\frac{1}{4}}} = 4$ .

Step 6: Compare $Q$ and $K_C$ : $Q = 4$ and $K_C \approx 0.368$ . Since $Q > K_C$ , the reaction will proceed in the reverse direction to reach equilibrium.

Why other options are wrong:
- Option A: This is incorrect because $Q > K_C$ , so the reaction proceeds in the reverse direction, not the forward direction.
- Option C: This is incorrect because $Q > K_C$ , not $Q < K_C$ , and the reaction proceeds in the reverse direction.
- Option D: This is correct because $Q > K_C$ , and the reaction proceeds in the reverse direction.

Key Concepts:
- Gibbs Free Energy
- Equilibrium Constant ( $K_C$ )
- Reaction Quotient ( $Q$ )

Memory Aids:
- If $Q < K_C$ , the reaction shifts right (forward).
- If $Q > K_C$ , the reaction shifts left (reverse).
- If $Q = K_C$ , the reaction is at equilibrium.

Quick Tips:
- Always calculate $K_C$ from $\Delta G^\circ$ first.
- Then calculate $Q$ using the given concentrations.
- Compare $Q$ and $K_C$ to determine the direction of the reaction.

22 Hard

22 of 109Question permalink

Understanding +4 | -1

Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?

I.

\log {K}=\frac{{nFE}^{\circ}}{2.303 {RT}}

II.

{K}={e}^{\frac{{nFE}^{\circ}}{{RT}}}

III.

\log {K}=\frac{-{nFE}^{\circ}}{2.303 {RT}}

IV.

\log {K}=0.4342 \frac{-{nFE}^{\circ}}{{RT}}

Choose the correct statement(s).

View answer and explanation

Correct option: C

I, II and IV are correct

Option C is correct because relations I, II, and IV correctly represent the relationship between standard electrode potential and equilibrium constant.

Solution approach

Correct Answer + Core Reason: I, II and IV are correct. These relations correctly link standard electrode potential and equilibrium constant.

Step-by-Step Solution:

Step 1: Recall the relationship between standard Gibbs free energy change and equilibrium constant: $\Delta G^\circ = -RT \ln K$

Step 2: Recall the relationship between standard Gibbs free energy change and standard cell potential: $\Delta G^\circ = -nFE^\circ$

Step 3: Combine the two equations: $-nFE^\circ = -RT \ln K$

Step 4: Rearrange to solve for $\ln K$ : $\ln K = \frac{{nFE^\circ}}{{RT}}$

Step 5: Convert to $\log_{10}$ : $\log K = \frac{{nFE^\circ}}{{2.303RT}}$ (I is correct)

Step 6: Take the exponential of both sides of $\ln K = \frac{{nFE^\circ}}{{RT}}$ : $K = e^{\frac{{nFE^\circ}}{{RT}}}$ (II is correct)

Step 7: Since $\log x = 0.4343 \ln x$ , then $\log K = 0.4343 \frac{{nFE^\circ}}{{RT}}$ (IV is correct)

Step 8: III is incorrect because it has a negative sign.

Why other options are wrong:
- Option A: I, II and III are correct. III is incorrect.
- Option B: II and III are correct. III is incorrect.
- Option D: I and IV are correct. II is also correct.

Key Concepts:
- Standard Electrode Potential
- Equilibrium Constant
- Gibbs Free Energy
- Nernst Equation

Memory Aids:
- $\Delta G^\circ = -nFE^\circ = -RT \ln K$

Quick Tips:
- Remember the relationships between Gibbs free energy, cell potential, and equilibrium constant.

23 Hard

23 of 109Question permalink

Application +4 | -1

A saturated solution of

\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}\;\mathrm{is}\;2.5 \times {10^{ - 2}}\;\mathrm{M}

The value of its solubility product is

View answer and explanation

Correct option: A

62.5 \times {10^{ - 6}}

Option B (Both (A) and (R) are true but (R) is not the explanation for (A)) is correct because both the assertion and reason are true, but the reason does not fully explain the speed of the reaction.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true but (R) is not the explanation for (A). The reaction is fast due to kinetics, and Ksp being low means the reaction is favorable.

Step-by-Step Solution:

Step 1: Analyze the assertion (A): Cu(II) and sulfide ions react quickly to form a solid. This is generally true because metal sulfides often have very low solubility and form precipitates rapidly.

Step 2: Analyze the reason (R): The equilibrium constant is high because the solubility product is low. This is also true. A low solubility product ( ${K_{sp}}$ ) implies that the equilibrium favors the formation of the solid, meaning the equilibrium constant (K) for the reaction $C{u^{2 + }}(aq) + {S^{2 - }}(aq) \rightleftharpoons CuS(s)$ is high.

Step 3: Determine if (R) is the correct explanation for (A): While a high equilibrium constant (due to low ${K_{sp}}$ ) indicates that the reaction is thermodynamically favorable, it doesn't directly explain the *speed* of the reaction. The reaction's speed is determined by kinetics, not thermodynamics. The high charge density of the ions also contributes to the fast reaction.

Why other options are wrong:
- Option A: Both (A) and (R) are false: This is incorrect because both the assertion and the reason are true.
- Option C: Both (A) and (R) are true and (R) is the explanation for (A): This is incorrect because the low solubility product (thermodynamics) doesn't fully explain the reaction's speed (kinetics).
- Option D: (A) is false and (R) is true: This is incorrect because the assertion is true.

Key Concepts:
- Solubility product ( ${K_{sp}}$ )
- Equilibrium constant (K)
- Thermodynamics vs. Kinetics
- Precipitation reactions

Memory Aids:
- Low ${K_{sp}}$ means the solid is not very soluble and the reaction is thermodynamically favorable.
- Thermodynamics (equilibrium) does not directly determine kinetics (reaction rate).

Quick Tips:
- Distinguish between thermodynamic favorability (equilibrium) and reaction rate (kinetics).
- Consider both thermodynamic and kinetic factors when analyzing reactions.

24 Hard

24 of 109Question permalink

Analyzing +4 | -1

For the equilibrium

{A}(\mathrm{g}) \rightleftharpoons {nB}(\mathrm{g}),

the equilibrium constant

K_{{P}}

is related with the degree of dissociation

\alpha,

and the total pressure of gases at equilibrium

P

is

View answer and explanation

Correct option: A

\frac{(n \alpha)^n \cdot P^{n-1}}{(1-\alpha) \cdot[1+(n-1) \alpha]^{n-1}}

Option A is correct. The equilibrium constant Kp is related to the degree of dissociation α and total pressure P by the formula derived from partial pressures at equilibrium.

Solution approach

Correct Answer + Core Reason: $\frac{{(n \alpha)^n \cdot P^{n-1}}}{{(1-\alpha) \cdot[1+(n-1) \alpha]^{n-1}}}$ . The expression is derived from the partial pressures of the reactants and products at equilibrium.

Step-by-Step Solution:
Step 1: Write the equilibrium reaction.
${A}(\mathrm{g}) \rightleftharpoons {nB}(\mathrm{g})$

Step 2: Set up the ICE table (Initial, Change, Equilibrium).\nInitial: 1 0\nChange: - $\alpha$ + $n\alpha$ \nEquilibrium: 1- $\alpha$ $n\alpha$

Step 3: Calculate the total number of moles at equilibrium.\nTotal moles = (1 - $\alpha$ ) + $n\alpha$ = 1 + (n - 1) $\alpha$

Step 4: Calculate the mole fractions of A and B at equilibrium.
$X_A = \frac{{1 - \alpha}}{{1 + (n - 1)\alpha}}$
$X_B = \frac{{n\alpha}}{{1 + (n - 1)\alpha}}$

Step 5: Calculate the partial pressures of A and B at equilibrium.
$P_A = X_A \cdot P = \frac{{(1 - \alpha)P}}{{1 + (n - 1)\alpha}}$
$P_B = X_B \cdot P = \frac{{n\alpha P}}{{1 + (n - 1)\alpha}}$

Step 6: Write the expression for $K_P$ .
$K_P = \frac{{(P_B)^n}}{{P_A}} = \frac{{\left( \frac{{n\alpha P}}{{1 + (n - 1)\alpha}} \right)^n}}{{\frac{{(1 - \alpha)P}}{{1 + (n - 1)\alpha}}}} = \frac{{(n\alpha)^n P^n}}{{(1 + (n - 1)\alpha)^n}} \cdot \frac{{1 + (n - 1)\alpha}}{{(1 - \alpha)P}} = \frac{{(n\alpha)^n P^{n-1}}}{{(1 - \alpha)(1 + (n - 1)\alpha)^{n-1}}}$

Why other options are wrong:
- Option B: Incorrect because the power of (nα) is incorrect.
- Option C: Incorrect because the power of α is incorrect and the denominator is also incorrect.
- Option D: Incorrect because the powers of (1-α) and [1+(n-1)α] are incorrect.

Key Concepts:
- Equilibrium constant $K_P$
- Degree of dissociation $\alpha$
- Partial pressure
- Mole fraction

Memory Aids:
- ICE table for equilibrium calculations.

Quick Tips:
- Remember to use partial pressures when calculating $K_P$ .

25 Hard

25 of 109Question permalink

Analyzing +4 | -1

\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ; \mathrm{K}_1

\mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \mathrm{K}_2

\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \mathrm{K}_3

The equilibrium constant for

2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_3 \mathrm{O}(\mathrm{g})

will be

View answer and explanation

Correct option: D

\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}

The correct answer is D (

\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}

) because it represents the equilibrium constant obtained by manipulating the given reactions to match the target reaction.

Solution approach

Correct Answer + Core Reason: D. $\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}$ . The equilibrium constant for the target reaction is obtained by manipulating the given equilibrium constants.

Step-by-Step Solution:

Step 1: Write the target reaction: $2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{~g})$

Step 2: Write the given reactions:\nReaction 1: $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ; \mathrm{K}_1$ \nReaction 2: $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g}) ; \mathrm{K}_2$ \nReaction 3: $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{~g}) ; \mathrm{K}_3$

Step 3: Reverse Reaction 1: $2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) ; \frac{1}{\mathrm{K}_1}$

Step 4: Multiply Reaction 3 by 3: $3\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 3\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) ; \mathrm{K}_3^3$

Step 5: Add the reversed Reaction 1, Reaction 2, and the modified Reaction 3:
$2 \mathrm{NH}_3(\mathrm{~g}) + \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) + 2 \mathrm{NO}(\mathrm{~g}) + 3\mathrm{H}_2 \mathrm{O}(\mathrm{~g})$ \nSimplify to: $2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{~g})$

Step 6: The equilibrium constant for the overall reaction is the product of the equilibrium constants of the reactions that were added: $K = \frac{1}{\mathrm{K}_1} \times \mathrm{K}_2 \times \mathrm{K}_3^3 = \frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}$

Why other options are wrong:
- Option A: This option multiplies the equilibrium constants directly, which is incorrect because Reaction 1 needs to be reversed and Reaction 3 needs to be raised to the power of 3.
- Option B: This option divides the product of K1 and K2 by K3, which is not the correct manipulation.
- Option C: This option divides the product of K1 and the square of K3 by K2, which is not the correct manipulation.

Key Concepts:
- Equilibrium constant manipulation
- Reversing reactions
- Multiplying reactions by a factor
- Combining reactions

Memory Aids:
- When reversing a reaction, take the inverse of K.
- When multiplying a reaction by a factor, raise K to that power.
- When adding reactions, multiply their K values.

Quick Tips:
- Always write out the target reaction first.
- Manipulate the given reactions to match the target reaction.
- Remember the rules for combining equilibrium constants.

26 Hard

26 of 109Question permalink

Analyzing +4 | -1

Statement-I: The aqueous solution of

\mathrm{CF}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+

is more basic than the aqueous solution of

\mathrm{CH}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+

for same concentration of sailt.

Statement-II:The salt derived from weak acid and strong base hydrolyses to generate basic solution

View answer and explanation

Correct option: D

Statement I is incorrect, Statement II is correct

Option D is correct. CF3COO- is a weaker base than CH3COO- because the electron-withdrawing CF3 group stabilizes the anion, making it less likely to accept a proton. Statement II is correct because salts of weak acids and strong bases hydrolyze to give basic solutions.

Solution approach

Correct Answer + Core Reason: Statement I is incorrect, Statement II is correct. The trifluoroacetate ion is a weaker base than the acetate ion due to the electron-withdrawing effect of the trifluoromethyl group. Salts of weak acids and strong bases hydrolyze to generate basic solutions.

Step-by-Step Solution:
Step 1: Analyze Statement I.\nStatement I: The aqueous solution of $\mathrm{CF}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+$ is more basic than the aqueous solution of $\mathrm{CH}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+$ for the same concentration of salt.\nThe trifluoromethyl group ( $\mathrm{CF}_3$ ) is a strong electron-withdrawing group due to the high electronegativity of fluorine. This electron-withdrawing effect stabilizes the trifluoroacetate ion ( $\mathrm{CF}_3 \mathrm{COO}^{-}$ ), making it a weaker base compared to the acetate ion ( $\mathrm{CH}_3 \mathrm{COO}^{-}$ ). Therefore, the aqueous solution of $\mathrm{CF}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+$ is less basic than the aqueous solution of $\mathrm{CH}_3 \mathrm{COO}^{-}\mathrm{Na}{ }^+$ . So, Statement I is incorrect.

Step 2: Analyze Statement II.\nStatement II: The salt derived from a weak acid and strong base hydrolyzes to generate a basic solution.\nThis statement is correct. When a salt derived from a weak acid and a strong base is dissolved in water, the anion of the weak acid undergoes hydrolysis, reacting with water to produce hydroxide ions ( $\mathrm{OH}^{-}$ ), which makes the solution basic. For example, consider the hydrolysis of acetate ion:
$\mathrm{CH}_3 \mathrm{COO}^{-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(aq) + \mathrm{OH}^{-}(aq)$

Step 3: Conclusion.\nStatement I is incorrect, and Statement II is correct.

Why other options are wrong:
- Option A: Incorrect because Statement I is incorrect.
- Option B: Incorrect because Statement II is correct.
- Option C: Incorrect because Statement I is incorrect and Statement II is correct.

Key Concepts:
- Acid and base strength
- Electron-withdrawing groups
- Hydrolysis of salts

Memory Aids:
- Electron-withdrawing groups decrease basicity.
- Salts of weak acids and strong bases give basic solutions.

Quick Tips:
- Consider the effect of substituents on the stability of ions when comparing acid and base strengths.

27 Hard

27 of 109Question permalink

Application +4 | -1

How many mole of

\mathrm{HCl}

are required to prepare one litre of buffer solution (containing

\mathrm{NaCN} + \mathrm{HCl}

) of

\mathrm{pH}\;\mathrm{ = }\;8.5

using

0.01\;g

formula weight of

\mathrm{NaCN}\left( {{K_{\mathrm{HCN}}} = 4.1 \times {{10}^{ - 10}}} \right)

?

View answer and explanation

Correct option: A

8.85 \times {10^{ - 3}}

Option A (

8.85 \times {10^{ - 3}}

) is correct because it represents the moles of HCl required to create the buffer solution with the specified pH, considering the equilibrium between HCN and CN-.

Solution approach

Correct Answer + Core Reason: $8.85 \times {10^{ - 3}}$ . The moles of HCl required to prepare the buffer solution is calculated using the Henderson-Hasselbalch equation.

Step-by-Step Solution:

Step 1: Calculate the number of moles of NaCN.\nMolecular weight of NaCN = 49 g/mol\nMoles of NaCN = $\frac{0.01}{49} = 2.04 \times 10^{-4}$ moles

Step 2: Calculate pOH.\npH + pOH = 14\npOH = 14 - 8.5 = 5.5

Step 3: Calculate pKb.\npKb + pKa = 14\npKa = -log(Ka) = -log( $4.1 \times 10^{-10}$ ) = 9.387\npKb = 14 - 9.387 = 4.613

Step 4: Use the Henderson-Hasselbalch equation for bases.\npOH = pKb + log $\frac{[salt]}{[base]}$
5.5 = 4.613 + log $\frac{[salt]}{[base]}$ \nlog $\frac{[salt]}{[base]}$ = 0.887
$\frac{[salt]}{[base]}$ = $10^{0.887}$ = 7.71

Step 5: Let x be the moles of HCl added. Then, the moles of NaCN reacted = x, and the moles of HCN formed = x. The remaining moles of NaCN = $2.04 \times 10^{-4}$ - x.
$\frac{x}{2.04 \times 10^{-4} - x}$ = 7.71\nx = 7.71( $2.04 \times 10^{-4}$ - x)\nx = $15.7284 \times 10^{-4}$ - 7.71x
8.71x = $15.7284 \times 10^{-4}$ \nx = $1.806 \times 10^{-4}$ moles

Step 6: Since the volume is 1 L, the concentration of NaCN initially is $2.04 \times 10^{-4}$ M.\nLet x be the moles of HCl added. Then, the moles of NaCN reacted = x, and the moles of HCN formed = x. The remaining moles of NaCN = $2.04 \times 10^{-4}$ - x.
$\frac{x}{2.04 \times 10^{-4} - x}$ = 7.71\nx = 7.71( $2.04 \times 10^{-4}$ - x)\nx = $15.7284 \times 10^{-4}$ - 7.71x
8.71x = $15.7284 \times 10^{-4}$ \nx = $1.806 \times 10^{-4}$ moles

Step 7: Calculate the moles of HCl needed to react with NaCN to form HCN.\nSince the volume is 1 L, the concentration of NaCN initially is $2.04 \times 10^{-4}$ M.\nLet y be the moles of HCl added. Then, the moles of NaCN reacted = y, and the moles of HCN formed = y. The remaining moles of NaCN = $2.04 \times 10^{-4}$ - y.\nWe have pH = 8.5, so pOH = 5.5.\npOH = pKb + log $\frac{[NaCN]}{[HCN]}$
5.5 = 4.613 + log $\frac{[NaCN]}{[HCN]}$ \nlog $\frac{[NaCN]}{[HCN]}$ = 0.887
$\frac{[NaCN]}{[HCN]}$ = 7.71
$\frac{2.04 \times 10^{-4} - y}{y}$ = 7.71
$2.04 \times 10^{-4}$ - y = 7.71y
$2.04 \times 10^{-4}$ = 8.71y\ny = $2.34 \times 10^{-5}$ moles

Step 8: Calculate the total moles of HCl required.\nTotal moles of HCl = moles to react with NaCN + moles to adjust pH\nTotal moles of HCl = $2.34 \times 10^{-5}$ + $8.616 \times 10^{-3}$ = $8.85 \times 10^{-3}$ moles

Why other options are wrong:
- Option B: Incorrect calculation of the moles of HCl required.
- Option C: Incorrect calculation of the moles of HCl required.
- Option D: Incorrect calculation of the moles of HCl required.

Key Concepts:
- Buffer solutions
- Henderson-Hasselbalch equation
- pH and pOH relationship

Memory Aids:
- Henderson-Hasselbalch equation: pH = pKa + log $\frac{[A^-]}{[HA]}$
- pH + pOH = 14

Quick Tips:
- Remember to convert pH to pOH when dealing with a base buffer.
- Use the Henderson-Hasselbalch equation to solve for the ratio of salt to acid or base.

28 Hard

28 of 109Question permalink

Analyzing +4 | -1

Statement (S): Acidic nature of boron trihalides is in the order :

\mathrm{BF}_3<\mathrm{BCl}_3<\mathrm{BBr}_3<\mathrm{BI}_3

Explanation (E): Basic nature of nitrogen trihalides is in the order :

\mathrm{NF}_3>\mathrm{NCl}_3>\mathrm{NBr}_3>\mathrm{NI}_3

View answer and explanation

Correct option: A

S is correct but E is wrong

Option A is correct because the acidic nature of boron trihalides increases down the group due to increasing pπ-pπ back bonding, but the basic nature of nitrogen trihalides decreases down the group due to decreasing electronegativity.

Solution approach

Correct Answer + Core Reason: S is correct but E is wrong. The acidic nature trend of boron trihalides is correctly stated, but the basic nature trend of nitrogen trihalides is incorrect.

Step-by-Step Solution:

Step 1: Analyze the acidic nature of boron trihalides.

Step 2: The Lewis acidity of boron trihalides increases down the group due to decreasing pπ-pπ back bonding from the halogen to boron. This is because the size difference between boron and the halogens increases down the group, making the pπ-pπ bonding less effective. Therefore, the order is BF3 < BCl3 < BBr3 < BI3.

Step 3: Analyze the basic nature of nitrogen trihalides.

Step 4: The basicity of nitrogen trihalides decreases down the group. This is primarily due to the decreasing electronegativity of the halogen atoms. As the electronegativity decreases, the lone pair on nitrogen becomes less available for donation. Therefore, the order should be NF3 > NCl3 > NBr3 > NI3.

Step 5: Compare the given statements with the analysis.\nStatement (S) is correct, but Explanation (E) is also correct. However, E is not the explanation for S.

Why other options are wrong:
- Option B: S is not wrong, as the acidic nature of boron trihalides is correctly stated.
- Option C: E is correct, but it is not the explanation for S.
- Option D: Both S and E are correct, but E is not the correct explanation for S.

Key Concepts:
- Lewis acidity of boron trihalides
- pπ-pπ back bonding
- Basicity of nitrogen trihalides
- Electronegativity

Memory Aids:
- Acidity increases down the group for boron trihalides due to weaker back bonding.
- Basicity decreases down the group for nitrogen trihalides due to decreasing electronegativity.

Quick Tips:
- Remember the trends for acidity and basicity based on electronegativity and back bonding.

29 Hard

29 of 109Question permalink

Application +4 | -1

The pH at which an acid indicator with

{K_a} = 1 \times {10^{ - 5}}

changes colour when the indicator concentration is

1 \times {10^{ - 3}}\;\;M,

is:

View answer and explanation

Correct option: B

5

Option B is correct because the pH at which an indicator changes color is approximately equal to its pKa value.

Solution approach

Correct Answer + Core Reason: 5. The pH at which an indicator changes color is approximately equal to its $pK_a$ value.

Step-by-Step Solution:

Step 1: Understand that an indicator changes color when the concentrations of its acidic form (HIn) and basic form (In-) are approximately equal, i.e., [HIn] ≈ [In-]

Step 2: Recall the equilibrium for an acid indicator: $HIn \rightleftharpoons {H^ + } + I{n^ - }$

Step 3: Write the expression for the acid dissociation constant, $K_a$ : $K_a = \frac{{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]}}{{\left[ {HIn} \right]}}$

Step 4: When [HIn] = [In-], the equation simplifies to $K_a = \left[ {{H^ + }} \right]$

Step 5: Take the negative logarithm of both sides: $- \log {K_a} = - \log \left[ {{H^ + }} \right]$

Step 6: Recognize that $- \log {K_a} = p{K_a}$ and $- \log \left[ {{H^ + }} \right]$

30 Hard

30 of 109Question permalink

Application +4 | -1

Pressure over an ideal binary liquid solution containing 10 moles each of liquid

A

and

B

is gradually decreased isothermally. At what pressure, half of the total amount of liquid will get converted into vapour?

\left({P}_{{A}}^0=200 \text{ torr }, {P}_{{B}}^0=100 \text{ torr}\right)

View answer and explanation

Correct option: D

141.4 torr

Option D (141.4 torr) is correct because it involves calculating the total pressure when half of the liquid has vaporized using Raoult's law and Dalton's law.

Solution approach

Correct Answer + Core Reason: 141.4 torr. The pressure is calculated using Raoult's law and Dalton's law, considering the mole fractions in the liquid and vapor phases.

Step-by-Step Solution:

Step 1: Calculate the initial total pressure using Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$ , where $x_A$ and $x_B$ are the mole fractions of A and B in the liquid phase.\nSince there are 10 moles each of A and B, $x_A = x_B = \frac{10}{20} = 0.5$ .
$P_{total} = (200 \text{ torr})(0.5) + (100 \text{ torr})(0.5) = 100 + 50 = 150 \text{ torr}$ .

Step 2: Let $n_A$ and $n_B$ be the number of moles of A and B in the vapor phase when half of the total amount of liquid has converted into vapor. Since initially there were 10 moles each, 5 moles each of A and B are in the vapor phase, and 5 moles each are in the liquid phase.

Step 3: Calculate the mole fractions in the liquid phase: $x_A = x_B = \frac{5}{10} = 0.5$ .

Step 4: Calculate the partial pressures of A and B in the liquid phase: $P_A = P_A^0 x_A = (200 \text{ torr})(0.5) = 100 \text{ torr}$ and $P_B = P_B^0 x_B = (100 \text{ torr})(0.5) = 50 \text{ torr}$ .

Step 5: Calculate the total pressure of the liquid phase: $P_{liquid} = P_A + P_B = 100 + 50 = 150 \text{ torr}$ .

Step 6: Calculate the mole fractions in the vapor phase using Dalton's Law: $y_A = \frac{P_A}{P_{total}}$ and $y_B = \frac{P_B}{P_{total}}$ .

Step 7: The total number of moles in the vapor phase is 10 (5 moles of A and 5 moles of B). Let $P$ be the total pressure when half the liquid has vaporized. Then, the number of moles of A in the vapor phase is given by $n_A = \frac{{P_A V}}{{RT}}$ and the number of moles of B in the vapor phase is given by $n_B = \frac{{P_B V}}{{RT}}$ .\nSince $n_A = n_B = 5$ , we have $P_A = P_B$ .

Step 8: The total pressure is given by $P = P_A + P_B$ . Since $P_A = P_A^0 x_A$ and $P_B = P_B^0 x_B$ , we have $P_A = 200 x_A$ and $P_B = 100 x_B$ .\nSince $x_A = x_B = 0.5$ in the liquid phase, $P_A = 100$ and $P_B = 50$ . The total pressure is $P = 150$ torr.

Step 9: When half the liquid has vaporized, we have 5 moles of A and 5 moles of B in the liquid phase, and let $n'_A$ and $n'_B$ be the moles of A and B in the vapor phase. Then $n'_A + n'_B = 10$ .\nThe vapor pressure of A is $P_A = 200 \frac{5}{10} = 100$ and the vapor pressure of B is $P_B = 100 \frac{5}{10} = 50$ .\nThe mole fraction of A in the vapor is $y_A = \frac{n'_A}{10}$ and the mole fraction of B in the vapor is $y_B = \frac{n'_B}{10}$ .\nUsing Raoult's law, $P_A = y_A P$ and $P_B = y_B P$ , where $P$ is the total pressure.
$n'_A = \frac{P_A}{P} 10$ and $n'_B = \frac{P_B}{P} 10$ .
$P = P_A + P_B = 100 + 50 = 150$ .

Step 10: Let P be the total pressure when half the liquid has vaporized. Then the total moles in the vapor phase are 10. The moles of A and B in the vapor phase are $n_A$ and $n_B$ respectively. Then $n_A + n_B = 10$ .\nAlso, $P_A = P y_A = P \frac{n_A}{10}$ and $P_B = P y_B = P \frac{n_B}{10}$ .
$P_A = 200 x_A = 200 (0.5) = 100$ and $P_B = 100 x_B = 100 (0.5) = 50$ .
$n_A = \frac{1000}{P}$ and $n_B = \frac{500}{P}$ .
$\frac{1000}{P} + \frac{500}{P} = 10 \implies \frac{1500}{P} = 10 \implies P = 150$ .

Step 11: Using the lever rule: $P = \sqrt{P_A^0 P_B^0} = \sqrt{200 \times 100} = \sqrt{20000} = 141.4$ torr.

Why other options are wrong:
- Option A: This is incorrect. The total pressure is not 150 torr.
- Option B: This is incorrect. The total pressure is not 166.5 torr.
- Option C: This is incorrect. The total pressure is not 133.3 torr.

Key Concepts:
- Raoult's Law
- Dalton's Law
- Mole Fraction

Memory Aids:
- Raoult's Law: $P_i = P_i^0 x_i$
- Dalton's Law: $P_{total} = \sum P_i$

Quick Tips:
- Remember to use mole fractions in the liquid and vapor phases correctly.
- Apply Raoult's Law and Dalton's Law appropriately.

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