In a gaseous reversible reaction, ${{\rm{N}}_2} + {{\rm{O}}_2}\leftrightharpoons2{\rm{NO}} + {\rm{heat}}$ If pressure is increased then the equilibrium constant would be

Sometimes increased, sometimes decreased

For the reaction, ${\rm{CuS}}{{\rm{O}}_4} \cdot 5{{\rm{H}}_2}{\rm{O}}\left( s \right)\leftrightharpoons{\rm{CuS}}{{\rm{O}}_4} \cdot 3{{\rm{H}}_2}{\rm{O}}\left( s \right) + 2{{\rm{H}}_2}{\rm{O}}\left( v \right)$ which one is correct representation?

${K_p} = {\left( {{P_{{{\rm{H}}_2}{\rm{O}}}}} \right)^2}$

${K_c} = {\left[ {{{\rm{H}}_2}{\rm{O}}} \right]^2}$

${K_p} = {K_c}{\left( {RT} \right)^2}$

The equilibrium constant for the reaction, ${\rm{HONO}} + {\rm{C}}{{\rm{N}}^ - } \to {\rm{HCN}} + {\rm{ON}}{{\rm{O}}^ - }$ is $1 \times {10^{ - 14}}.$ The magnitude of the equilibrium constant suggests that:

${\rm{HCN}}$ is stronger acid than ${\rm{HONO}}$

${\rm{C}}{{\rm{N}}^ - }$ is stronger base than ${\rm{ON}}{{\rm{O}}^ - }$

${\rm{ON}}{{\rm{O}}^ - }$ is the conjugate base of ${\rm{HONO}}$

The conjugate acid of ${\rm{C}}{{\rm{N}}^ - }$ is ${\rm{HCN}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

In the following reversible reaction, $2{\rm{S}}{{\rm{O}}_2} + {{\rm{O}}_2}\leftrightharpoons2{\rm{S}}{{\rm{O}}_3} + Q{\kern 1pt} {\kern 1pt} {\rm{cal}}$ Most suitable condition for the higher production of ${\rm{S}}{{\rm{O}}_3}$ is

Low temperature and low pressure

High temperature and high pressure

High temperature and low pressure

For the reactions, ${{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)\leftrightharpoons{\rm{CO}}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)$ if the initial concentration of $[{{\rm{H}}_2}\left] = \right[{\rm{C}}{{\rm{O}}_2}]$ and $x$ mol/L of hydrogen is consumed at equilibrium, the correct expression of${K_p}$ is

$\frac{{{x^2}}}{{{{\left( {2 + x} \right)}^2}}}$

$\frac{{{{\left( {1 + x} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}}$

Equilibrium MCQs for NEET UG CHEMISTRY — Easy

1 Easy

1 of 144

Remembering +4 | -1

Phenolphthalein shows … in acid medium.

View answer and explanation

Correct option: D

No colour

Option D (No colour) is correct because phenolphthalein is colorless in acidic solutions and turns pink in basic solutions.

Solution approach

Correct Answer + Core Reason: No colour. Phenolphthalein is colorless in acidic medium.

Step-by-Step Solution:

Step 1: Recall the color change of phenolphthalein indicator.

Step 2: Phenolphthalein is a common acid-base indicator that is colorless in acidic solutions and pink in basic solutions.

Step 3: Therefore, in an acidic medium, phenolphthalein shows no color.

Why other options are wrong:
- Option A: Red colour: Phenolphthalein does not turn red in acidic medium.
- Option B: Yellow colour: Phenolphthalein does not turn yellow in acidic medium.
- Option C: Pink colour: Phenolphthalein turns pink in basic medium, not acidic medium.

Key Concepts:
- Acid-base indicators
- Phenolphthalein
- Color change of indicators

Memory Aids:
- Phenolphthalein: Pink in Phenols (basic conditions).

Quick Tips:
- Remember the color changes of common acid-base indicators.

2 Easy

2 of 144

Analyzing +4 | -1

Which can act as buffer?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the given options contain a weak acid/base and its salt with a strong base/acid, which are the components of a buffer solution.

Solution approach

Correct Answer + Core Reason: All of the above. Buffer solutions are mixtures of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Step-by-Step Solution:

Step 1: Define a buffer solution. A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Step 2: Analyze option A: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + \mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ . This is a mixture of a weak base ( $\mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ ) and its salt ( $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ ) with a strong acid.

Step 3: Analyze option B: $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} + \mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ . This is a mixture of a weak acid ( $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ ) and its salt ( $\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ ) with a strong base.

Step 4: Analyze option C: $40\;\mathrm{mL}\;of\;0.1\;M\;\mathrm{NaCN}\,\; + \;20\;\mathrm{mL}\;\;of\;\;0.1\;M\;\mathrm{HCl}$ . Here, moles of NaCN = 0.004 and moles of HCl = 0.002. HCl will react with NaCN to form HCN and NaCl. After the reaction, we will have 0.002 moles of HCN and 0.002 moles of NaCN. This is a mixture of a weak acid (HCN) and its salt (NaCN).

Step 5: Conclusion: All options represent buffer solutions.

Why other options are wrong:
- Option A: This option is a valid buffer system, so it cannot be the only correct answer.
- Option B: This option is a valid buffer system, so it cannot be the only correct answer.
- Option C: This option is a valid buffer system, so it cannot be the only correct answer.

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

Memory Aids:
- A buffer contains a weak acid/base and its salt.

Quick Tips:
- Identify weak acids/bases and their corresponding salts to determine if a solution is a buffer.

3 Easy

3 of 144

Remembering +4 | -1

The

pH

of gastric juice is normally:

View answer and explanation

Correct option: C

Greater than 1 and less than 3

Option C (Greater than 1 and less than 3) is correct because the normal pH of gastric juice is highly acidic, typically ranging from 1 to 3 due to the presence of hydrochloric acid (HCl).

Solution approach

Correct Answer + Core Reason: Greater than 1 and less than 3. The pH of gastric juice is highly acidic due to the presence of hydrochloric acid (HCl).

Step-by-Step Solution:

Step 1: Define gastric juice.\nGastric juice is a digestive fluid formed in the stomach. It contains hydrochloric acid (HCl), enzymes, and mucus.

Step 2: Explain the acidity of gastric juice.\nThe presence of HCl makes gastric juice highly acidic. The pH of gastric juice is typically between 1 and 3.

Step 3: Determine the correct answer.\nTherefore, the pH of gastric juice is normally greater than 1 and less than 3.

Why other options are wrong:
- Option A: Greater than $1.5$ and less than $1.2$ is not a valid range, as 1.5 is greater than 1.2.
- Option B: Less than $1.5$ is too broad and doesn't accurately represent the typical pH range of gastric juice.
- Option D: Less than 1 and greater than zero is too acidic for normal gastric juice.

Key Concepts:
- Gastric juice
- pH
- Hydrochloric acid (HCl)

Memory Aids:
- Gastric juice is acidic due to HCl.

Quick Tips:
- Remember the typical pH range of common biological fluids.

4 Easy

4 of 144

Application +4 | -1

A buffer mixture of acetic acid and potassium acetate has

\mathrm{pH} = 5.24.

The ratio of

\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }} \right]/\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}} \right]

in this buffer is,

\left( {\mathrm{p}{K_a} = 4.74} \right):

View answer and explanation

Correct option: A

3:1

Option A (

3:1

) is correct because using the Henderson-Hasselbalch equation, the ratio of the conjugate base to the acid is calculated to be approximately 3:1.

Solution approach

Correct Answer + Core Reason: $3:1$ . The ratio is calculated using the Henderson-Hasselbalch equation.

Step-by-Step Solution:

Step 1: Use the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }]}{[\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}]})$

Step 2: Given $pH = 5.24$ and $pK_a = 4.74$ .

Step 3: $5.24 = 4.74 + log(\frac{[\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }]}{[\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}]})$

Step 4: $log(\frac{[\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }]}{[\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}]}) = 5.24 - 4.74 = 0.5$

Step 5: $\frac{[\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }]}{[\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}]} = 10^{0.5} \approx 3.16 \approx 3$

Step 6: Therefore, the ratio is approximately $3:1$ .

Why other options are wrong:
- Option B: This is incorrect. The correct ratio is approximately 3:1.
- Option C: This is incorrect. The correct ratio is approximately 3:1.
- Option D: This is incorrect. The correct ratio is approximately 3:1.

Key Concepts:
- Henderson-Hasselbalch equation
- Buffer solutions

Memory Aids:
- $pH = pK_a + log(\frac{[A^-]}{[HA]})$

Quick Tips:
- Remember the Henderson-Hasselbalch equation.

5 Easy

5 of 144

Understanding +4 | -1

Relation between hydrolysis constant and dissociation constant are given. Which is the correct formula for

\mathrm{MgC}{\mathrm{l}_2}

?

View answer and explanation

Correct option: B

{K_h} = \frac{{{K_w}}}{{{K_b}}}

Option B (

{K_h} = \frac{{{K_w}}}{{{K_b}}}

) is correct because

MgCl_2

is a salt of a strong acid (HCl) and a weak base (Mg(OH)2), so its hydrolysis constant is related to the base dissociation constant.

Solution approach

Correct Answer + Core Reason: ${K_h} = \frac{{{K_w}}}{{{K_b}}}$ . $MgCl_2$ is a salt of a strong acid and a weak base, so the hydrolysis constant is related to $K_b$ .

Step-by-Step Solution:

Step 1: Determine the nature of the salt $\mathrm{MgC}{\mathrm{l}_2}$ .

Step 2: $\mathrm{MgC}{\mathrm{l}_2}$ is formed from a strong acid (HCl) and a weak base (Mg(OH)2).

Step 3: Recall the relationship between the hydrolysis constant ( $K_h$ ), the dissociation constant of water ( $K_w$ ), and the dissociation constant of the weak base ( $K_b$ ) for a salt of a strong acid and a weak base.

Step 4: The correct relationship is ${K_h} = \frac{{{K_w}}}{{{K_b}}}$ .

Why other options are wrong:
- Option A: ${K_h} = \frac{{{K_w}}}{{{K_a}}}$ : This is the relationship for a salt of a weak acid and a strong base.
- Option C: ${K_h} = \frac{{{K_w}}}{{{K_a} \times {K_b}}}$ : This relationship is not correct for a simple salt hydrolysis.
- Option D: ${K_w} = \frac{{{K_h}}}{{{K_b}}}$ : This is a rearrangement of the correct formula, but it's not the standard way to express the relationship.

Key Concepts:
- Salt hydrolysis
- Hydrolysis constant ( $K_h$ )
- Dissociation constant of water ( $K_w$ )
- Dissociation constant of acid ( $K_a$ ) and base ( $K_b$ )

Memory Aids:
- Strong Acid/Weak Base: $K_h = K_w/K_b$
- Weak Acid/Strong Base: $K_h = K_w/K_a$

Quick Tips:
- Identify whether the salt is derived from a strong acid/weak base or a weak acid/strong base to determine the correct formula.

6 Easy

6 of 144

Understanding +4 | -1

The salt that forms neutral solution in water is

View answer and explanation

Correct option: B

NaCl

Option B (

NaCl

) is correct because it is the salt of a strong acid (HCl) and a strong base (NaOH), resulting in a neutral solution when dissolved in water.

Solution approach

Correct Answer + Core Reason: $NaCl$ . Sodium chloride is the salt of a strong acid and a strong base, forming a neutral solution.

Step-by-Step Solution:

Step 1: Analyze each option:\nA. $N{H_4}Cl$ : Salt of a weak base ( $NH_3$ ) and a strong acid (HCl), so it forms an acidic solution.\nB. $NaCl$ : Salt of a strong acid (HCl) and a strong base (NaOH), so it forms a neutral solution.\nC. $N{a_2}C{O_3}$ : Salt of a strong base (NaOH) and a weak acid ( $H_2CO_3$ ), so it forms a basic solution.\nD. ${K_3}B{O_3}$ : Salt of a strong base (KOH) and a weak acid ( $H_3BO_3$ ), so it forms a basic solution.

Step 2: Conclusion: $NaCl$ forms a neutral solution.

Why other options are wrong:
- Option A: $N{H_4}Cl$ forms an acidic solution due to the hydrolysis of $NH_4^+$ .
- Option C: $N{a_2}C{O_3}$ forms a basic solution due to the hydrolysis of $CO_3^{2-}$ .
- Option D: ${K_3}B{O_3}$ forms a basic solution due to the hydrolysis of $BO_3^{3-}$ .

Key Concepts:
- Hydrolysis of salts
- Salts of strong/weak acids/bases

Memory Aids:
- Strong acid + Strong base = Neutral salt
- Strong acid + Weak base = Acidic salt
- Weak acid + Strong base = Basic salt

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

7 Easy

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Understanding +4 | -1

In a gaseous reversible reaction,

{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO} + \mathrm{heat}

If pressure is increased then the equilibrium constant would be

View answer and explanation

Correct option: A

Unchanged

Option A (Unchanged) is correct because increasing the pressure does not affect the equilibrium constant, only the equilibrium position, for a reaction with an equal number of moles of gas on both sides.

Solution approach

Correct Answer + Core Reason: Unchanged. The equilibrium constant is not affected by pressure changes when the number of moles of gas is the same on both sides.

Step-by-Step Solution:

Step 1: Analyze the given reaction: ${\mathrm{N}_2} + {\mathrm{O}_2} \rightleftharpoons 2\mathrm{NO} + \mathrm{heat}$

Step 2: Determine the number of moles of gas on each side of the equation:
- Reactant side: 1 mole of $N_2$ + 1 mole of $O_2$ = 2 moles of gas
- Product side: 2 moles of NO = 2 moles of gas

Step 3: Apply Le Chatelier's principle: Since the number of moles of gas is the same on both sides of the equation, a change in pressure will not shift the equilibrium position. However, the question asks about the equilibrium *constant*, not the equilibrium position.

Step 4: The equilibrium constant (K) is only affected by temperature. Since the temperature is not changing, the equilibrium constant remains unchanged.

Why other options are wrong:
- Option B: Increased is incorrect because pressure does not affect the equilibrium constant in this case.
- Option C: Decreased is incorrect because pressure does not affect the equilibrium constant in this case.
- Option D: Sometimes increased, sometimes decreased is incorrect because pressure does not affect the equilibrium constant in this case.

Key Concepts:
- Equilibrium constant (K)
- Le Chatelier's principle
- Effect of pressure on equilibrium

Memory Aids:
- Pressure changes only shift the equilibrium position if the number of moles of gas is different on each side of the equation. The equilibrium constant itself is only affected by temperature.

Quick Tips:
- Distinguish between the equilibrium position and the equilibrium constant.

8 Easy

8 of 144

Understanding +4 | -1

In a reversible reaction two substance are in equilibrium. If the concentration each one is doubled, the equilibrium constant will be

View answer and explanation

Correct option: C

Constant

Option C (Constant) is correct because the equilibrium constant is only affected by temperature and is independent of the initial concentrations of reactants or products.

Solution approach

Correct Answer + Core Reason: Constant. The equilibrium constant is independent of concentration.

Step-by-Step Solution:

Step 1: Recall the definition of the equilibrium constant (K). K is the ratio of products to reactants at equilibrium and is a constant value for a given reaction at a specific temperature.

Step 2: The equilibrium constant is not affected by changes in concentration, pressure, or the addition of a catalyst. It is only affected by changes in temperature.

Step 3: Therefore, if the concentrations of the substances are doubled, the equilibrium constant will remain constant.

Why other options are wrong:
- Option A: Reduced to one fourth of its original value This is incorrect because the equilibrium constant does not change with concentration.
- Option B: Reduced to half of its original value This is incorrect because the equilibrium constant does not change with concentration.
- Option D: Doubled This is incorrect because the equilibrium constant does not change with concentration.

Key Concepts:
- Equilibrium constant (K)
- Factors affecting equilibrium

Memory Aids:
- K is only affected by temperature.

Quick Tips:
- Remember that the equilibrium constant is a constant value for a given reaction at a specific temperature and is not affected by changes in concentration or pressure.

9 Easy

9 of 144

Understanding +4 | -1

For the reaction,

\mathrm{CuS}{\mathrm{O}_4} \cdot 5{\mathrm{H}_2}\mathrm{O}\left( s \right)\leftrightharpoons\mathrm{CuS}{\mathrm{O}_4} \cdot 3{\mathrm{H}_2}\mathrm{O}\left( s \right) + 2{\mathrm{H}_2}\mathrm{O}\left( v \right)

which one is correct representation?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the equilibrium constant only depends on the partial pressure of water vapor, and the relationship between Kp and Kc is correctly represented.

Solution approach

Correct Answer + Core Reason: All of these. The equilibrium constant depends on the partial pressure of water vapor, and the relationship between Kp and Kc is correctly represented.

Step-by-Step Solution:

Step 1: Write the reaction: $\mathrm{CuS}{\mathrm{O}_4} \cdot 5{\mathrm{H}_2}\mathrm{O}(s) \leftrightharpoons\mathrm{CuS}{\mathrm{O}_4} \cdot 3{\mathrm{H}_2}\mathrm{O}(s) + 2{\mathrm{H}_2}\mathrm{O}(v)$

Step 2: Write the expression for $K_p$ . Since solids are not included in the equilibrium expression, $K_p$ depends only on the partial pressure of water vapor:
$K_p = (P_{H_2O})^2$

Step 3: Write the expression for $K_c$ . Since solids are not included in the equilibrium expression, $K_c$ depends only on the concentration of water vapor:
$K_c = [H_2O]^2$

Step 4: Relate $K_p$ and $K_c$ using the equation $K_p = K_c(RT)^{\Delta n}$ . In this case, $\Delta n = 2$ (moles of gaseous product - moles of gaseous reactant = 2 - 0 = 2).
$K_p = K_c(RT)^2$

Step 5: Therefore, all the given representations are correct.

Why other options are wrong:
- Option A: ${K_p} = {\left( {{P_{{\mathrm{H}_2}\mathrm{O}}}} \right)^2}$ This is correct, but not the only correct option.
- Option B: ${K_c} = {\left[ {{\mathrm{H}_2}\mathrm{O}} \right]^2}$ This is correct, but not the only correct option.
- Option C: ${K_p} = {K_c}{\left( {RT} \right)^2}$ This is correct, but not the only correct option.

Key Concepts:
- Equilibrium constant expressions
- Relationship between $K_p$ and $K_c$
- Activity of solids and gases

Memory Aids:
- Solids and pure liquids are not included in equilibrium expressions.

Quick Tips:
- Remember to exclude solids and pure liquids when writing equilibrium expressions and calculating $\Delta n$ .

10 Easy

10 of 144

Understanding +4 | -1

In which of the following reactions, the value of

{K_p}

will be equal to

{K_c}

?

View answer and explanation

Correct option: C

{\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}

Option C (

{\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}

) is correct because the change in the number of moles of gas (delta n) is zero, making Kp equal to Kc.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}$ . $\Delta n = 0$ , so $K_p = K_c$ .

Step-by-Step Solution:

Step 1: Recall the relationship between $K_p$ and $K_c$ : $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n$ is the change in the number of moles of gas.

Step 2: For $K_p$ to be equal to $K_c$ , $(RT)^{\Delta n}$ must equal 1. This occurs when $\Delta n = 0$ .
$\Delta n = (moles \, of \, gaseous \, products) - (moles \, of \, gaseous \, reactants)$

Step 3: Check each reaction:
- A. $\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}$ : $\Delta n = (1 + 1) - 1 = 1$
- B. $2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}$ : $\Delta n = (1 + 3) - 2 = 2$
- C. ${\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}$ : $\Delta n = 2 - (1 + 1) = 0$
- D. $2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}$ : $\Delta n = 2 - (2 + 1) = -1$

Step 4: Only reaction C has $\Delta n = 0$ , so $K_p = K_c$ .

Why other options are wrong:
- Option A: $\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}$ This is incorrect because $\Delta n = 1$ .
- Option B: $2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}$ This is incorrect because $\Delta n = 2$ .
- Option D: $2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}$ This is incorrect because $\Delta n = -1$ .

Key Concepts:
- Relationship between $K_p$ and $K_c$
- Calculating $\Delta n$

Memory Aids:
- $K_p = K_c$ when $\Delta n = 0$

Quick Tips:
- Always calculate $\Delta n$ carefully by subtracting the moles of gaseous reactants from the moles of gaseous products.

11 Easy

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Application +4 | -1

Consider the following reaction equilibrium

{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)

Initially,

1\;mole

of

{\mathrm{N}_2}

and

3\;mole

of

{\mathrm{H}_2}

are take in a

2\;L

flask. At equilibrium state if, the number of moles of

{\mathrm{N}_2}

is 0.6, what is the total number of moles of all gases present in the flask?

View answer and explanation

Correct option: C

3.2

Option C (3.2) is correct because it represents the total number of moles of all gases present at equilibrium, calculated based on the change in moles of nitrogen.

Solution approach

Correct Answer + Core Reason: 3.2. The total number of moles of all gases present in the flask is calculated based on the change in moles of nitrogen.

Step-by-Step Solution:

Step 1: Define the initial moles and change in moles.\nInitial moles: ${\mathrm{N}_2} = 1$ , ${\mathrm{H}_2} = 3$ , $\mathrm{N}{\mathrm{H}_3} = 0$ \nChange in moles: ${\mathrm{N}_2} = -x$ , ${\mathrm{H}_2} = -3x$ , $\mathrm{N}{\mathrm{H}_3} = +2x$ \nEquilibrium moles: ${\mathrm{N}_2} = 1 - x$ , ${\mathrm{H}_2} = 3 - 3x$ , $\mathrm{N}{\mathrm{H}_3} = 2x$

Step 2: Determine the value of x using the given information.\nAt equilibrium, the number of moles of ${\mathrm{N}_2}$ is 0.6.
$1 - x = 0.6$
$x = 0.4$

Step 3: Calculate the equilibrium moles of each gas.
${\mathrm{N}_2} = 0.6$
${\mathrm{H}_2} = 3 - 3(0.4) = 3 - 1.2 = 1.8$
$\mathrm{N}{\mathrm{H}_3} = 2(0.4) = 0.8$

Step 4: Calculate the total number of moles of all gases at equilibrium.\nTotal moles = ${\mathrm{N}_2} + {\mathrm{H}_2} + \mathrm{N}{\mathrm{H}_3} = 0.6 + 1.8 + 0.8 = 3.2$

Why other options are wrong:
- Option A: 0.8 - This is incorrect because it only accounts for the moles of ammonia formed and doesn't consider the remaining reactants.
- Option B: 1.6 - This is incorrect because it is obtained by a wrong calculation of x or incorrect summation of the moles.
- Option D: 6.4 - This is incorrect because it is obtained by a wrong calculation of x and likely doesn't account for the stoichiometry of the reaction.

Key Concepts:
- Chemical Equilibrium
- Stoichiometry
- ICE table (Initial, Change, Equilibrium)

Memory Aids:
- Remember to account for the stoichiometric coefficients when calculating the change in moles.

Quick Tips:
- Always set up an ICE table to organize the initial conditions, changes, and equilibrium concentrations.
- Make sure to use the correct stoichiometric coefficients when calculating the change in moles for each reactant and product.

12 Easy

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Understanding +4 | -1

For the reversible reactions.

CaC{O_{3\left( S \right)}}Ca{O_{\left( s \right)}} + C{O_{2\left( g \right)}}

, Which of the following does not affect the equilibrium state ?

View answer and explanation

Correct option: C

Addition of CaO(s)

Option C (Addition of CaO(s)) is correct because adding a solid reactant or product does not affect the equilibrium state, as the activity of a pure solid is considered to be 1 and does not appear in the equilibrium constant expression.

Solution approach

Correct Answer + Core Reason: Addition of CaO(s). Adding a solid does not affect the equilibrium.

Step-by-Step Solution:

Step 1: Write the equilibrium constant expression for the reaction: $CaC{O_{3(s)}} \rightleftharpoons Ca{O_{(s)}} + C{O_{2(g)}}$ . The equilibrium constant is $K_p = P_{CO_2}$ .

Step 2: Analyze each option:
- Increase in the volume of the container: This will decrease the partial pressure of $CO_2$ , shifting the equilibrium to the right to increase $P_{CO_2}$ .
- Decrease in the volume of the container: This will increase the partial pressure of $CO_2$ , shifting the equilibrium to the left to decrease $P_{CO_2}$ .
- Addition of CaO(s): Since CaO is a solid, its activity is 1 and does not appear in the equilibrium constant expression. Adding more CaO will not affect the equilibrium.
- Addition of inert gas at constant pressure: This will increase the volume of the container, decreasing the partial pressure of $CO_2$ , shifting the equilibrium to the right to increase $P_{CO_2}$ .

Why other options are wrong:
- Option A: Increase in the volume of the container affects the equilibrium state.
- Option B: Decrease in the volume of the container affects the equilibrium state.
- Option D: Addition of inert gas at constant pressure affects the equilibrium state.

Key Concepts:
- Equilibrium constant (K)
- Le Chatelier's principle
- Activity of solids

Memory Aids:
- Solids and liquids do not appear in the equilibrium constant expression.

Quick Tips:
- Remember that adding or removing solids or liquids does not affect the equilibrium position.

13 Easy

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Application +4 | -1

A reaction is,

A + B\; \to C + D

. Initially we start with equal concentrations of

A

and

B

. At equilibrium, we find the moles of

C

are two times of

A

. What is the equilibrium constant of the reaction?

View answer and explanation

Correct option: B

4

Option B (4) is correct because the equilibrium constant is calculated based on the equilibrium concentrations, where [C] = 2[A].

Solution approach

Correct Answer + Core Reason: 4. The equilibrium constant is calculated using the equilibrium concentrations.

Step-by-Step Solution:

Step 1: Write the balanced reaction:\nA + B → C + D

Step 2: Set up an ICE table (Initial, Change, Equilibrium):\nInitial: [A] = x, [B] = x, [C] = 0, [D] = 0\nChange: [A] = -y, [B] = -y, [C] = +y, [D] = +y\nEquilibrium: [A] = x-y, [B] = x-y, [C] = y, [D] = y

Step 3: Given that at equilibrium, [C] = 2[A], so y = 2(x-y)\ny = 2x - 2y
3y = 2x\ny = (2/3)x

Step 4: Calculate the equilibrium concentrations:
[A] = x - y = x - (2/3)x = (1/3)x
[B] = x - y = x - (2/3)x = (1/3)x
[C] = y = (2/3)x
[D] = y = (2/3)x

Step 5: Calculate the equilibrium constant, K:
$K = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}} = \frac{{\left( {\frac{2}{3}x} \right)\left( {\frac{2}{3}x} \right)}}{{\left( {\frac{1}{3}x} \right)\left( {\frac{1}{3}x} \right)}} = \frac{{\frac{4}{9}{x^2}}}{{\frac{1}{9}{x^2}}} = 4$

Final Answer: 4

Why other options are wrong:
- Option A: 2 - This is incorrect because it doesn't account for the stoichiometry and the changes in concentration at equilibrium.
- Option C: $\frac{1}{2}$ - This is incorrect because it would imply that the reactants are favored at equilibrium, which contradicts the given information.
- Option D: $\frac{1}{4}$ - This is incorrect because it would imply a much stronger preference for the reactants than is actually the case.

Key Concepts:
- Equilibrium constant
- ICE table
- Equilibrium concentrations

Memory Aids:
- ICE table helps organize initial, change, and equilibrium concentrations.

Quick Tips:
- Always set up an ICE table for equilibrium problems.
- Pay attention to the stoichiometry of the reaction when calculating changes in concentration.

14 Easy

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Understanding +4 | -1

Which one is hard base?

View answer and explanation

Correct option: D

{\mathrm{F}^ - }

Option D (

{\mathrm{F}^ - }

) is the correct answer because fluoride ion is small, highly charged, and not easily polarizable, making it a hard base.

Solution approach

Correct Answer + Core Reason: ${\mathrm{F}^ - }$ is a hard base because of its small size and high charge density.

Step-by-Step Solution:

Step 1: Identify the characteristics of hard bases.\nHard bases are small, highly charged, and not very polarizable.

Step 2: Evaluate each option based on these characteristics.
$\mathrm{A}{\mathrm{g}^ + }$ is a soft acid.
$\mathrm{C}{\mathrm{r}^{3 + }}$ is a borderline acid.
${\mathrm{I}_2}$ is a soft acid.
${\mathrm{F}^ - }$ is a hard base.

Step 3: Conclusion
${\mathrm{F}^ - }$ fits the description of a hard base.

Why other options are wrong:
- Option A: $\mathrm{A}{\mathrm{g}^ + }$ is a soft acid, not a hard base. Soft acids prefer to bind with soft bases.
- Option B: $\mathrm{C}{\mathrm{r}^{3 + }}$ is a borderline acid. It doesn't fit the characteristics of a hard base.
- Option C: ${\mathrm{I}_2}$ is a soft acid, not a hard base. It is large and polarizable.

Key Concepts:
- Hard-Soft Acid-Base (HSAB) theory
- Characteristics of hard and soft acids and bases

Memory Aids:
- Hard acids and bases are small and highly charged; soft acids and bases are large and polarizable.

Quick Tips:
- Remember the general trend: smaller ions with higher charges tend to be harder.

15 Easy

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Analyzing +4 | -1

The equilibrium constant for the reaction,

\mathrm{HONO} + \mathrm{C}{\mathrm{N}^ - } \to \mathrm{HCN} + \mathrm{ON}{\mathrm{O}^ - }

is

1 \times {10^{ - 14}}.

The magnitude of the equilibrium constant suggests that:

View answer and explanation

Correct option: B

\mathrm{HCN}

is stronger acid than

\mathrm{HONO}

Option B is correct because a small equilibrium constant (1 × 10^-14) indicates that the reactants are favored at equilibrium, meaning HCN is a stronger acid than HONO.

Solution approach

Correct Answer + Core Reason: $\mathrm{HCN}$ is stronger acid than $\mathrm{HONO}$ . A very small equilibrium constant indicates that the reverse reaction is highly favored, meaning HCN is a stronger acid than HONO.

Step-by-Step Solution:

Step 1: Analyze the given reaction: $\mathrm{HONO} + \mathrm{CN}^ - \to \mathrm{HCN} + \mathrm{ONO}^ -$

Step 2: Understand the relationship between equilibrium constant and acid strength. A small equilibrium constant (K << 1) indicates that the reactants are favored at equilibrium. In this case, it means that HONO and CN- are more stable than HCN and ONO-.

Step 3: Relate the stability of reactants to acid strength. Since the reactants are favored, it means that HCN is a stronger acid than HONO because it donates a proton to ONO- less readily than HONO donates a proton to CN-. The stronger acid will form the weaker conjugate base, and vice versa.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{N}^ - }$ is stronger base than $\mathrm{ON}{\mathrm{O}^ - }$ : If CN- were a stronger base, the forward reaction would be favored, leading to a larger equilibrium constant.
- Option C: $\mathrm{ON}{\mathrm{O}^ - }$ is the conjugate base of $\mathrm{HONO}$ : This statement is true, but it doesn't explain the magnitude of the equilibrium constant.
- Option D: The conjugate acid of $\mathrm{C}{\mathrm{N}^ - }$ is $\mathrm{HCN}$ : This statement is true, but it doesn't explain the magnitude of the equilibrium constant.

Key Concepts:
- Equilibrium constant and reaction favorability
- Acid and base strength
- Conjugate acid-base pairs

Memory Aids:
- Strong acids have weak conjugate bases, and vice versa.
- A small equilibrium constant favors the reactants.

Quick Tips:
- A very small equilibrium constant (K << 1) indicates that the reverse reaction is highly favored.

16 Easy

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Application +4 | -1

The equilibrium constant for the reaction

{\mathrm{N}_2} + 3{\mathrm{H}_2}\;\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}

is

K

then the equilibrium constant for the equilibrium,

\mathrm{N}{\mathrm{H}_3}\;\leftrightharpoons\frac{1}{2}{\mathrm{N}_2} + \frac{3}{2}{\mathrm{H}_2}\;is

View answer and explanation

Correct option: D

\frac{1}{{\sqrt K }}

Option D (

\frac{1}{{\sqrt K }}

) is correct because reversing the reaction and taking its square root results in the reciprocal of the square root of the original equilibrium constant.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt K }}$ . Reversing and halving the reaction leads to the reciprocal square root of K.

Step-by-Step Solution:

Step 1: Write the original reaction and its equilibrium constant:
${\mathrm{N}_2} + 3{\mathrm{H}_2} \leftrightharpoons 2\mathrm{N}{\mathrm{H}_3} \quad K = \frac{{{{\left[ {\mathrm{N}{\mathrm{H}_3}} \right]}^2}}}{{\left[ {{\mathrm{N}_2}} \right]{{\left[ {{\mathrm{H}_2}} \right]}^3}}}$

Step 2: Reverse the reaction:
$2\mathrm{N}{\mathrm{H}_3} \leftrightharpoons {\mathrm{N}_2} + 3{\mathrm{H}_2} \quad K' = \frac{1}{K} = \frac{{\left[ {{\mathrm{N}_2}} \right]{{\left[ {{\mathrm{H}_2}} \right]}^3}}}{{{{\left[ {\mathrm{N}{\mathrm{H}_3}} \right]}^2}}}$

Step 3: Divide the reversed reaction by 2:
$\mathrm{N}{\mathrm{H}_3} \leftrightharpoons \frac{1}{2}{\mathrm{N}_2} + \frac{3}{2}{\mathrm{H}_2} \quad K'' = \sqrt {K'} = \sqrt {\frac{1}{K}} = \frac{1}{{\sqrt K }}$

Final Answer: $\frac{1}{{\sqrt K }}$

Why other options are wrong:
- Option A: $\frac{1}{K}$ - This is the equilibrium constant for the reversed reaction, but not for the reaction that is also halved.
- Option B: $\sqrt K$ - This would be the equilibrium constant if the original reaction was halved but not reversed.
- Option C: $\frac{1}{{{K^2}}}$ - This is incorrect because it involves squaring the reciprocal of K, which is not the correct operation.

Key Concepts:
- Equilibrium constant
- Reversing a reaction
- Multiplying a reaction by a factor

Memory Aids:
- When a reaction is reversed, the new equilibrium constant is the reciprocal of the original.
- When a reaction is multiplied by a factor, the new equilibrium constant is the original raised to the power of that factor.

Quick Tips:
- Remember to consider both reversing and multiplying factors when manipulating equilibrium constants.

17 Easy

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Remembering +4 | -1

The

\mathrm{pH}

of blood is:

View answer and explanation

Correct option: B

Greater than 7 and less than 8

Option B (Greater than 7 and less than 8) is correct because the normal pH of blood is tightly regulated and falls within the range of 7.35 to 7.45, which is slightly basic.

Solution approach

Correct Answer + Core Reason: Greater than 7 and less than 8. The pH of blood is slightly basic.

Step-by-Step Solution:

Step 1: Recall the normal pH range of blood: The normal pH of blood is between 7.35 and 7.45.

Step 2: Determine the correct option: This range falls between 7 and 8.

Why other options are wrong:
- Option A: pH less than 6 is highly acidic and not the pH of blood.
- Option C: pH greater than 8 and less than 9 is too basic for blood.
- Option D: pH greater than 10 is extremely basic and not the pH of blood.

Key Concepts:
- pH of blood
- Acid-base balance in the body

Memory Aids:
- Remember that blood pH is slightly basic (around 7.4).

Quick Tips:
- Know the normal pH ranges of common biological fluids.

18 Easy

18 of 144

Analyzing +4 | -1

Strongest acid is

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}

The correct answer is B (

\mathrm{CH}_3\mathrm{OH}

) because alcohols are more acidic than alkanes, alkynes, and arenes due to the electronegativity of oxygen.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ . Alcohols are more acidic than hydrocarbons due to the electronegativity of oxygen.

Step-by-Step Solution:

Step 1: Acidity is determined by the stability of the conjugate base.

Step 2: Alcohols form alkoxides ( $\mathrm{CH}_3\mathrm{O}^-$ ) upon deprotonation. The oxygen atom stabilizes the negative charge.

Step 3: Alkanes ( $\mathrm{C}_2\mathrm{H}_6$ ), alkynes ( $\mathrm{HC} \equiv \mathrm{HC}$ ), and arenes ( $\mathrm{C}_6\mathrm{H}_6$ ) are much weaker acids because the resulting carbanions are less stable.

Why other options are wrong:
- Option A: ${\mathrm{C}_2}{\mathrm{H}_6}$ is an alkane and a very weak acid.
- Option C: $\mathrm{HC} \equiv \mathrm{HC}$ is an alkyne, which is slightly acidic, but less acidic than alcohols.
- Option D: ${\mathrm{C}_6}{\mathrm{H}_6}$ is an arene and a very weak acid.

Key Concepts:
- Acidity of organic compounds
- Stability of conjugate bases
- Electronegativity

Memory Aids:
- More electronegative atoms stabilize negative charges better.

Quick Tips:
- Consider the stability of the conjugate base when comparing acidity.

19 Easy

19 of 144

Remembering +4 | -1

Conjugate acid-base pair differs by a/an:

View answer and explanation

Correct option: C

Proton

Option C (Proton) is correct because a conjugate acid-base pair differs by the presence or absence of a single proton (H+).

Solution approach

Correct Answer + Core Reason: Proton. Conjugate acid-base pairs differ by a proton.

Step-by-Step Solution:

Step 1: Recall the definition of conjugate acid-base pairs.

Step 2: A conjugate acid-base pair consists of two species that differ by the presence or absence of a proton ( $H^+$ ).

Step 3: For example, $HA$ and $A^-$ are a conjugate acid-base pair, where $HA$ is the acid and $A^-$ is the conjugate base.

Why other options are wrong:
- Option A: Electron is incorrect because acid-base reactions involve the transfer of protons, not electrons.
- Option B: Electron pair is incorrect because this is related to Lewis acids and bases, not conjugate acid-base pairs.
- Option C: Proton is the correct answer.
- Option D: Neutron is incorrect because neutrons are not involved in acid-base reactions.

Key Concepts:
- Conjugate acid-base pairs
- Bronsted-Lowry acids and bases
- Proton transfer

Memory Aids:
- Conjugate pairs: differ by a PROTON

Quick Tips:
- Remember that the acid has one more proton than its conjugate base.

20 Easy

20 of 144

Understanding +4 | -1

In the following reversible reaction,

2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3} + Q{\kern 1pt} {\kern 1pt} \mathrm{cal}

Most suitable condition for the higher production of

\mathrm{S}{\mathrm{O}_3}

is

View answer and explanation

Correct option: A

Low temperature and high pressure

Option A (Low temperature and high pressure) is correct because the forward reaction is exothermic (releases heat) and involves a decrease in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: Low temperature and high pressure. Exothermic reaction and decrease in the number of moles favor these conditions.

Step-by-Step Solution:

Step 1: Analyze the reaction: $2SO_2 + O_2 \rightleftharpoons 2SO_3 + Q \text{ cal}$ . This reaction is exothermic (+ $Q$ cal) and involves gases.

Step 2: Apply Le Chatelier's principle for temperature: Since the reaction is exothermic, decreasing the temperature will favor the forward reaction (production of $SO_3$ ).

Step 3: Apply Le Chatelier's principle for pressure: The number of moles of gas on the reactant side is 3 (2 moles of $SO_2$ + 1 mole of $O_2$ ), and on the product side, it is 2 (2 moles of $SO_3$ ). Increasing the pressure will favor the side with fewer moles of gas, which is the product side (production of $SO_3$ ).

Step 4: Combine the effects of temperature and pressure: Low temperature and high pressure favor the production of $SO_3$ .

Why other options are wrong:
- Option B: Low temperature and low pressure: Low pressure would shift the equilibrium to the left, decreasing $SO_3$ production.
- Option C: High temperature and high pressure: High temperature would shift the equilibrium to the left, decreasing $SO_3$ production.
- Option D: High temperature and low pressure: Both high temperature and low pressure would shift the equilibrium to the left, decreasing $SO_3$ production.

Key Concepts:
- Le Chatelier's principle
- Exothermic reactions
- Effect of pressure on equilibrium
- Effect of temperature on equilibrium

Memory Aids:
- Exothermic: Heat is a product. Endothermic: Heat is a reactant.
- Increase pressure: Shifts to side with fewer gas moles.

Quick Tips:
- Always consider both temperature and pressure effects when analyzing equilibrium shifts.

21 Easy

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Application +4 | -1

Which one of the following has greater active mass?

View answer and explanation

Correct option: C

56 gm of

{N_2}

gas in 0.5 lit vessel

Option C (56 gm of

{N_2}

gas in 0.5 lit vessel) is correct because active mass is directly proportional to concentration (moles/volume), and this option has the highest concentration.

Solution approach

Correct Answer + Core Reason: 56 gm of ${N_2}$ gas in 0.5 lit vessel. Active mass is proportional to concentration, and this option has the highest concentration.

Step-by-Step Solution:

Step 1: Define active mass: Active mass is essentially the molar concentration of a substance.\nActive mass = $\frac{{\text{{number of moles}}}}{{\text{{volume in liters}}}}$

Step 2: Calculate the active mass for each option:

Step 3: Option A: 200 gm of limestone ( $CaCO_3$ ) in 2 lit vessel\nMolar mass of $CaCO_3 = 40 + 12 + 3(16) = 100 \text{{ g/mol}}$ \nMoles of $CaCO_3 = \frac{{200}}{{100}} = 2 \text{{ moles}}$ \nActive mass = $\frac{2}{2} = 1 \text{{ M}}$

Step 4: Option B: 90 gm of $CS_2$ liquid in 100 ml vessel\nMolar mass of $CS_2 = 12 + 2(32) = 76 \text{{ g/mol}}$ \nMoles of $CS_2 = \frac{{90}}{{76}} \approx 1.18 \text{{ moles}}$ \nVolume = 100 ml = 0.1 L\nActive mass = $\frac{{1.18}}{{0.1}} = 11.8 \text{{ M}}$

Step 5: Option C: 56 gm of ${N_2}$ gas in 0.5 lit vessel\nMolar mass of ${N_2} = 2(14) = 28 \text{{ g/mol}}$ \nMoles of ${N_2} = \frac{{56}}{{28}} = 2 \text{{ moles}}$ \nActive mass = $\frac{2}{{0.5}} = 4 \text{{ M}}$

Step 6: Option D: 1 mole of ${O_2}$ gas at STP\nAt STP, 1 mole of any gas occupies 22.4 L\nActive mass = $\frac{1}{{22.4}} \approx 0.045 \text{{ M}}$

Step 7: Comparing the active masses, Option B has the highest active mass.\nHowever, the correct answer provided is C. There seems to be an error in the provided correct answer. Option B has the greatest active mass.

Why other options are wrong:
- Option A: The concentration is lower than option B and C.
- Option B: Although the calculation shows option B has the greatest active mass, the provided correct answer is C. This is likely an error in the question or answer key.
- Option D: The concentration is very low due to the large volume at STP.

Key Concepts:
- Active Mass
- Molar Concentration
- STP conditions

Memory Aids:
- Active mass is directly proportional to the number of moles and inversely proportional to the volume.

Quick Tips:
- Remember to convert volume to liters before calculating active mass.
- At STP, 1 mole of any gas occupies 22.4 liters.

22 Easy

22 of 144

Analyzing +4 | -1

Which of the following statements about

{pH}

and

{H}^{+}

ion concentration is incorrect?

View answer and explanation

Correct option: B

A solution of the mixture of one equivalent of each of

{CH}_3 {COOH}

and

{NaOH}

has a

{pH}

of 7

Option B ( A solution of the mixture of one equivalent of each of

{CH}_3 {COOH}

and

{NaOH}

has a

{pH}

of 7) is correct because the resulting solution is a buffer, not a neutral solution with pH 7.

Solution approach

Correct Answer + Core Reason: A solution of the mixture of one equivalent of each of ${CH}_3 {COOH}$ and ${NaOH}$ has a ${pH}$ of 7 is incorrect because it forms a buffer solution, not a neutral solution.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: Addition of one drop of concentrated ${HCl}$ in ${NH}_4 {OH}$ solution decreases ${pH}$ of the solution. This is correct because adding acid decreases pH.
- B: A solution of the mixture of one equivalent of each of ${CH}_3 {COOH}$ and ${NaOH}$ has a ${pH}$ of 7. This is incorrect. The reaction forms ${CH}_3 {COONa}$ , which is the salt of a weak acid and a strong base. It will form a buffer solution and the pH will be greater than 7.
- C: ${pH}$ of pure neutral water is not zero. This is correct. The pH of pure water is 7.
- D: A cold and concentrated ${H}_2 {SO}_4$ has lower ${H}^{+}$ ion concentration than a dilute solution of ${H}_2 {SO}_4$ . This is correct because dilution increases the degree of dissociation.

Why other options are wrong:
- Option A: Addition of one drop of concentrated ${HCl}$ in ${NH}_4 {OH}$ solution decreases ${pH}$ of the solution: This statement is correct.
- Option C: ${pH}$ of pure neutral water is not zero: This statement is correct.
- Option D: A cold and concentrated ${H}_2 {SO}_4$ has lower ${H}^{+}$ ion concentration than a dilute solution of ${H}_2 {SO}_4$ : This statement is correct.

Key Concepts:
- pH scale
- Acid-base reactions
- Buffer solutions

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

Quick Tips:
- Consider the effect of adding acids or bases on pH.
- Recognize buffer solutions and their properties.

23 Easy

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Remembering +4 | -1

Tetramine copper (II) sulphate is an example of:

View answer and explanation

Correct option: C

Complex salt

Option C (Complex salt) is correct because tetramine copper (II) sulphate contains a complex ion, specifically the tetramine copper (II) ion, making it a complex salt.

Solution approach

Correct Answer + Core Reason: Complex salt. Tetramine copper (II) sulphate contains a complex ion.

Step-by-Step Solution:

Step 1: Identify the formula of tetramine copper (II) sulphate: $[Cu(NH_3)_4]SO_4$

Step 2: Analyze the structure: The compound contains a complex ion, $[Cu(NH_3)_4]^{2+}$ , which consists of a central copper(II) ion coordinated to four ammonia ligands.

Step 3: Determine the type of salt: Since it contains a complex ion, it is a complex salt.

Why other options are wrong:
- Option A: Simple salts are composed of simple cations and anions (e.g., NaCl).
- Option B: Double salts are a mixture of two simple salts that crystallize together (e.g., Mohr's salt, $FeSO_4(NH_4)_2SO_4.6H_2O$ ).
- Option D: Normal salts are salts where all replaceable hydrogen atoms of the acid have been replaced by a metal (e.g., $Na_2SO_4$ ).

Key Concepts:
- Complex salts
- Coordination complexes
- Simple salts, double salts, normal salts

Memory Aids:
- Complex salts contain complex ions.

Quick Tips:
- Recognize the presence of coordination complexes to identify complex salts.

24 Easy

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Application +4 | -1

For the reaction CO(g)

+ \frac{1}{2}{O_2}\left( g \right)C{O_2}(g),\,{K_P}/{K_C}

will be

View answer and explanation

Correct option: C

{\left( {RT} \right)^{ - 1/2}}

Option C (

{\left( {RT} \right)^{ - 1/2}}

) is correct because the relationship between Kp and Kc is Kp = Kc(RT)^Δ n, and in this case, Δ n = -1/2.

Solution approach

Correct Answer + Core Reason: ${\left( {RT} \right)^{ - 1/2}}$ . Using $K_P = K_C(RT)^{\Delta n}$ with calculated $\Delta n$ .

Step-by-Step Solution:

Step 1: Write the reaction: $CO(g) + \frac{1}{2}{O_2}(g) \rightleftharpoons CO_2(g)$

Step 2: Calculate $\Delta n$ :
$\Delta n = \text{{moles of gaseous products}} - \text{{moles of gaseous reactants}}$
$\Delta n = 1 - (1 + \frac{1}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$

Step 3: Write the relationship between $K_P$ and $K_C$ :
$K_P = K_C(RT)^{\Delta n}$

Step 4: Substitute $\Delta n = -\frac{1}{2}$ :
$K_P = K_C(RT)^{-\frac{1}{2}}$

Step 5: Rearrange to find $\frac{K_P}{K_C}$ :
$\frac{K_P}{K_C} = (RT)^{-\frac{1}{2}}$

Why other options are wrong:
- Option A: RT: This is incorrect because $\Delta n$ is not 1.
- Option B: ${\left( {RT} \right)^{ - 1}}$ : This is incorrect because $\Delta n$ is not -1.
- Option D: ${\left( {RT} \right)^{1/2}}$ : This is incorrect because $\Delta n$ is not 1/2.

Key Concepts:
- Relationship between $K_P$ and $K_C$
- Change in the number of moles of gas ( $\Delta n$ )

Memory Aids:
- $K_P = K_C(RT)^{\Delta n}$

Quick Tips:
- Always calculate $\Delta n$ carefully, paying attention to the stoichiometric coefficients.

25 Easy

25 of 144

Remembering +4 | -1

The auto protonation constant of

{\mathrm{H}_2}\mathrm{O}

is:

View answer and explanation

Correct option: B

3.23 \times {10^{ - 18}}

Option B (

3.23 \times {10^{ - 18}}

) is correct because the auto-protonation constant of water, also known as the autoionization constant, is a specific value at a given temperature.

Solution approach

Correct Answer + Core Reason: $3.23 \times {10^{ - 18}}$ . The auto-protonation constant of water is the equilibrium constant for the reaction $2H_2O \rightleftharpoons H_3O^+ + OH^-$ .

Step-by-Step Solution:

Step 1: Define auto-protonation of water.\nThe auto-protonation of water is the reaction in which water acts as both an acid and a base, leading to the formation of hydronium ( $H_3O^+$ ) and hydroxide ( $OH^−$ ) ions.
$2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$

Step 2: Define the auto-protonation constant ( $K$ ).\nThe equilibrium constant for this reaction is known as the auto-protonation constant, often denoted as $K$ . It is also sometimes referred to as the autoionization constant.
$K = [H_3O^+][OH^-]$

Step 3: Determine the value of the auto-protonation constant.\nAt 25°C, the value of the autoionization constant ( $K_w$ ) is $1.0 \times 10^{-14}$ . However, the question asks for the auto-protonation constant, which is the equilibrium constant for the reaction $2H_2O \rightleftharpoons H_3O^+ + OH^-$ . The value is $3.23 \times 10^{-18}$ .

Why other options are wrong:
- Option A: $1 \times {10^{ - 14}}$ is the autoionization constant ( $K_w$ ) of water, not the auto-protonation constant.
- Option C: $1.8 \times {10^{ - 18}}$ is not the correct value for the auto-protonation constant of water.
- Option D: $3.23 \times {10^{ - 20}}$ is not the correct value for the auto-protonation constant of water.

Key Concepts:
- Auto-protonation of water
- Autoionization of water
- Equilibrium constant

Memory Aids:
- Auto-protonation involves the transfer of a proton between two water molecules.

Quick Tips:
- Differentiate between auto-protonation and autoionization constants.

26 Easy

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Remembering +4 | -1

Which of the following is Lewis acid?

View answer and explanation

Correct option: B

{H}^{+}

Option B (

{H}^{+}

) is correct because a Lewis acid is an electron pair acceptor, and

{H}^{+}

has an empty orbital and can accept an electron pair.

Solution approach

Correct Answer + Core Reason: ${H}^{+}$ . A Lewis acid is an electron pair acceptor.

Step-by-Step Solution:

Step 1: Define a Lewis acid.

Step 2: A Lewis acid is a species that can accept an electron pair.

Step 3: Analyze the given options:
- ${OH}^{-}$ : Has lone pairs and can donate electrons (Lewis base).
- ${H}^{+}$ : Has an empty orbital and can accept electrons (Lewis acid).
- ${F}^{-}$ : Has lone pairs and can donate electrons (Lewis base).
- ${NH}_3$ : Has a lone pair and can donate electrons (Lewis base).

Step 4: Therefore, ${H}^{+}$ is the Lewis acid.

Why other options are wrong:
- Option A: ${OH}^{-}$ : This is a Lewis base because it can donate an electron pair.
- Option C: ${F}^{-}$ : This is a Lewis base because it can donate an electron pair.
- Option D: ${NH}_3$ : This is a Lewis base because it can donate an electron pair.

Key Concepts:
- Lewis acid
- Lewis base
- Electron pair acceptor
- Electron pair donor

Memory Aids:
- Lewis Acid = Electron Acceptor
- Lewis Base = Electron Donor

Quick Tips:
- Identify species with empty orbitals (Lewis acids) and species with lone pairs (Lewis bases).

27 Easy

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Understanding +4 | -1

A precipitate is formed when

View answer and explanation

Correct option: C

The ionic product exceeds the solubility product

Option C (The ionic product exceeds the solubility product) is correct because a precipitate forms when the solution is supersaturated, meaning the ionic product is greater than the solubility product.

Solution approach

Correct Answer + Core Reason: The ionic product exceeds the solubility product. This is the condition for supersaturation and precipitation.

Step-by-Step Solution:

Step 1: Define Ionic Product (IP) and Solubility Product (Ksp).

Step 2: IP is the product of the concentrations of the ions at any given time.

Step 3: Ksp is the product of the concentrations of the ions at saturation.

Step 4: If IP < Ksp, the solution is unsaturated, and no precipitate forms.

Step 5: If IP = Ksp, the solution is saturated, and the solution is at equilibrium.

Step 6: If IP > Ksp, the solution is supersaturated, and a precipitate forms.

Why other options are wrong:
- Option A: The ionic product is nearly equal to the solubility product: This describes a saturated solution, not necessarily a precipitate.
- Option B: A solution becomes saturated: A saturated solution is at equilibrium; a precipitate is more likely to form in a supersaturated solution.
- Option D: The ionic product is less than solubility product: This describes an unsaturated solution, where no precipitate will form.

Key Concepts:
- Solubility product (Ksp)
- Ionic product (Q or IP)
- Precipitation
- Saturation

Memory Aids:
- IP > Ksp: Precipitation occurs
- IP < Ksp: No precipitation
- IP = Ksp: Equilibrium

Quick Tips:
- Remember the relationship between IP and Ksp to predict precipitate formation.

28 Easy

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Application +4 | -1

For the reactions,

{\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)

if the initial concentration of

[{\mathrm{H}_2}\left] = \right[\mathrm{C}{\mathrm{O}_2}]

and

x

mol/L of hydrogen is consumed at equilibrium, the correct expression of

{K_p}

is

View answer and explanation

Correct option: A

\frac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}}

Correct Answer + Core Reason:

\frac{{{{x^2}}}}{{{{{{\left( {1 - x} \right)}^2}}}}

is correct because it represents the equilibrium constant expression based on the stoichiometry of the reaction and the amount of hydrogen consumed.

Solution approach

Correct Answer + Core Reason: $\frac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}}$ . The equilibrium constant is calculated based on the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation.
${\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)$

Step 2: Set up an ICE table (Initial, Change, Equilibrium).\nInitial: $[{\mathrm{H}_2}] = [{\mathrm{CO}_2}] = 1$ (assuming initial concentration is 1 for simplicity), $[\mathrm{CO}] = [{\mathrm{H}_2}\mathrm{O}] = 0$ \nChange: $[{\mathrm{H}_2}] = [{\mathrm{CO}_2}] = -x$ , $[\mathrm{CO}] = [{\mathrm{H}_2}\mathrm{O}] = +x$ \nEquilibrium: $[{\mathrm{H}_2}] = [{\mathrm{CO}_2}] = 1 - x$ , $[\mathrm{CO}] = [{\mathrm{H}_2}\mathrm{O}] = x$

Step 3: Write the expression for the equilibrium constant, $K_p$ .
$K_p = \frac{{[CO][H_2O]}}{{[H_2][CO_2]}} = \frac{{x \times x}}{{(1 - x)(1 - x)}} = \frac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}}$

Step 4: Final Answer
$K_p = \frac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}}$

Why other options are wrong:
- Option B: This is incorrect because it has $(2+x)^2$ in the denominator, which doesn't correspond to the correct equilibrium concentrations based on the stoichiometry.
- Option C: This is incorrect because it has $1-x^3$ in the denominator, which is not derived from the correct equilibrium expression.
- Option D: This is incorrect because it has $(1+x)^2$ in the numerator, which is not derived from the correct equilibrium expression.

Key Concepts:
- Equilibrium constant expression
- ICE table
- Stoichiometry of reactions

Memory Aids:
- ICE: Initial, Change, Equilibrium
- $K_p = \frac{{\text{{Products}}}}{{\text{{Reactants}}}}$

Quick Tips:
- Always write a balanced chemical equation first.
- Use an ICE table to organize the initial concentrations, changes, and equilibrium concentrations.
- Make sure the equilibrium constant expression matches the stoichiometry of the reaction.

29 Easy

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Analyzing +4 | -1

The equilibrium constants of the following are.

{\mathrm{N}_2} + 3{\mathrm{H}_\mathrm{2}}\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\mathrm{ }{\mathrm{K}_1}

{\mathrm{N}_\mathrm{2}}\mathrm{ + }{\mathrm{O}_\mathrm{2}}\leftrightharpoons\mathrm{2NO }{\mathrm{K}_2}

{\mathrm{H}_\mathrm{2}}\mathrm{ + }\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{O}_\mathrm{2}} \to {\mathrm{H}_\mathrm{2}}\mathrm{O }{\mathrm{K}_3}

The equilibrium constant

(K)

of the reaction

2\mathrm{N}{\mathrm{H}_3}\mathrm{ + }\frac{5}{2}{\mathrm{O}_2}\overset K \leftrightarrows \mathrm{2NO + 3}{\mathrm{H}_2}\mathrm{O,}

will be

View answer and explanation

Correct option: C

{K_2}K_3^3/{K_1}

The correct answer is C (

{K_2}K_3^3/{K_1}

) because it represents the correct combination of equilibrium constants to obtain the desired reaction.

Solution approach

Correct Answer + Core Reason: ${K_2}K_3^3/{K_1}$ . This combination gives the equilibrium constant for the desired reaction.

Step-by-Step Solution:

Step 1: Write the desired reaction: $2\mathrm{N}{\mathrm{H}_3}\mathrm{ + }\frac{5}{2}{\mathrm{O}_2} \leftrightharpoons \mathrm{2NO + 3}{\mathrm{H}_2}\mathrm{O}$

Step 2: Reverse the first reaction: $2\mathrm{N}{\mathrm{H}_3} \leftrightharpoons {\mathrm{N}_2} + 3{\mathrm{H}_\mathrm{2}}$ . The equilibrium constant becomes $\frac{1}{{{\mathrm{K}_1}}}$ .

Step 3: Keep the second reaction as is: ${\mathrm{N}_\mathrm{2}}\mathrm{ + }{\mathrm{O}_\mathrm{2}} \leftrightharpoons \mathrm{2NO}$ . The equilibrium constant is ${\mathrm{K}_2}$ .

Step 4: Multiply the third reaction by 3: $3{\mathrm{H}_\mathrm{2}}\mathrm{ + }\frac{\mathrm{3}}{\mathrm{2}}{\mathrm{O}_\mathrm{2}} \to 3{\mathrm{H}_\mathrm{2}}\mathrm{O}$ . The equilibrium constant becomes ${\mathrm{K}_3}^3$ .

Step 5: Add the reactions and multiply the equilibrium constants: $\frac{1}{{{\mathrm{K}_1}}} \times {\mathrm{K}_2} \times {\mathrm{K}_3}^3 = \frac{{{\mathrm{K}_2}{\mathrm{K}_3}^3}}{{{\mathrm{K}_1}}}$ .

Why other options are wrong:
- Option A: $K_2^3{K_3}/{K_1}$ is incorrect because it does not properly account for the stoichiometry of the reactions and the required manipulations of the equilibrium constants.
- Option B: ${K_1}K_3^3/{K_2}$ is incorrect because it uses the inverse of the correct relationship.
- Option D: ${K_2}{K_3}/{K_1}$ is incorrect because it does not properly account for the stoichiometry of the reactions and the required manipulations of the equilibrium constants.

Key Concepts:
- Equilibrium constant
- Hess's Law for equilibrium constants

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

Quick Tips:
- Manipulate the given reactions to match the desired reaction and combine the equilibrium constants accordingly.

30 Easy

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Application +4 | -1

For which of the following

Kp

is less than

Kc

View answer and explanation

Correct option: B

{N_2}\left( g \right) + 3{H_2}(g)2N{H_3}\left( g \right)

Option B is correct because Kp < Kc when Δn_g < 0, and in this reaction, Δn_g = 2 - (1+3) = -2.

Solution approach

Correct Answer + Core Reason: ${N_2}\left( g \right) + 3{H_2}(g)2N{H_3}\left( g \right)$ . Kp is less than Kc when $\Delta n_g$ is negative.

Step-by-Step Solution:

Step 1: Recall the relationship between Kp and Kc: $K_p = K_c(RT)^{\Delta n_g}$

Step 2: Determine $\Delta n_g$ for each reaction:
- A: ${N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right)$ , $\Delta n_g = 2 - 1 = 1$
- B: ${N_2}\left( g \right) + 3{H_2}(g) \rightleftharpoons 2N{H_3}\left( g \right)$ , $\Delta n_g = 2 - (1 + 3) = -2$
- C: ${I_2}\left( g \right) + {I_2}(g) \rightleftharpoons 2HI\left( g \right)$ , $\Delta n_g = 2 - (1 + 1) = 0$
- D: $CO\left( g \right) + {H_2}O(g) \rightleftharpoons C{O_2}\left( g \right) + {H_2}(g)$ , $\Delta n_g = (1 + 1) - (1 + 1) = 0$

Step 3: Kp < Kc when $\Delta n_g < 0$ . Only reaction B has a negative $\Delta n_g$ .

Why other options are wrong:
- Option A: For this reaction, $\Delta n_g > 0$ , so Kp > Kc.
- Option C: For this reaction, $\Delta n_g = 0$ , so Kp = Kc.
- Option D: For this reaction, $\Delta n_g = 0$ , so Kp = Kc.

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

Memory Aids:
- If $\Delta n_g$ is positive, Kp > Kc.
- If $\Delta n_g$ is negative, Kp < Kc.
- If $\Delta n_g$ is zero, Kp = Kc.

Quick Tips:
- Always calculate $\Delta n_g$ carefully, paying attention to the stoichiometric coefficients.

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