In what manner will increase of pressure affect the following equation? ${\rm{C}}\left( s \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{CO}}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)$

Increase in the yield of hydrogen

Pure ammonia is placed in a vessel at temperature where its dissociation constant (${\rm{\alpha }}$) is appreciable. At equilibrium,

Concentration of ${{\rm{H}}_2}$ is less than that of ${{\rm{N}}_2}$

${K_p}$ does not change significantly with pressure

${\rm{\alpha }}$ does not change with pressure

Concentration of ${\rm{N}}{{\rm{H}}_3}$ does not change with pressure

For the gaseous reaction, ${{\rm{C}}_2}{{\rm{H}}_4} + {{\rm{H}}_2}\leftrightharpoons{{\rm{C}}_2}{{\rm{H}}_6},\;\Delta H = - 130\;{\rm{kJ}}\;{\rm{ mo}}{{\rm{l}}^{ - 1}}$ carried in a closed vessel, the equilibrium concentration of the ${{\rm{C}}_2}{{\rm{H}}_6}$ can definitely be increased by

Increasing temperature and decreasing pressure

Increasing temperature and pressure both

Decreasing temperature and pressure only

Equilibrium MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

91 of 144

Application +4 | -1

In what manner will increase of pressure affect the following equation?

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

View answer and explanation

Correct option: A

Shift in the reverse direction

Option A (Shift in the reverse direction) is correct because increasing pressure favors the side with fewer moles of gas, which is the reactant side in this case.

Solution approach

Correct Answer + Core Reason: Shift in the reverse direction. Increase in pressure favors the side with fewer moles of gas.

Step-by-Step Solution:

Step 1: Write the balanced reaction:\nC(s) + H₂O(g) \leftrightharpoons CO(g) + H₂(g)

Step 2: Determine the number of moles of gas on each side:\nReactant side: 1 mole of gas (H₂O)\nProduct side: 2 moles of gas (CO + H₂)

Step 3: Apply Le Chatelier's principle:\nIncreasing the pressure will shift the equilibrium to the side with fewer moles of gas. In this case, the reactant side has fewer moles of gas.

Final Answer: Shift in the reverse direction

Why other options are wrong:
- Option B: Shift in the forward direction - This is incorrect because increasing pressure favors the side with *fewer* moles of gas, not more.
- Option C: Increase in the yield of hydrogen - This is incorrect because increasing pressure shifts the equilibrium to the reverse direction, reducing the yield of hydrogen.
- Option D: No effect - This is incorrect because pressure changes affect equilibria where there is a difference in the number of moles of gas between reactants and products.

Key Concepts:
- Le Chatelier's principle
- Effect of pressure on equilibrium
- Moles of gas

Memory Aids:
- Le Chatelier's principle: \

92 Easy

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Application +4 | -1

At a given temperature the

{K_c}

for the reaction,

\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)

is

2.4 \times {10^{ - 3}}

. At the same temperature, the

{K_c}

for the reaction

\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)

is :

View answer and explanation

Correct option: C

4.2 \times {10^{ - 2}}

Option C is correct because the second reaction is the reverse of the first, so its equilibrium constant is the inverse of the first: Kc' = 1/Kc = 1/(2.4 × 10^-3) = 416.67. However, the provided answer is incorrect. The correct answer should be approximately 416.67, which is not among the options.

Solution approach

Correct Answer + Core Reason: The provided answer is incorrect. The correct answer should be approximately 416.67. The second reaction is the reverse of the first, so its equilibrium constant is the inverse of the first.

Step-by-Step Solution:

Step 1: Recognize that the second reaction is the reverse of the first.

Step 2: Recall that if a reaction is reversed, the new equilibrium constant is the inverse of the original.
$K_{c, reverse} = \frac{1}{K_{c, forward}}$

Step 3: Calculate the new equilibrium constant:
$K_c' = \frac{1}{2.4 \times 10^{-3}} = \frac{1000}{2.4} = 416.67$

Why other options are wrong:
- Option A: This is the equilibrium constant for the forward reaction, not the reverse.
- Option B: Equilibrium constants cannot be negative.
- Option D: This value is not the inverse of the given equilibrium constant.

Key Concepts:
- Equilibrium constant
- Reversing reactions

Memory Aids:
- Reversing a reaction inverts the equilibrium constant.

Quick Tips:
- Always check if the reaction is reversed or multiplied by a factor.

93 Easy

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Analyzing +4 | -1

Which equilibrium can be described as Lewis acid-base reaction but not Bronsted acid-base reaction?

View answer and explanation

Correct option: D

{\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}

Option D is correct because the reaction involves the donation of electron pairs from NH3 to Cu2+, which is a Lewis acid-base interaction, but no proton transfer, thus not a Bronsted acid-base reaction.

Solution approach

Correct Answer + Core Reason: ${\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}$ . This reaction involves the formation of a coordinate covalent bond, which is a Lewis acid-base reaction, but no proton transfer, so it is not a Bronsted acid-base reaction.

Step-by-Step Solution:

Step 1: Understand the definitions of Lewis and Bronsted acids and bases.
- Lewis acid: electron pair acceptor
- Lewis base: electron pair donor
- Bronsted acid: proton donor
- Bronsted base: proton acceptor

Step 2: Analyze each reaction to determine if it involves proton transfer (Bronsted) and/or electron pair donation/acceptance (Lewis).

Step 3: Evaluate option A: ${\mathrm{H}_2}\mathrm{O} + \mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\leftrightharpoons{\mathrm{H}_3}{\mathrm{O}^ + } + \mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }$ . Acetic acid donates a proton to water. This is both a Bronsted and Lewis acid-base reaction.

Step 4: Evaluate option B: $2\mathrm{N}{\mathrm{H}_3} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}\leftrightharpoons2\mathrm{NH}_4^ + + \mathrm{SO}_4^{2 - }$ . Sulfuric acid donates protons to ammonia. This is both a Bronsted and Lewis acid-base reaction.

Step 5: Evaluate option C: $\mathrm{N}{\mathrm{H}_3} + \mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\leftrightharpoons\mathrm{NH}_4^ + + \mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }$ . Acetic acid donates a proton to ammonia. This is both a Bronsted and Lewis acid-base reaction.

Step 6: Evaluate option D: ${\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}$ . Ammonia donates electron pairs to the copper ion, forming coordinate covalent bonds. There is no proton transfer. This is a Lewis acid-base reaction but not a Bronsted acid-base reaction.

Why other options are wrong:
- Option A: This reaction involves proton transfer from acetic acid to water, making it both a Bronsted and Lewis acid-base reaction.
- Option B: This reaction involves proton transfer from sulfuric acid to ammonia, making it both a Bronsted and Lewis acid-base reaction.
- Option C: This reaction involves proton transfer from acetic acid to ammonia, making it both a Bronsted and Lewis acid-base reaction.
- Option D: This reaction involves the formation of coordinate covalent bonds, which is a Lewis acid-base reaction, but no proton transfer, so it is not a Bronsted acid-base reaction.

Key Concepts:
- Lewis acids and bases
- Bronsted acids and bases
- Proton transfer
- Electron pair donation
- Coordinate covalent bonds

Memory Aids:
- Lewis: "LEwis = Electron"; Bronsted: "Bronsted = Proton"

Quick Tips:
- Look for proton transfer to identify Bronsted acid-base reactions.
- Look for electron pair donation/acceptance to identify Lewis acid-base reactions.
- Reactions involving coordinate covalent bonds are often Lewis acid-base reactions but not Bronsted acid-base reactions.

94 Easy

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Analyzing +4 | -1

Hydrogen and oxygen were heated together in a closed vessel. The equilibrium constant is found to decrease after

2000^\circ C

. Which is responsible for this?

View answer and explanation

Correct option: D

It is a property of the system, hence no reason for lower value

Option D (It is a property of the system, hence no reason for lower value) is the correct answer because the equilibrium constant is temperature-dependent, and its decrease at higher temperatures indicates an exothermic reaction.

Solution approach

Correct Answer + Core Reason: It is a property of the system, hence no reason for lower value. The equilibrium constant is temperature-dependent.

Step-by-Step Solution:

Step 1: Recall that the equilibrium constant, $K$ , is temperature-dependent.

Step 2: If $K$ decreases with increasing temperature, it indicates that the reaction is exothermic (heat is released). According to Le Chatelier's principle, increasing the temperature will shift the equilibrium towards the reactants to counteract the increase in temperature.

Step 3: The question states that the equilibrium constant decreases after 2000°C. This is an inherent property of the system and is due to the exothermic nature of the reaction.

Why other options are wrong:
- Option A: Backward reaction predominates: While it is true that the backward reaction is favored at higher temperatures for an exothermic reaction, this doesn't explain *why* the equilibrium constant decreases. The decrease in K is the fundamental observation.
- Option B: Forward reaction predominates: This is incorrect because if the forward reaction predominated, the equilibrium constant would increase with temperature.
- Option C: Both forward and backward reactions have the same rate: If both reactions have the same rate, there would be no shift in equilibrium, and the equilibrium constant would not change.

Key Concepts:
- Le Chatelier's Principle
- Temperature dependence of equilibrium constant
- Exothermic and Endothermic Reactions

Memory Aids:
- Remember that for exothermic reactions, increasing temperature decreases K, and for endothermic reactions, increasing temperature increases K.

Quick Tips:
- Always consider the temperature dependence of the equilibrium constant when analyzing equilibrium shifts.

95 Easy

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Understanding +4 | -1

Which is a buffer solution

View answer and explanation

Correct option: A

C{H_3}COOH + C{H_3}COONa

Option A (

C{H_3}COOH + C{H_3}COONa

) is correct because it contains a weak acid (acetic acid) and its salt (sodium acetate), which is the definition of a buffer solution.

Solution approach

Correct Answer + Core Reason: $C{H_3}COOH + C{H_3}COONa$ . This is a mixture of a weak acid and its salt, which forms a buffer solution.

Step-by-Step Solution:

Step 1: Define a buffer solution.

Step 2: A buffer solution is a mixture of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).

Step 3: Analyze the given options:
- $C{H_3}COOH + C{H_3}COONa$ : Weak acid ( $C{H_3}COOH$ ) and its salt ( $C{H_3}COONa$ ) - Buffer.
- $C{H_3}COOH + C{H_3}COON{H_4}$ : Weak acid ( $C{H_3}COOH$ ) and another salt, but not a conjugate base - Not a buffer.
- $C{H_3}COOH + N{H_4}Cl$ : Weak acid ( $C{H_3}COOH$ ) and a salt of a weak base and strong acid - Not a buffer.
- $NaOH + HCl$ : Strong acid and strong base - Not a buffer.

Step 4: Therefore, $C{H_3}COOH + C{H_3}COONa$ is a buffer solution.

Why other options are wrong:
- Option B: $C{H_3}COOH + C{H_3}COON{H_4}$ : This is not a standard buffer because $C{H_3}COON{H_4}$ is the salt of a weak acid and weak base, and its effect on pH depends on the relative strengths of the acid and base.
- Option C: $C{H_3}COOH + N{H_4}Cl$ : This is not a buffer because $N{H_4}Cl$ is the salt of a weak base and a strong acid, and it will make the solution acidic.
- Option D: $NaOH + HCl$ : This is a strong acid and strong base, which will neutralize each other and not form a buffer.

Key Concepts:
- Buffer solution
- Weak acid/base
- Conjugate acid/base

Memory Aids:
- Buffer = Weak Acid + Its Salt (or Weak Base + Its Salt)

Quick Tips:
- Look for mixtures of weak acids/bases and their corresponding salts to identify buffer solutions.

96 Easy

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Understanding +4 | -1

The reaction quotient (Q) for the reaction

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

is given by

Q = \frac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}

The reaction will proceed towards right side if

View answer and explanation

Correct option: D

Q < Kc

Option D (

Q < Kc

) is correct because when Q is less than Kc, the ratio of products to reactants is smaller than at equilibrium, so the reaction will shift right to form more products.

Solution approach

Correct Answer + Core Reason: $Q < Kc$ . The reaction will proceed towards the right side to reach equilibrium.

Step-by-Step Solution:

Step 1: Define the reaction quotient (Q) and the equilibrium constant (Kc). Q is a measure of the relative amount of products and reactants present in a reaction at any given time. Kc is the value of Q at equilibrium.

Step 2: Compare Q and Kc:
- If Q < Kc, the ratio of products to reactants is less than that at equilibrium. The reaction will proceed in the forward direction (towards the right) to form more products and reach equilibrium.
- If Q > Kc, the ratio of products to reactants is greater than that at equilibrium. The reaction will proceed in the reverse direction (towards the left) to form more reactants and reach equilibrium.
- If Q = Kc, the system is at equilibrium, and there is no net change in the concentrations of reactants and products.

Step 3: In this case, $Q = \frac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$ . If Q < Kc, the reaction will proceed towards the right side.

Why other options are wrong:
- Option A: $Q > Kc$ If Q > Kc, the reaction will proceed towards the left side.
- Option B: $Q = 0$ If Q = 0, there are no products, and the reaction will proceed towards the right side, but this is not the general condition for proceeding to the right.
- Option C: $Q = Kc$ If Q = Kc, the system is at equilibrium, and there is no net change.

Key Concepts:
- Reaction quotient (Q)
- Equilibrium constant (Kc)
- Le Chatelier's principle

Memory Aids:
- Q < K: Shift Right; Q > K: Shift Left; Q = K: Equilibrium

Quick Tips:
- Always compare Q and Kc to determine the direction in which the reaction will proceed to reach equilibrium.

97 Easy

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Understanding +4 | -1

Which of the following equilibrium will shift in forward direction on applying pressure?

View answer and explanation

Correct option: A

S{O_{2\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}}S{O_{3\left( g \right)}}

Option A (

S{O_{2(g)}} + \frac{1}{2}{O_{2(g)}} \rightleftharpoons S{O_{3(g)}}

) is correct because the forward reaction decreases the number of moles of gas, so increasing pressure favors the forward direction.

Solution approach

Correct Answer + Core Reason: $S{O_{2(g)}} + \frac{1}{2}{O_{2(g)}} \rightleftharpoons S{O_{3(g)}}$ . Applying pressure shifts the equilibrium towards the side with fewer moles of gas.

Step-by-Step Solution:

Step 1: State Le Chatelier's principle.\nLe Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Step 2: Analyze the effect of pressure on equilibrium.\nIncreasing the pressure favors the side with fewer moles of gas.

Step 3: Analyze each reaction.
- Option A: $S{O_{2(g)}} + \frac{1}{2}{O_{2(g)}} \rightleftharpoons S{O_{3(g)}}$ .\nMoles of gaseous reactants = $1 + \frac{1}{2} = 1.5$ .\nMoles of gaseous products = $1$ .\nIncreasing pressure favors the forward direction.
- Option B: $PC{l_{5(g)}} \rightleftharpoons PC{l_{3(g)}} + C{l_2(g)}$ .\nMoles of gaseous reactants = $1$ .\nMoles of gaseous products = $1 + 1 = 2$ .\nIncreasing pressure favors the reverse direction.
- Option C: ${H_{2(g)}} + C{l_{2(g)}} \rightleftharpoons 2HC{l_{(g)}}$ .\nMoles of gaseous reactants = $1 + 1 = 2$ .\nMoles of gaseous products = $2$ .\nPressure has no effect.
- Option D: $CaC{O_{3(s)}} \rightleftharpoons Ca{O_{(s)}} + C{O_{2(g)}}$ .\nMoles of gaseous reactants = $0$ .\nMoles of gaseous products = $1$ .\nIncreasing pressure favors the reverse direction.

Step 4: Identify the reaction that shifts forward with increasing pressure.\nOnly reaction A shifts in the forward direction when pressure is applied.

Why other options are wrong:
- Option B: Increasing pressure shifts the equilibrium in the reverse direction because there are more moles of gas on the product side.
- Option C: Pressure has no effect on this equilibrium because the number of moles of gas is the same on both sides.
- Option D: Increasing pressure shifts the equilibrium in the reverse direction because there are more moles of gas on the product side.

Key Concepts:
- Le Chatelier's Principle
- Effect of Pressure on Equilibrium

Memory Aids:
- Increasing pressure favors the side with fewer gas molecules.

Quick Tips:
- Count the number of moles of gas on each side of the equation to determine the effect of pressure.
- Solids and liquids do not affect the equilibrium position when considering pressure changes.

98 Easy

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Application +4 | -1

Degree of dissociation of

0.1\;\mathrm{ }N

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

is (dissociation constant

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

)

View answer and explanation

Correct option: D

{10^{ - 2}}

Option D (

{10^{ - 2}}

) is correct because it is calculated using the formula for the degree of dissociation of a weak acid.

Solution approach

Correct Answer + Core Reason: ${10^{ - 2}}$ . The degree of dissociation is calculated using the formula $\alpha = \sqrt{\frac{K_a}{C}}$ .

Step-by-Step Solution:

Step 1: Identify the given values:
- Concentration (C) = 0.1 N = 0.1 M
- Dissociation constant (Ka) = $1 \times 10^{-5}$

Step 2: Recall the formula for the degree of dissociation ( $\alpha$ ) of a weak acid:
$\alpha = \sqrt{\frac{K_a}{C}}$

Step 3: Substitute the values into the formula:
$\alpha = \sqrt{\frac{1 \times 10^{-5}}{0.1}}$

Step 4: Calculate the degree of dissociation:
$\alpha = \sqrt{10^{-4}}$
$\alpha = 10^{-2}$

Why other options are wrong:
- Option A: ${10^{ - 5}}$ : This is incorrect because it is the value of Ka, not the degree of dissociation.
- Option B: ${10^{ - 4}}$ : This is incorrect because it is the value inside the square root, not the final answer.
- Option C: ${10^{ - 3}}$ : This is incorrect because it is not the correct calculation of the square root.

Key Concepts:
- Degree of dissociation
- Dissociation constant
- Weak acids

Memory Aids:
- Remember the formula: $\alpha = \sqrt{\frac{K_a}{C}}$

Quick Tips:
- Make sure to use the correct units for concentration (Molarity).

99 Easy

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Remembering +4 | -1

Value of equilibrium constant depends on

View answer and explanation

Correct option: C

Temperature

Option C (Temperature) is correct because the equilibrium constant (K) is temperature-dependent, as described by the van't Hoff equation.

Solution approach

Correct Answer + Core Reason: Temperature. The equilibrium constant is temperature-dependent.

Step-by-Step Solution:

Step 1: Define the equilibrium constant (K): It is a value that relates the amounts of reactants and products at equilibrium.

Step 2: Recall the van't Hoff equation: $\frac{{d(\ln K)}}{{dT}} = \frac{{\Delta H^\circ}}{{RT^2}}$ , where ΔH° is the standard enthalpy change, R is the gas constant, and T is the temperature.

Step 3: The van't Hoff equation shows that the equilibrium constant (K) is dependent on temperature (T). Changes in concentration and pressure may shift the equilibrium position, but they do not change the value of the equilibrium constant itself.

Why other options are wrong:
- Option A: Concentration is incorrect because changing the concentration of reactants or products will shift the equilibrium position, but it will not change the value of the equilibrium constant.
- Option B: Pressure is incorrect because changing the pressure will shift the equilibrium position for gaseous reactions, but it will not change the value of the equilibrium constant.
- Option D: All is incorrect because only temperature directly affects the value of the equilibrium constant.

Key Concepts:
- Equilibrium constant (K)
- van't Hoff equation
- Temperature dependence of equilibrium

Memory Aids:
- K is temperature-dependent.

Quick Tips:
- Remember that only temperature changes the value of the equilibrium constant.

100 Easy

100 of 144

Analyzing +4 | -1

Pure ammonia is placed in a vessel at temperature where its dissociation constant (

\mathrm{\alpha }

) is appreciable. At equilibrium,

View answer and explanation

Correct option: D

Concentration of

{\mathrm{H}_2}

is less than that of

{\mathrm{N}_2}

Option D (Concentration of H2 is less than that of N2) is correct because the stoichiometry of the ammonia dissociation produces 3 moles of H2 for every 1 mole of N2, but the reaction doesn't go to completion.

Solution approach

Correct Answer + Core Reason: Concentration of $H_2$ is less than that of $N_2$ . The stoichiometry of the reaction dictates the relative amounts of $H_2$ and $N_2$ produced.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the dissociation of ammonia:
$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$

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

| | $2NH_3$ | $N_2$ | $3H_2$ |
|-------------|---------|-------|--------|
| Initial | a | 0 | 0 |
| Change | -2x | +x | +3x |
| Equilibrium | a-2x | x | 3x |

Step 3: Analyze the equilibrium concentrations:
$[N_2] = x$
$[H_2] = 3x$

Step 4: Since the dissociation constant ( $\alpha$ ) is appreciable, the reaction proceeds to some extent, but not to completion. Therefore, $x$ will be a positive value, but not equal to $a/2$ .

Step 5: Compare the concentrations of $H_2$ and $N_2$ :
Since $[H_2] = 3x$ and $[N_2] = x$ , we would expect $[H_2] > [N_2]$

101 Easy

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Application +4 | -1

For the reaction,

{\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{NO}\left( \mathrm{g} \right)

, the value of

{K_c}

at

800^\circ C

is 0.1. When the equilibrium concentration of both the reactants is 0.5 mole, what is the value of

{K_p}

at the same temperature?

View answer and explanation

Correct option: B

0.1

Option B (0.1) is correct because

{K_p} = {K_c}

when the change in the number of moles of gas is zero.

Solution approach

Correct Answer + Core Reason: 0.1. Since the number of moles of gaseous reactants equals the number of moles of gaseous products, ${K_p} = {K_c}$ .

Step-by-Step Solution:

Step 1: Identify the given information: ${K_c} = 0.1$ at $800^\circ C$ .

Step 2: Write the relationship between ${K_p}$ and ${K_c}$ : ${K_p} = {K_c}{(RT)^{\Delta n_g}}$ , where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.

Step 3: Calculate $\Delta n_g$ : $\Delta n_g = (2) - (1 + 1) = 0$ .

Step 4: Substitute $\Delta n_g = 0$ into the equation: ${K_p} = {K_c}{(RT)^0} = {K_c} \times 1 = {K_c}$ .

Step 5: Therefore, ${K_p} = 0.1$ .

Why other options are wrong:
- Option A: This is incorrect because it assumes ${K_p}$ is equal to the concentration of reactants, which is not true.
- Option C: This is incorrect because it might be a result of incorrect calculation or misunderstanding of the relationship between ${K_p}$ and ${K_c}$ .
- Option D: This is incorrect because it might be a result of incorrect calculation or misunderstanding of the relationship between ${K_p}$ and ${K_c}$ .

Key Concepts:
- Equilibrium constant
- Relationship between ${K_p}$ and ${K_c}$

Memory Aids:
- If $\Delta n_g = 0$ , then ${K_p} = {K_c}$ .

Quick Tips:
- Always check the change in the number of moles of gaseous reactants and products before calculating ${K_p}$ .

102 Easy

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Understanding +4 | -1

Conjugate acid and bases for

\mathrm{H}_2 \mathrm{PO}_4^{-}

ion respectively are

View answer and explanation

Correct option: D

\mathrm{H}_3 \mathrm{PO}_4

and

\mathrm{HPO}_4^{2-}

Option D (

\mathrm{H}_3 \mathrm{PO}_4

and

\mathrm{HPO}_4^{2-}

) is correct because

\mathrm{H}_3 \mathrm{PO}_4

is formed by adding a proton to

\mathrm{H}_2 \mathrm{PO}_4^{-}

and

\mathrm{HPO}_4^{2-}

is formed by removing a proton from

\mathrm{H}_2 \mathrm{PO}_4^{-}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{HPO}_4^{2-}$ . A conjugate acid is formed by adding a proton, and a conjugate base is formed by removing a proton.

Step-by-Step Solution:

Step 1: Define conjugate acid and conjugate base.\nA conjugate acid is formed when a base accepts a proton ( $\mathrm{H}^+$ ). A conjugate base is formed when an acid donates a proton ( $\mathrm{H}^+$ ).

Step 2: Determine the conjugate acid of $\mathrm{H}_2 \mathrm{PO}_4^{-}$ .\nAdding a proton ( $\mathrm{H}^+$ ) to $\mathrm{H}_2 \mathrm{PO}_4^{-}$ gives $\mathrm{H}_3 \mathrm{PO}_4$ .

Step 3: Determine the conjugate base of $\mathrm{H}_2 \mathrm{PO}_4^{-}$ .\nRemoving a proton ( $\mathrm{H}^+$ ) from $\mathrm{H}_2 \mathrm{PO}_4^{-}$ gives $\mathrm{HPO}_4^{2-}$ .

Step 4: Combine the results.\nThe conjugate acid is $\mathrm{H}_3 \mathrm{PO}_4$ , and the conjugate base is $\mathrm{HPO}_4^{2-}$ .

Why other options are wrong:
- Option A: $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{H}_3 \mathrm{PO}_2$ - $\mathrm{H}_3 \mathrm{PO}_2$ is not formed by removing a proton from $\mathrm{H}_2 \mathrm{PO}_4^{-}$ .
- Option B: $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{PO}_4^{3-}$ - $\mathrm{PO}_4^{3-}$ is formed by removing two protons from $\mathrm{H}_2 \mathrm{PO}_4^{-}$ , not one.
- Option C: $\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{HPO}_4^{2-}$ - $\mathrm{H}_3 \mathrm{PO}_3$ is not formed by adding a proton to $\mathrm{H}_2 \mathrm{PO}_4^{-}$ .

Key Concepts:
- Conjugate Acid
- Conjugate Base
- Bronsted-Lowry Acid-Base Theory

Memory Aids:
- Acid donates a proton, forming a conjugate base.
- Base accepts a proton, forming a conjugate acid.

Quick Tips:
- Remember that the conjugate acid has one more proton than the original species, and the conjugate base has one less proton.

103 Easy

103 of 144

Analyzing +4 | -1

Which one of the following statements is not true?

View answer and explanation

Correct option: C

The pH of

1 \times {10^{ - 8}}\;M\;\mathrm{HCl}

is 8

Option C is correct because the pH of a very dilute strong acid like

1 \times 10^{-8} M \;HCl

cannot be 8. It must be less than 7, considering the contribution of water's autoionization.

Solution approach

Correct Answer + Core Reason: The pH of $1 \times {10^{ - 8}}\;M\;\mathrm{HCl}$ is 8. The pH of a dilute acid cannot be greater than 7.

Step-by-Step Solution:

Step 1: Recognize that HCl is a strong acid and completely dissociates in water.

Step 2: The concentration of $H^+$ ions from HCl is $1 \times 10^{-8} M$ .

Step 3: Consider the autoionization of water: ${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$ , where $[H^+] = [OH^-] = 1 \times 10^{-7} M$ .

Step 4: The total $[H^+]$ is the sum of $[H^+]$ from HCl and water: $1 \times 10^{-8} + 1 \times 10^{-7} = 1.1 \times 10^{-7} M$ .

Step 5: Calculate the pH: $pH = -log[H^+] = -log(1.1 \times 10^{-7}) \approx 6.96$ .

Step 6: The pH should be less than 7, not 8.

Why other options are wrong:
- Option A: The conjugate base of ${\mathrm{H}_2}\mathrm{PO}_4^ - \;is\;\mathrm{HPO}_4^{2 - }$ is correct. ${\mathrm{HPO}_4^{2 - }}$ is formed by removing a proton from ${\mathrm{H}_2}\mathrm{PO}_4^ -$ .
- Option B: $\mathrm{pH} + \mathrm{pOH} = 14$ for all aqueous solutions is correct at 25°C.
- Option D: 96,500 C of electricity when passed through a $\mathrm{CuS}{\mathrm{O}_4}$ solution deposits 1 g equivalent of copper at the cathode is correct based on Faraday's laws of electrolysis.

Key Concepts:
- pH calculation
- Strong acids
- Autoionization of water
- Conjugate acid-base pairs
- Faraday's laws of electrolysis

Memory Aids:
- pH = -log[H+]
- Strong acids completely dissociate

Quick Tips:
- Always consider the autoionization of water when dealing with very dilute acid or base solutions.
- The pH of an acidic solution must be less than 7.

104 Easy

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Understanding +4 | -1

For which reaction

{K_p}

is less than

{K_c}

?

View answer and explanation

Correct option: C

2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}

Option C (

2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}

) is correct because in this reaction, Δn < 0, which makes Kp < Kc.

Solution approach

Correct Answer + Core Reason: $2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}$ . Kp < Kc when Δn < 0.

Step-by-Step Solution:

Step 1: Recall the relationship between Kp and Kc: $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n = n_{products} - n_{reactants}$ (gaseous moles).

Step 2: If $\Delta n < 0$ , then $K_p < K_c$ .

Step 3: If $\Delta n > 0$ , then $K_p > K_c$ .

Step 4: If $\Delta n = 0$ , then $K_p = K_c$ .

Step 5: Analyze each reaction:
- A. ${\mathrm{N}_2}{\mathrm{O}_4}\leftrightharpoons2\mathrm{N}{\mathrm{O}_2}$ : $\Delta n = 2 - 1 = 1$
- B. $2\mathrm{HI}\leftrightharpoons{\mathrm{H}_2} + {\mathrm{I}_2}$ : $\Delta n = 1 + 1 - 2 = 0$
- C. $2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}$ : $\Delta n = 2 - (2 + 1) = -1$
- D. ${\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}$ : $\Delta n = 2 - (1 + 1) = 0$

Step 6: Only reaction C has $\Delta n < 0$ .

Why other options are wrong:
- Option A: ${\mathrm{N}_2}{\mathrm{O}_4}\leftrightharpoons2\mathrm{N}{\mathrm{O}_2}$ : Here, Δn = 1, so Kp > Kc.
- Option B: $2\mathrm{HI}\leftrightharpoons{\mathrm{H}_2} + {\mathrm{I}_2}$ : Here, Δn = 0, so Kp = Kc.
- Option D: ${\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}$ : Here, Δn = 0, so Kp = Kc.

Key Concepts:
- Relationship between Kp and Kc
- $\Delta n = n_{products} - n_{reactants}$

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

Quick Tips:
- Calculate Δn carefully, considering only gaseous species.

105 Easy

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Understanding +4 | -1

Which of the following can act both as Bronsted acid and Bronsted base?

View answer and explanation

Correct option: B

HCO_3^ -

Option B (

HCO_3^ -

) is correct because it can both donate a proton to become

CO_3^{2-}

(acting as an acid) and accept a proton to become

H_2CO_3

(acting as a base).

Solution approach

Correct Answer + Core Reason: $HCO_3^-$ . It can both donate and accept a proton.

Step-by-Step Solution:

Step 1: Recall the definition of Bronsted acids and bases.

Step 2: A Bronsted acid is a proton donor, and a Bronsted base is a proton acceptor.

Step 3: Analyze each option:
- $Cl^-$ can only accept a proton.
- $HCO_3^-$ can donate a proton to form $CO_3^{2-}$ and accept a proton to form $H_2CO_3$ .
- $H_3O^+$ can only donate a proton.
- $OH^-$ can only accept a proton.

Step 4: Therefore, $HCO_3^-$ can act both as a Bronsted acid and a Bronsted base.

Why other options are wrong:
- Option A: $Cl^-$ can only act as a Bronsted base by accepting a proton.
- Option B: $HCO_3^-$ is the correct answer because it can act as both a Bronsted acid and a Bronsted base.
- Option C: $H_3O^+$ can only act as a Bronsted acid by donating a proton.
- Option D: $OH^-$ can only act as a Bronsted base by accepting a proton.

Key Concepts:
- Bronsted acids and bases
- Amphoteric species
- Proton donation and acceptance

Memory Aids:
- Amphoteric: Can be both acid and base.

Quick Tips:
- Look for species with both a removable proton and a lone pair of electrons.

106 Easy

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Application +4 | -1

The

pH

of a

0.0001\mathrm{ }N

solution of

KOH

will be

View answer and explanation

Correct option: C

10

Option C (10) is correct because a 0.0001 N solution of KOH is a 10^-4 M solution of a strong base, so the pOH is 4 and the pH is 14 - 4 = 10.

Solution approach

Correct Answer + Core Reason: 10. KOH is a strong base, so pOH can be calculated directly from concentration, and then pH from pOH.

Step-by-Step Solution:

Step 1: Recognize that KOH is a strong base, meaning it completely dissociates in water to form $K^+$ and $OH^-$ ions.

Step 2: The concentration of the KOH solution is 0.0001 N, which is equivalent to $10^{-4}$ N. Since KOH is a monoprotic base, its normality is equal to its molarity. Therefore, the concentration of $KOH$ is $10^{-4}$ M.

Step 3: Since KOH completely dissociates, the concentration of $OH^-$ ions is also $10^{-4}$ M.

Step 4: Calculate the pOH: $pOH = -log[OH^-]$

107 Easy

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Application +4 | -1

40% of a mixture of 0.2 mole of

{\mathrm{N}_2}

and 0.6 mole of

{\mathrm{H}_2}

react to give

\mathrm{N}{\mathrm{H}_3}

according to the equation,

{\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)

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is :

View answer and explanation

Correct option: A

4\mathrm{ }:\mathrm{ }5

Option A (

4\mathrm{ }:\mathrm{ }5

) is correct because the ratio of final to initial volume is calculated based on the stoichiometry of the reaction and the percentage conversion.

Solution approach

Correct Answer + Core Reason: $4\mathrm{ }:\mathrm{ }5$ . The ratio of final volume to initial volume is calculated based on the change in moles due to the reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: ${\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$

Step 2: Initial moles: ${\mathrm{N}_2} = 0.2$ , ${\mathrm{H}_2} = 0.6$

Step 3: Calculate moles reacted: 40% of the mixture reacts. Since the limiting reactant is N2, 40% of 0.2 moles of N2 reacts, which is $0.40 \times 0.2 = 0.08$ moles.

Step 4: Moles of ${\mathrm{H}_2}$ reacted: $3 \times 0.08 = 0.24$ moles.

Step 5: Moles of $\mathrm{N}{\mathrm{H}_3}$ formed: $2 \times 0.08 = 0.16$ moles.

Step 6: Moles at equilibrium:
${\mathrm{N}_2} = 0.2 - 0.08 = 0.12$ moles
${\mathrm{H}_2} = 0.6 - 0.24 = 0.36$ moles
$\mathrm{N}{\mathrm{H}_3} = 0.16$ moles

Step 7: Total moles at equilibrium: $0.12 + 0.36 + 0.16 = 0.64$ moles.

Step 8: Initial total moles: $0.2 + 0.6 = 0.8$ moles.

Step 9: Ratio of final volume to initial volume: $\frac{{0.64}}{{0.8}} = \frac{{64}}{{80}} = \frac{{4}}{{5}}$

Why other options are wrong:
- Option B: This is the inverse of the correct ratio. It's likely a calculation error where the initial and final moles were flipped.
- Option C: This option doesn't account for the stoichiometry of the reaction and the percentage conversion correctly.
- Option D: This option is also likely a result of incorrect calculations or misunderstanding of the mole ratios.

Key Concepts:
- Stoichiometry
- Limiting Reactant
- Equilibrium
- Mole concept

Memory Aids:
- Remember to balance the chemical equation before calculating mole ratios.

Quick Tips:
- Always identify the limiting reactant when dealing with reactions involving multiple reactants.
- Pay attention to the stoichiometry of the reaction when calculating changes in moles.

108 Easy

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Remembering +4 | -1

The pH of a solution is 4. The hydrogen ion concentration of the solution in mol/L is

View answer and explanation

Correct option: B

{10^{ - 4}}

Option B (

{10^{ - 4}}

) is correct because pH is defined as the negative logarithm of the hydrogen ion concentration, so if pH = 4, then [H+] = 10^-4 M.

Solution approach

Correct Answer + Core Reason: ${10^{ - 4}}$ . The hydrogen ion concentration is calculated from the pH value.

Step-by-Step Solution:

Step 1: Recall the definition of pH: $pH = -\log[H^+]$

Step 2: Rearrange the equation to solve for $[H^+]$ : $[H^+] = 10^{-pH}$

Step 3: Substitute the given pH value: $[H^+] = 10^{-4} \text{{ mol/L}}$

Why other options are wrong:
- Option A: Incorrect calculation of hydrogen ion concentration.
- Option C: Incorrect calculation of hydrogen ion concentration.
- Option D: Incorrect calculation of hydrogen ion concentration.

Key Concepts:
- pH definition
- Hydrogen ion concentration

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

Quick Tips:
- Remember the relationship between pH and hydrogen ion concentration.

109 Easy

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Understanding +4 | -1

The aqueous solution of disodium hydrogen phosphate is:

View answer and explanation

Correct option: C

Basic

Option C (Basic) is correct because disodium hydrogen phosphate (

Na_2HPO_4

) is a salt of a weak acid and a strong base, resulting in a basic aqueous solution.

Solution approach

Correct Answer + Core Reason: Basic. The aqueous solution of disodium hydrogen phosphate is basic.

Step-by-Step Solution:

Step 1: Identify the formula: Disodium hydrogen phosphate is $Na_2HPO_4$ .

Step 2: Consider the hydrolysis: $HPO_4^{2-}$ is the conjugate base of a weak acid ( $H_2PO_4^-$ ). Therefore, it will undergo hydrolysis in water:
$HPO_4^{2-}(aq) + H_2O(l) \rightleftharpoons H_2PO_4^-(aq) + OH^-(aq)$

Step 3: Determine the pH: The production of $OH^-$ ions indicates that the solution is basic.

Why other options are wrong:
- Option A: Acidic solutions have a pH less than 7.
- Option B: Neutral solutions have a pH of 7.
- Option D: The solution is basic, so this option is incorrect.

Key Concepts:
- Hydrolysis of salts
- Acid-base properties of ions

Memory Aids:
- Salts of weak acids and strong bases are basic.

Quick Tips:
- Consider the hydrolysis of the anion to determine the pH of the solution.

110 Easy

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Analyzing +4 | -1

If

{K_1}

and

{K_2}

are equilibrium constants for reactions (I) and (II) respectively for,

{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}\,\,\,\,\, \ldots \left( \mathrm{i} \right)

\frac{1}{2}{\mathrm{N}_2} + \frac{1}{2}{\mathrm{O}_2}\leftrightharpoons\mathrm{NO}\,\,\,\, \ldots \left( {\mathrm{ii}} \right)

Then:

View answer and explanation

Correct option: B

{K_2} = \sqrt {{K_1}}

Option B (

{K_2} = \sqrt {{K_1}}

) is correct because reaction (II) is half of reaction (I), so its equilibrium constant is the square root of the first.

Solution approach

Correct Answer + Core Reason: ${K_2} = \sqrt {{K_1}}$ . The equilibrium constant for the second reaction is the square root of the first because the second reaction is half of the first.

Step-by-Step Solution:

Step 1: Write the equilibrium constant for reaction (I): ${\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}$
$K_1 = \frac{{[NO]^2}}{{[N_2][O_2]}}$

Step 2: Write the equilibrium constant for reaction (II): $\frac{1}{2}{\mathrm{N}_2} + \frac{1}{2}{\mathrm{O}_2}\leftrightharpoons\mathrm{NO}$
$K_2 = \frac{{[NO]}}{{[N_2]^{1/2}[O_2]^{1/2}}}$

Step 3: Relate $K_1$ and $K_2$ : Notice that $K_1 = K_2^2$ . Therefore, $K_2 = \sqrt{K_1}$ .

Why other options are wrong:
- Option A: ${K_2} = {K_1}$ : This is incorrect because the reactions are not identical, and the equilibrium constants are related by a square root.
- Option C: ${K_1} = 2{K_2}$ : This is incorrect because the relationship is not linear.
- Option D: ${K_1} = \frac{1}{2}{K_2}$ : This is incorrect because the relationship is not linear.

Key Concepts:
- Equilibrium Constant
- Relationship between equilibrium constants for related reactions

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

Quick Tips:
- If a reaction is reversed, the new equilibrium constant is the inverse of the original. If a reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power of n.

111 Easy

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Analyzing +4 | -1

Assertion : For the reaction,

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

, the unit of

K_p

will be atm.

Reason : Unit of

K_p

is

(a t m)^{\Delta n}

.

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D is correct because the assertion is false as the unit of Kp is atm², and the reason is true as it correctly states the general formula for the unit of Kp.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. The assertion is incorrect because the unit of $K_p$ is atm $^2$ , not atm. The reason is correct as it states the general formula for the unit of $K_p$ .

Step-by-Step Solution:

Step 1: Analyze the given reaction: $2 {NH}_3({~g}) \rightleftharpoons {N}_2({~g})+3 {H}_2({~g})$

Step 2: Calculate the change in the number of moles of gas, $\Delta n = n_{\text{products}} - n_{\text{reactants}}$ .
$\Delta n = (1 + 3) - 2 = 2$

Step 3: Determine the unit of $K_p$ . The unit of $K_p$ is $(atm)^{\Delta n}$ .\nUnit of $K_p = (atm)^2$

Step 4: Evaluate the assertion. The assertion states that the unit of $K_p$ is atm, which is incorrect. Therefore, the assertion is false.

Step 5: Evaluate the reason. The reason correctly states that the unit of $K_p$ is $(atm)^{\Delta n}$ . Therefore, the reason is true.

Why other options are wrong:
- Option A: If both assertion and reason are true and the reason is the correct explanation of the assertion: The assertion is false.
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: The assertion is false.
- Option C: If assertion is true but reason is false: The assertion is false and the reason is true.

Key Concepts:
- Equilibrium constant $K_p$
- Change in the number of moles of gas, $\Delta n$
- Units of $K_p$

Memory Aids:
- Remember that $\Delta n$ is products minus reactants.

Quick Tips:
- Always calculate $\Delta n$ correctly to determine the unit of $K_p$ .

112 Easy

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Analyzing +4 | -1

Consider the following gaseous equilibrium with equilibrium constants

{K_1}

and

{K_2}

respectively

\begin{array}{l} S{O_2}(g) + \frac{1}{2}{O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}S{O_3}(g)\\ 2S{O_3}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}2S{O_2}(g) + {O_2}(g) \end{array}

The equilibrium constants are related as

View answer and explanation

Correct option: A

K_1^2 = \frac{1}{{{K_2}}}

Option A (

K_1^2 = \frac{1}{{{K_2}}}

) is correct because the second reaction is the reverse of the first reaction multiplied by 2, so the equilibrium constant is the inverse square of the first reaction's equilibrium constant.

Solution approach

Correct Answer + Core Reason: $K_1^2 = \frac{1}{{{K_2}}}$ . The second reaction is the reverse of the first reaction multiplied by 2.

Step-by-Step Solution:

Step 1: Write the given reactions and their equilibrium constants:
Reaction 1: $S{O_2}(g) + \frac{1}{2}{O_2}(g) \rightleftharpoons S{O_3}(g)$ ${K_1}$
Reaction 2: $2S{O_3}(g) \rightleftharpoons 2S{O_2}(g) + {O_2}(g)$ ${K_2}$

Step 2: Relate the two reactions. Notice that Reaction 2 is the reverse of Reaction 1 multiplied by 2.

Step 3: Express the equilibrium constant for the reverse reaction. If we reverse Reaction 1, the new equilibrium constant is $\frac{1}{{{K_1}}}$ .
$S{O_3}(g) \rightleftharpoons S{O_2}(g) + \frac{1}{2}{O_2}(g)$ $\frac{1}{{{K_1}}}$

Step 4: Multiply the reversed reaction by 2:
$2S{O_3}(g) \rightleftharpoons 2S{O_2}(g) + {O_2}(g)$ $(\frac{1}{{{K_1}}})^2 = \frac{1}{{{K_1^2}}}$

Step 5: Compare the equilibrium constant of the modified Reaction 1 with Reaction 2. We have:
${K_2} = \frac{1}{{{K_1^2}}}$

Step 6: Rearrange the equation to find the relationship between ${K_1}$ and ${K_2}$ :
$K_1^2 = \frac{1}{{{K_2}}}$

Why other options are wrong:
- Option B: $2{K_1} = K_2^2$ is incorrect because it doesn't follow the rules for manipulating equilibrium constants when reactions are reversed and multiplied.
- Option C: ${K_2} = \frac{2}{{K_1^2}}$ is incorrect because the factor of 2 is not justified by the relationship between the reactions.
- Option D: $K_2^2 = \frac{1}{{{K_1}}}$ is incorrect because the exponents are not consistent with the relationship between the reactions.

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

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

Quick Tips:
- Always pay attention to how the reactions are related to each other.
- Remember the rules for manipulating equilibrium constants.

113 Easy

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Analyzing +4 | -1

Which of the following has lowest

{P^H}

value

View answer and explanation

Correct option: D

N{H_4}Cl

Option D (

N{H_4}Cl

) is correct because it is a salt of a strong acid and a weak base, resulting in an acidic solution with the lowest pH.

Solution approach

Correct Answer + Core Reason: $N{H_4}Cl$ has the lowest pH because it is a salt of a strong acid and a weak base.

Step-by-Step Solution:

Step 1: Analyze each option:
- $C{H_3}COONa$ : Salt of a weak acid ( $C{H_3}COOH$ ) and a strong base ( $NaOH$ ), so it is basic.
- $NaCl$ : Salt of a strong acid ( $HCl$ ) and a strong base ( $NaOH$ ), so it is neutral.
- $C{H_3}COON{H_4}$ : Salt of a weak acid ( $C{H_3}COOH$ ) and a weak base ( $N{H_4}OH$ ), so its pH depends on the relative strengths of the acid and base. It is close to neutral but slightly acidic.
- $N{H_4}Cl$ : Salt of a strong acid ( $HCl$ ) and a weak base ( $N{H_4}OH$ ), so it is acidic.

Step 2: Compare the pH values: Acidic solutions have lower pH values. Among the given options, $N{H_4}Cl$ is the most acidic.

Why other options are wrong:
- Option A: $C{H_3}COONa$ : This is a basic salt and will have a higher pH.
- Option B: $NaCl$ : This is a neutral salt and will have a pH of 7.
- Option C: $C{H_3}COON{H_4}$ : This is a salt of a weak acid and a weak base, and its pH will be closer to neutral than $N{H_4}Cl$ .

Key Concepts:
- pH scale
- Salt hydrolysis
- Acidic and basic salts

Memory Aids:
- Strong Acid + Weak Base = Acidic Salt
- Strong Base + Weak Acid = Basic Salt
- Strong Acid + Strong Base = Neutral Salt

Quick Tips:
- Identify the acid and base that form the salt to determine its acidity or basicity.

114 Easy

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Understanding +4 | -1

The stronger Bronsted base is:

View answer and explanation

Correct option: A

\mathrm{Cl}{\mathrm{O}^ - }

Option A (

\mathrm{Cl}{\mathrm{O}^ - }

) is the strongest Bronsted base because it is the conjugate base of the weakest acid (HClO) among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}{\mathrm{O}^ - }$ . The conjugate base of the weakest acid is the strongest base.

Step-by-Step Solution:

Step 1: Identify the conjugate acids of the given bases:
- $\mathrm{Cl}{\mathrm{O}^ - }$ has HClO as its conjugate acid.
- $\mathrm{ClO}_2^ -$ has HClO $_2$ as its conjugate acid.
- $\mathrm{ClO}_3^ -$ has HClO $_3$ as its conjugate acid.
- $\mathrm{ClO}_4^ -$ has HClO $_4$ as its conjugate acid.

Step 2: Determine the relative acid strengths of the conjugate acids. The acid strength increases with the number of oxygen atoms attached to the central chlorine atom. Therefore, the order of acid strength is: HClO < HClO $_2$ < HClO $_3$ < HClO $_4$ .

Step 3: Relate acid strength to base strength. The stronger the acid, the weaker its conjugate base, and vice versa. Therefore, the order of base strength is: $\mathrm{Cl}{\mathrm{O}^ - }$ > $\mathrm{ClO}_2^ -$ > $\mathrm{ClO}_3^ -$ > $\mathrm{ClO}_4^ -$ .

Step 4: Identify the strongest Bronsted base. Based on the order of base strength, $\mathrm{Cl}{\mathrm{O}^ - }$ is the strongest Bronsted base.

Why other options are wrong:
- Option B: $\mathrm{ClO}_2^ -$ : This is a weaker base than $\mathrm{Cl}{\mathrm{O}^ - }$ because its conjugate acid (HClO $_2$ ) is stronger than HClO.
- Option C: $\mathrm{ClO}_3^ -$ : This is a weaker base than $\mathrm{Cl}{\mathrm{O}^ - }$ and $\mathrm{ClO}_2^ -$ because its conjugate acid (HClO $_3$ ) is stronger than HClO and HClO $_2$ .
- Option D: $\mathrm{ClO}_4^ -$ : This is the weakest base among the given options because its conjugate acid (HClO $_4$ ) is the strongest acid.

Key Concepts:
- Bronsted-Lowry acid-base theory
- Conjugate acid-base pairs
- Acid strength and base strength

Memory Aids:
- Strong acid = weak conjugate base, and vice versa.

Quick Tips:
- When comparing the strengths of oxyacids, the acid strength generally increases with the number of oxygen atoms.

115 Easy

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Application +4 | -1

For the gaseous reaction,

{\mathrm{C}_2}{\mathrm{H}_4} + {\mathrm{H}_2}\leftrightharpoons{\mathrm{C}_2}{\mathrm{H}_6},\;\Delta H = - 130\;\mathrm{kJ}\;\mathrm{ mo}{\mathrm{l}^{ - 1}}

carried in a closed vessel, the equilibrium concentration of the

{\mathrm{C}_2}{\mathrm{H}_6}

can definitely be increased by

View answer and explanation

Correct option: B

Decreasing temperature and increasing pressure

Option B (Decreasing temperature and increasing pressure) is correct because the reaction is exothermic (

\Delta H < 0

) and involves a decrease in the number of moles of gas, both of which favor the formation of C2H6.

Solution approach

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

Step-by-Step Solution:

Step 1: Analyze the reaction: ${C_2}{H_4} + {H_2} \rightleftharpoons {C_2}{H_6}; \Delta H = -130 \text{ kJ/mol}$ . This reaction is exothermic ( $\Delta H < 0$ ).

Step 2: Apply Le Chatelier's principle for temperature: Since the reaction is exothermic, decreasing the temperature will favor the forward reaction (formation of $C_2H_6$ ).

Step 3: Apply Le Chatelier's principle for pressure: The number of moles of gas on the reactant side is 2 (1 mole of $C_2H_4$ + 1 mole of $H_2$ ), and on the product side, it is 1 (1 mole of $C_2H_6$ ). Increasing the pressure will favor the side with fewer moles of gas, which is the product side (formation of $C_2H_6$ ).

Step 4: Combine the effects of temperature and pressure: Decreasing temperature and increasing pressure favor the formation of $C_2H_6$ .

Why other options are wrong:
- Option A: Increasing temperature and decreasing pressure: Increasing temperature would shift the equilibrium to the left, and decreasing pressure would also shift the equilibrium to the left.
- Option C: Increasing temperature and pressure both: Increasing temperature would shift the equilibrium to the left, while increasing pressure would shift it to the right. The temperature effect is stronger in this case.
- Option D: Decreasing temperature and pressure only: Decreasing temperature would shift the equilibrium to the right, but decreasing pressure would shift it to the left.

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.

116 Easy

116 of 144

Application +4 | -1

If the equilibrium constant for

{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,\mathrm{2NO}\left( \mathrm{g} \right)\,

is k, the equilibrium constant for

\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,\frac{\mathrm{1}}{\mathrm{2}}\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\leftrightharpoons\,\,\mathrm{NO}\left( \mathrm{g} \right)

will be:

View answer and explanation

Correct option: C

\,{\mathrm{K}^{\frac{1}{2}}}

Option C (

\sqrt{{K}}

) is correct because when the reaction is halved, the equilibrium constant is raised to the power of 1/2.

Solution approach

Correct Answer + Core Reason: $\,{\mathrm{K}^{\frac{1}{2}}}$ . When a reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor.

Step-by-Step Solution:

Step 1: Original reaction: ${\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\,\,\leftrightharpoons\,\,\,\,\mathrm{2NO}\left( \mathrm{g} \right)\,; K$

Step 2: Modified reaction: $\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,\frac{\mathrm{1}}{\mathrm{2}}\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\,\,\leftrightharpoons\,\,\,\,\mathrm{NO}\left( \mathrm{g} \right)$

Step 3: The modified reaction is the original reaction multiplied by $\frac{1}{2}$ .

Step 4: Therefore, the new equilibrium constant is ${K^{\frac{1}{2}}} = \sqrt{K}$

Why other options are wrong:
- Option A (K): This is incorrect because it assumes the equilibrium constant remains unchanged, which is not true when the reaction is modified.
- **Option B ( ${\mathrm{K}^\mathrm{2}}$ ): This is incorrect because it squares the equilibrium constant, which would be the case if the reaction was multiplied by 2, not 1/2.
- Option D ( $\frac{1}{2}\mathrm{K}$ ): This is incorrect because it multiplies the equilibrium constant by 1/2, which is not the correct way to adjust the equilibrium constant when the reaction is modified.

Key Concepts:
- Equilibrium constant
- Effect of multiplying a reaction on the equilibrium constant

Memory Aids:**
- If a reaction is multiplied by n, then the new equilibrium constant is $K^n$ .

Quick Tips:
- Remember that reversing a reaction takes the inverse of K, multiplying by n raises K to the power of n, and adding reactions multiplies their K values.

117 Easy

117 of 144

Application +4 | -1

\mathrm{20}\;\mathrm{mL}

of a

0.1\;N\;\mathrm{HCl}

is mixed with

\mathrm{20}\;\mathrm{mL}

of a

0.1\;N\;\mathrm{KOH}

solution. The

\mathrm{pH}

of the solution would be:

View answer and explanation

Correct option: B

7

Option B is correct because equal volumes and concentrations of a strong acid and strong base will neutralize each other, resulting in a neutral solution with a pH of 7.

Solution approach

Correct Answer + Core Reason: 7. Equal moles of strong acid and strong base neutralize each other, resulting in a neutral solution.

Step-by-Step Solution:

Step 1: Calculate the number of moles of HCl: $n_{HCl} = 0.1 \;N \times 0.02 \;L = 0.002 \;moles$

Step 2: Calculate the number of moles of KOH: $n_{KOH} = 0.1 \;N \times 0.02 \;L = 0.002 \;moles$

Step 3: Since the number of moles of HCl and KOH are equal, they will completely neutralize each other.

Step 4: The resulting solution will be neutral, with a pH of 7.

Why other options are wrong:
- Option A: Zero is incorrect because a pH of 0 indicates a very strong acidic solution, which is not the case after neutralization.
- Option C: 2 is incorrect because a pH of 2 indicates an acidic solution, which is not the case after neutralization.
- Option D: 9 is incorrect because a pH of 9 indicates a basic solution, which is not the case after neutralization.

Key Concepts:
- Neutralization reaction
- pH scale
- Strong acid and strong base reactions

Memory Aids:
- Strong acid + Strong base -> Salt + Water (pH = 7)

Quick Tips:
- When a strong acid and a strong base react in equal molar amounts, the resulting solution is neutral (pH = 7).

118 Easy

118 of 144

Analyzing +4 | -1

In which of the following reactions, increases in the volume at constant temperature do not affect the number of moles at equilibrium?

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because in all given reactions, the number of moles of gaseous reactants and products are different, so a change in volume will affect the equilibrium.

Solution approach

Correct Answer + Core Reason: None of the above. Changes in volume at constant temperature affect the number of moles at equilibrium when the number of moles of gaseous reactants and products are different.

Step-by-Step Solution:

Step 1: Analyze reaction A: $2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}$ . 2 moles of gas react to form 4 moles of gas. Change in volume will affect equilibrium.

Step 2: Analyze reaction B: $\mathrm{C}\left( s \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right)$ . 0.5 moles of gas react to form 1 mole of gas. Change in volume will affect equilibrium.

Step 3: Analyze reaction C: ${\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}{\mathrm{O}_2}\left( \mathrm{g} \right)$ . 2 moles of gas react to form 1 mole of gas. Change in volume will affect equilibrium.

Step 4: Conclusion: Since all the reactions are affected by volume change, the correct answer is None of the above.

Why other options are wrong:
- Option A: $2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}$ : Incorrect because the number of moles of gas changes during the reaction.
- Option B: $\mathrm{C}\left( s \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right)$ : Incorrect because the number of moles of gas changes during the reaction.
- Option C: ${\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}{\mathrm{O}_2}\left( \mathrm{g} \right)$ : Incorrect because the number of moles of gas changes during the reaction.

Key Concepts:
- Le Chatelier's Principle
- Effect of volume change on equilibrium

Memory Aids:
- Remember that only gaseous species are considered when determining the effect of volume change on equilibrium.

Quick Tips:
- If the number of moles of gaseous reactants equals the number of moles of gaseous products, then a change in volume will not affect the equilibrium.

119 Easy

119 of 144

Application +4 | -1

pH

of

0.05\;\;M\;\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}

is:

View answer and explanation

Correct option: A

13

Option A (13) is correct because Mg(OH)2 is a strong base that dissociates to give 2 OH- ions, leading to a high pOH and thus a pH close to 13.

Solution approach

Correct Answer + Core Reason: 13. Mg(OH)2 is a strong base and its pH is close to 13.

Step-by-Step Solution:

Step 1: Calculate the hydroxide ion concentration.\nSince Mg(OH)2 dissociates into Mg2+ and 2OH-, the concentration of OH- ions is 2 times the concentration of Mg(OH)2.
$[OH^-] = 2 \times 0.05 M = 0.1 M$

Step 2: Calculate the pOH.
$pOH = -log[OH^-] = -log(0.1) = -log(10^{-1}) = 1$

Step 3: Calculate the pH.
$pH + pOH = 14$
$pH = 14 - pOH = 14 - 1 = 13$

Step 4: Final Answer
$pH = 13$

Why other options are wrong:
- Option B: This is incorrect because a pH of 10 would correspond to a much lower hydroxide concentration than what is present in the solution.
- Option C: This is incorrect because a pH of 1 indicates a very acidic solution, not a basic solution like Mg(OH)2.
- Option D: This is incorrect because a pH of 0 is highly acidic, which is not the case for a solution of Mg(OH)2.

Key Concepts:
- pH and pOH relationship
- Dissociation of strong bases
- Calculation of pH from hydroxide concentration

Memory Aids:
- pH + pOH = 14
- Strong bases have high pH values

Quick Tips:
- Remember that strong bases completely dissociate in water.
- Always consider the stoichiometry of the base when calculating the hydroxide concentration.

120 Easy

120 of 144

Application +4 | -1

At

600\,,\,{K_p}

for the following reaction is 1 atm.

X\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( \mathrm{g} \right) + Z\left( \mathrm{g} \right)

At equilibrium,

50\,\%

of

X\left( \mathrm{g} \right)

is dissociated. The total pressure of the equilibrium system is

p

atm. What is the partial pressure (in atm) of

X\left( \mathrm{g} \right)

at equilibrium?

View answer and explanation

Correct option: A

1

Option A is correct because, at equilibrium, the partial pressure of X is equal to 1 atm, given the total pressure and the degree of dissociation.

Solution approach

Correct Answer + Core Reason: 1. The partial pressure of X at equilibrium is 1 atm.

Step-by-Step Solution:

Step 1: Write the reaction: $X(g) \rightleftharpoons Y(g) + Z(g)$

Step 2: Set up an ICE table:\nInitial: 1 0 0\nChange: -0.5 +0.5 +0.5\nEquilibrium: 0.5 0.5 0.5
(Assuming initial moles of X is 1)

Step 3: Calculate the total number of moles at equilibrium: $0.5 + 0.5 + 0.5 = 1.5$

Step 4: Calculate the mole fractions:\nX: $0.5/1.5 = 1/3$ \nY: $0.5/1.5 = 1/3$ \nZ: $0.5/1.5 = 1/3$

Step 5: Relate Kp to partial pressures:
$K_p = \frac{P_Y P_Z}{P_X} = 1$

Step 6: Let the total pressure be P. Then, $P_X = (1/3)P$ , $P_Y = (1/3)P$ , $P_Z = (1/3)P$ .\nSo, $1 = \frac{((1/3)P)((1/3)P)}{(1/3)P} = \frac{1}{3}P$ \nTherefore, $P = 3$ atm.

Step 7: Calculate the partial pressure of X: $P_X = (1/3)P = (1/3)(3) = 1$ atm.

Why other options are wrong:
- Option B: This is incorrect. It seems to be a miscalculation of the partial pressure.
- Option C: This is incorrect. It seems to be a miscalculation of the partial pressure.
- Option D: This is incorrect. It seems to be a miscalculation of the partial pressure.

Key Concepts:
- Equilibrium constant Kp
- Partial pressures
- ICE tables

Memory Aids:
- $P_i = x_i P_{total}$ , where $x_i$ is the mole fraction of component i.

Quick Tips:
- Use ICE tables to organize equilibrium calculations.

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