Question 4

Question: For the reaction $A(g) \rightleftharpoons 2B(g)$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given :] R = 0.0831 L atm mol⁻¹ K⁻¹; $K_p$ for the reaction at 1000 K is.

Options:

A. 0.033

B. 0.021

C. 83.1

D. $2.077 \times 10^5$

Correct Answer: A

Year: NEET 2025

Solution: The equilibrium constant in terms of concentrations $(K_c)$ is given by the ratio of the forward reaction rate constant $(k_f)$ to the backward reaction rate constant $(k_b)$: $K_c = \frac{k_f}{k_b} = \frac{1}{2500}$. To convert $K_c$ to $K_p$, use the relation: $K_p = K_c(RT)^{\Delta n_g}$. $\Delta n_g = 2 - 1 = 1$. Substituting the values: $K_p = \frac{1}{2500} \times 0.0831 \times 1000 = 0.033$.

Step Solution:

1.  Identify the relationship between rate constants and equilibrium constant: $K_c = \frac{k_f}{k_b}$.

2.  Substitute the given ratio: $K_c = \frac{1}{2500}$.

3.  Calculate the change in gaseous moles: $\Delta n_g = \text{moles of product} - \text{moles of reactant} = 2 - 1 = 1$.

4.  Apply the formula $K_p = K_c(RT)^{\Delta n_g}$.

5.  Perform the calculation: $K_p = \frac{1}{2500} \times (0.0831 \times 1000)^1 = \frac{83.1}{2500} = 0.03324 \approx 0.033$.

Difficulty Level: Medium

Concept Name: Relation between $K_p$ and $K_c$

Short cut solution: Since $K_p = K_c \times RT$ (as $\Delta n = 1$), directly multiply the ratio by $R \times \frac{T}{1000}$ to simplify: $K_p = \frac{83.1}{2500}$.

Question 5

Question: At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:

$3A_2 + B_2 \rightleftharpoons 2A_3B, K_1$

$A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, K_2$

The relation between $K_1$ and $K_2$ is.

Options:

A. $K_1^2 = 2K_2$

B. $K_2 = \frac{K_1}{2}$

C. $K_1 = \frac{1}{\sqrt{K_2}}$

D. $K_2 = \frac{1}{\sqrt{K_1}}$

Correct Answer: D

Year: NEET 2024 Re

Solution: For the reaction $3A_2 + B_2 \rightleftharpoons 2A_3B$ with constant $K_1$, the reverse reaction $2A_3B \rightleftharpoons 3A_2 + B_2$ has a constant $K' = \frac{1}{K_1}$. Multiplying this reversed reaction by $\frac{1}{2}$ gives the second reaction $A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2$ with constant $K_2 = \sqrt{K'} = \frac{1}{\sqrt{K_1}}$.

Step Solution:

1.  Write the first reaction: $3A_2 + B_2 \rightleftharpoons 2A_3B$ ($K_1$).

2.  Reverse the first reaction: $2A_3B \rightleftharpoons 3A_2 + B_2$, so the new constant is $\frac{1}{K_1}$.

3.  Compare the reversed reaction to the second reaction: $A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2$.

4.  Observe that the second reaction is exactly half of the reversed reaction.

5.  Apply the rule that if a reaction is multiplied by $n$, the new constant is $K^n$: $K_2 = (\frac{1}{K_1})^{1/2} = \frac{1}{\sqrt{K_1}}$.

Difficulty Level: Easy

Concept Name: Properties of Equilibrium Constant

Short cut solution: The second reaction is the reverse and half of the first, so $K_2$ must be the reciprocal and square root of $K_1$.

Question 9

Question: In which of the following equilibria, $K_p$ and $K_c$ are NOT equal?

Options:

A. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

B. $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

C. $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$

D. $2BrCl_{(g)} \rightleftharpoons Br_{2(g)} + Cl_{2(g)}$

Correct Answer: A

Year: NEET 2024

Solution: $K_p = K_c(RT)^{\Delta n_g}$. For $K_p \neq K_c$, the change in gaseous moles $\Delta n_g$ must be non-zero ($\Delta n_g \neq 0$). In option A, $\Delta n_g = (1 + 1) - 1 = 1$, so $K_p \neq K_c$. In all other options, $\Delta n_g = 0$.

Step Solution:

1.  Recall the formula: $K_p = K_c(RT)^{\Delta n_g}$.

2.  Understand that $K_p = K_c$ only when $\Delta n_g = 0$.

3.  Calculate $\Delta n_g$ for each option:

       A: $2 - 1 = 1$

       B: $2 - (1 + 1) = 0$

       C: $(1 + 1) - (1 + 1) = 0$

       D: $(1 + 1) - 2 = 0$

4.  Identify that only option A has $\Delta n_g \neq 0$.

5.  Conclude $K_p$ and $K_c$ are not equal for option A.

Difficulty Level: Easy

Concept Name: Relation between $K_p$ and $K_c$

Short cut solution: Look for the reaction where the total number of gaseous reactant moles is different from the gaseous product moles.

Question 16

Question: $3O_2(g) \rightleftharpoons 2O_3(g)$. For the above reaction at 298K, $K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is 0.040M then concentration of $O_3$ in M is.

Options:

A. $4.38 \times 10^{-32}$

B. $1.9 \times 10^{-83}$

C. $2.4 \times 10^{31}$

D. $1.2 \times 10^{21}$

Correct Answer: A

Year: NEET-2022

Solution: The equilibrium expression is $K_c = \frac{[O_3]^2}{[O_2]^3}$. Substituting the given values: $[O_3]^2 = K_c \times [O_2]^3 = 3 \times 10^{-59} \times (0.04)^3$. $[O_3]^2 = 1.9 \times 10^{-63} = 19 \times 10^{-64}$. Taking the square root, $[O_3] = 4.38 \times 10^{-32}$.

Step Solution:

1.  Set up the equilibrium constant expression: $K_c = \frac{[O_3]^2}{[O_2]^3}$.

2.  Rearrange to solve for the product concentration: $[O_3]^2 = K_c \times [O_2]^3$.

3.  Substitute values: $[O_3]^2 = (3.0 \times 10^{-59}) \times (0.04)^3$.

4.  Calculate $[O_3]^2 = 3.0 \times 10^{-59} \times 0.000064 = 1.92 \times 10^{-63}$.

5.  Find the square root: $[O_3] = \sqrt{19.2 \times 10^{-64}} \approx 4.38 \times 10^{-32}$.

Difficulty Level: Medium

Concept Name: Equilibrium Constant Expression

Short cut solution: Since $(0.04)^3 = 64 \times 10^{-6}$, $[O_3]^2 = 3 \times 64 \times 10^{-65} \approx 192 \times 10^{-65} = 19.2 \times 10^{-64}$. The square root of 19.2 is between 4 and 5.

Question 18

Question: $K_p$ for the following reaction is 3.0 at 1000K: $CO_2(g) + C(s) \rightleftharpoons 2CO(g)$. What will be the value of $K_c$ for the reaction at the same temperature? (Given $R = 0.083$ L bar $K^{-1} mol^{-1}$).

Options:

A. 3.6

B. 0.36

C. $3.6 \times 10^{-2}$

D. $3.6 \times 10^{-3}$

Correct Answer: C

Year: NEET Re-2022

Solution: $K_p = K_c(RT)^{\Delta n_g}$. Here, $\Delta n_g = 2 (\text{gaseous products}) - 1 (\text{gaseous reactant}) = 1$ (Note: C is solid and its activity is ignored). Substituting values: $3 = K_c(0.083 \times 1000)$. $K_c = \frac{3}{83} = 3.6 \times 10^{-2}$.

Step Solution:

1.  Determine $\Delta n_g$ for the reaction: $\Delta n_g = 2 - 1 = 1$ (excluding solid Carbon).

2.  Use the relation: $K_p = K_c(RT)^{\Delta n_g}$.

3.  Substitute $K_p = 3.0$, $R = 0.083$, and $T = 1000$.

4.  Solve for $K_c$: $3.0 = K_c \times (0.083 \times 1000)^1$.

5.  Calculate: $K_c = \frac{3.0}{83} \approx 0.03614 = 3.6 \times 10^{-2}$.

Difficulty Level: Medium

Concept Name: Relation between $K_p$ and $K_c$

Short cut solution: $K_c = \frac{K_p}{RT} = \frac{3}{83}$. Since $\frac{3}{75}$ is $0.04$, $\frac{3}{83}$ must be slightly less than $0.04$.

Question 31

Question: The equilibrium constants of the following are:

$N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$

$N_2 + O_2 \rightleftharpoons 2NO; K_2$

$H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$

The equilibrium constant ($K$) of the reaction: $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ will be.

Options:

A. $K_2K_3^3 / K_1$

B. $K_2K_3 / K_1$

C. $K_2^3K_3^3 / K_1$

D. $K_1K_3^3 / K_2$

Correct Answer: A

Year: NEET 2017, 2007, 2003

Solution: To get the target reaction, manipulate the given equations:

(i) Reverse reaction 1: $2NH_3 \rightleftharpoons N_2 + 3H_2$; constant $= \frac{1}{K_1}$.

(ii) Use reaction 2 as is: $N_2 + O_2 \rightleftharpoons 2NO$; constant $= K_2$.

(iii) Multiply reaction 3 by 3: $3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O$; constant $= K_3^3$.

Adding (i), (ii), and (iii) gives the target reaction, so $K = \frac{K_2 \times K_3^3}{K_1}$.

Step Solution:

1.  Analyze the target reaction: $2NH_3$ is on the left, so reverse reaction 1 ($1/K_1$).

2.  Target $2NO$ is on the right, so use reaction 2 as is ($K_2$).

3.  Target $3H_2O$ is on the right, so multiply reaction 3 by 3 ($K_3^3$).

4.  Combine the modified reactions.

5.  Multiply the individual equilibrium constants: $K = (\frac{1}{K_1}) \times K_2 \times K_3^3 = \frac{K_2K_3^3}{K_1}$.

Difficulty Level: Medium

Concept Name: Combining Equilibrium Constants

Short cut solution: Identify the placement and coefficients of reactants/products in the final equation relative to the initial ones; reactants in the final from products in initial means reciprocal, coefficients become powers.

Question 41

Question: If the equilibrium constant for $N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)$ is $K$, the equilibrium constant for $\frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons NO(g)$ will be

Options:

A. $\frac{1}{2}K$

B. $K$

C. $K^{2}$

D. $K^{1/2}$

Correct Answer: D

Year: 2015

Solution: If the reaction is multiplied by $\frac{1}{2}$ then the new equation constant, $K' = K^{1/2}$.

Step Solution:

1.  Identify the original reaction: $N_{2} + O_{2} \rightleftharpoons 2NO$ with constant $K$.

2.  Analyze the target reaction: $\frac{1}{2}N_{2} + \frac{1}{2}O_{2} \rightleftharpoons NO$.

3.  Note that the target reaction is the original reaction multiplied by $n = 1/2$.

4.  Apply the rule: If a reaction is multiplied by a factor $n$, the new equilibrium constant becomes $K^n$.

5.  Substitute the value: $K_{new} = K^{1/2}$.

Difficulty Level: Easy

Concept Name: Properties of Equilibrium Constant

Short cut solution: The exponent of the new constant is always the factor by which the original equation is multiplied.

Question 42

Question: If the value of equilibrium constant for a particular reaction is $1.6 \times 10^{12}$, then at equilibrium the system will contain

Options:

A. mostly products

B. similar amounts of reactants and products

C. all reactants

D. mostly reactants

Correct Answer: A

Year: 2015 Cancelled

Solution: The value of $K$ is high which means the reaction proceeds almost to completion, i.e., the system will contain mostly products.

Step Solution:

1.  Check the magnitude of the given equilibrium constant: $K = 1.6 \times 10^{12}$.

2.  Recall the definition: $K = \frac{[Products]}{[Reactants]}$.

3.  Observe that $K \gg 1$ (specifically $K > 10^3$).

4.  Understand that a very large $K$ implies the numerator (Products) is much larger than the denominator (Reactants).

5.  Conclude that the equilibrium mixture consists mostly of products.

Difficulty Level: Easy

Concept Name: Significance of the Magnitude of Equilibrium Constant

Short cut solution: Large $K$ ($>10^3$) means products dominate; small $K$ ($<10^{-3}$) means reactants dominate.

Question 45

Question: For a given exothermic reaction, $K_p$ and $K'_p$ are the equilibrium constants at temperatures $T_1$ and $T_2$, respectively. Assuming that heat of reaction is constant in temperature range between $T_1$ and $T_2$, it is readily observed that

Options:

A. $K_p > K'_p$

B. $K_p < K'_p$

C. $K_p = K'_p$

D. $K_p \cdot K'_p = 1$

Correct Answer: A

Year: 2014

Solution: $\log \frac{K'_p}{K_p} = -\frac{\Delta H}{2.303R} \left[ \frac{1}{T_2} - \frac{1}{T_1} \right]$. For an exothermic reaction, $\Delta H$ is negative. If $T_2 > T_1$, then $\left( \frac{1}{T_2} - \frac{1}{T_1} \right)$ is negative. Thus, $\log K_p > \log K'_p$ or $K_p > K'_p$.

Step Solution:

1.  Use the van't Hoff equation: $\log \frac{K'_p}{K_p} = \frac{\Delta H}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.

2.  Set the condition for exothermic reaction: $\Delta H < 0$.

3.  Assume temperature increases ($T_2 > T_1$), making the temperature term positive.

4.  Multiply the negative $\Delta H$ by the positive temperature term, resulting in a negative value for $\log \frac{K'_p}{K_p}$.

5.  Since the log of the ratio is negative, $K'_p < K_p$, meaning $K_p > K'_p$.

Difficulty Level: Medium

Concept Name: van't Hoff Equation

Short cut solution: For exothermic reactions, temperature and $K$ are inversely related; as $T$ increases, $K$ decreases.

Question 55

Question: Given the reaction between 2 gases represented by $A_2$ and $B_2$ to give the compound $AB(g)$: $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$. At equilibrium the concentration of $A_2 = 3.0 \times 10^{-3} M$, $B_2 = 4.2 \times 10^{-3} M$, $AB = 2.8 \times 10^{-3} M$. If the reaction takes place in a sealed vessel at $527^\circ C$, then the value of $K_c$ will be

Options:

A. 2.0

B. 1.9

C. 0.62

D. 4.5

Correct Answer: C

Year: 2012 Mains

Solution: $K_c = \frac{[AB]^2}{[A_2][B_2]} = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} = \frac{2.8 \times 2.8}{3.0 \times 4.2} = 0.62$.

Step Solution:

1.  Write the equilibrium expression: $K_c = \frac{[AB]^2}{[A_2][B_2]}$.

2.  Plug in the values: $K_c = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$.

3.  Cancel the $10^{-3}$ terms: $\frac{2.8 \times 2.8 \times 10^{-6}}{3.0 \times 4.2 \times 10^{-6}}$.

4.  Simplify the numeric fraction: $\frac{7.84}{12.6}$.

5.  Calculate the final value: $0.622 \dots \approx 0.62$.

Difficulty Level: Easy

Concept Name: Equilibrium Constant Expression

Short cut solution: Since all concentrations have the same $10^{-3}$ factor and $\Delta n_g = 0$, ignore the powers of 10 and calculate $\frac{2.8 \times 2.8}{3 \times 4.2}$.

Question 56

Question: Given that the equilibrium constant for the reaction, $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ has a value of 278 at a particular temperature. What is the value of the equilibrium constant for $SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$ at the same temperature?

Options:

A. $1.8 \times 10^{-3}$

B. $3.6 \times 10^{-3}$

C. $6.0 \times 10^{-2}$

D. $1.3 \times 10^{-5}$

Correct Answer: C

Year: 2012 Mains

Solution: For the reverse reaction $2SO_3 \rightleftharpoons 2SO_2 + O_2$, $K' = 1/K = 1/278$. For $SO_3 \rightleftharpoons SO_2 + \frac{1}{2}O_2$, $K'' = \sqrt{K'} = \sqrt{1/278} = 0.0599 \approx 0.06$.

Step Solution:

1.  Identify the original constant: $K = 278$.

2.  Reverse the reaction to get $2SO_3 \rightleftharpoons 2SO_2 + O_2$, so $K_{rev} = \frac{1}{278}$.

3.  Recognize the target equation is half of the reversed equation.

4.  Apply the square root for the half-coefficient: $K_{target} = \sqrt{\frac{1}{278}}$.

5.  Compute: $\sqrt{0.003597} \approx 0.0599 \approx 6.0 \times 10^{-2}$.

Difficulty Level: Medium

Concept Name: Properties of Equilibrium Constant

Short cut solution: Target is the reverse and half of the original; $K_{new} = (1/K)^{1/2} = \frac{1}{\sqrt{278}}$.

Question 60

Question: For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$. What is $K$ for the reaction: $NO_2(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + O_{2}(g)$?

Options:

A. $\frac{1}{2K_1 K_2}$

B. $\frac{1}{4K_1 K_2}$

C. $(\frac{1}{K_1 K_2})^{1/2}$

D. $\frac{1}{K_1 K_2}$

Correct Answer: C

Year: 2011

Solution: $K_1 = \frac{[NO]^2}{[N_2][O_2]}$, $K_2 = \frac{[NO_2]^2}{[NO]^2[O_2]}$. $K = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$. By manipulation, $K = \sqrt{\frac{1}{K_1 K_2}}$.

Step Solution:

1.  Add the two given reactions: $(N_2 + O_2) + (2NO + O_2) \rightleftharpoons 2NO + 2NO_2$.

2.  Simplify: $N_2 + 2O_2 \rightleftharpoons 2NO_2$ with constant $K' = K_1 \times K_2$.

3.  Reverse this combined reaction: $2NO_2 \rightleftharpoons N_2 + 2O_2$, constant $K'' = \frac{1}{K_1 K_2}$.

4.  Multiply by $1/2$ to reach the target: $NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2$.

5.  Apply the power of $1/2$: $K_{final} = (K'')^{1/2} = \sqrt{\frac{1}{K_1 K_2}}$.

Difficulty Level: Hard

Concept Name: Combining Equilibrium Constants

Short cut solution: The target reaction is the reverse and half of the sum of the first two reactions. Therefore, $K = \sqrt{\frac{1}{K_1 K_2}}$.

Question 64

Question: The reaction, $2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)$ is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression:.

Options:

A. $[(0.75)^3 (0.25)] \div [(1.00)^2 (1.00)]$

B. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)]$

C. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.25)]$

D. $[(0.75)^3 (0.25)] \div [(0.75)^2 (0.25)]$

Correct Answer: B

Year: 2010 Mains

Solution: Initial moles of A and B are 1 each. At equilibrium, if [D] = 0.25, then by stoichiometry, the moles of A consumed are $2 \times 0.25 = 0.50$ and B consumed is 0.25. Moles of C formed are $3 \times 0.25 = 0.75$. Thus, equilibrium concentrations are $[A] = 1 - 0.5 = 0.5$, $[B] = 1 - 0.25 = 0.75$, $[C] = 0.75$, and $[D] = 0.25$. The expression is $K = \frac{[C]^3 [D]}{[A]^2 [B]}$.

Step Solution:

1.  Define changes: Let $x$ be the change in concentration of D. At equilibrium, $[D] = x = 0.25$ M.

2.  Calculate equilibrium [A]: $[A] = [A]_0 - 2x = 1.00 - 2(0.25) = 0.50$ M.

3.  Calculate equilibrium [B] and [C]: $[B] = [B]_0 - x = 1.00 - 0.25 = 0.75$ M; $[C] = 3x = 3(0.25) = 0.75$ M.

4.  Write $K_c$ expression: $K_c = \frac{[C]^3[D]}{[A]^2[B]}$.

5.  Substitute values: $K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$.

Difficulty Level: Medium

Concept Name: ICE Table / Equilibrium Stoichiometry

Short cut solution: Use the stoichiometric ratio directly from $D = 0.25$ to find $C = 3D$, $B = 1-D$, and $A = 1-2D$.

Question 66

Question: In which of the following equilibrium $K_p$ and $K_c$ are not equal?.

Options:

A. $2NO(g) \rightleftharpoons N_2(g) + O_2(g)$

B. $SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$

C. $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

D. $2C(s) + O_2(g) \rightleftharpoons 2CO(g)$

Correct Answer: D

Year: 2010

Solution: $K_p = K_c(RT)^{\Delta n}$. If $\Delta n = 0$, then $K_p = K_c$. In option D, $\Delta n = 2 (\text{moles of } CO) - 1 (\text{mole of } O_2) = 1$ (excluding solid Carbon). Thus $K_p = K_c(RT) \neq K_c$.

Step Solution:

1.  Recall formula: $K_p = K_c(RT)^{\Delta n_g}$.

2.  Identify condition: $K_p \neq K_c$ implies $\Delta n_g \neq 0$.

3.  Calculate $\Delta n_g$ for gas reactions: For A, B, and C, products and reactants have equal gaseous moles, so $\Delta n_g = 0$.

4.  Analyze D: Reaction is $2C(s) + O_2(g) \rightleftharpoons 2CO(g)$. Only $O_2$ and $CO$ are gaseous.

5.  Calculate $\Delta n_g$ for D: $\Delta n_g = 2 - 1 = 1$. Since $1 \neq 0$, $K_p$ and $K_c$ are not equal.

Difficulty Level: Easy

Concept Name: Relation between $K_p$ and $K_c$

Short cut solution: Quickly scan for the reaction where the sum of gaseous coefficients on the left doesn't equal the sum on the right (remembering to ignore solids).

Question 72

Question: The dissociation constants for acetic acid and HCN at $25^\circ C$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$ respectively. The equilibrium constant for the equilibrium, $CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-$ would be:.

Options:

A. $3.0 \times 10^{-5}$

B. $3.0 \times 10^{-4}$

C. $3.0 \times 10^4$

D. $3.0 \times 10^5$

Correct Answer: C

Year: 2009

Solution: $K_1$ (Acetic acid) $= 1.5 \times 10^{-5}$ and $K_2$ (HCN) $= 4.5 \times 10^{-10}$. The target reaction constant $K = \frac{[HCN][CH_3COO^-]}{[CN^-][CH_3COOH]}$. This is equivalent to $K = \frac{K_1}{K_2} = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} \approx 0.33 \times 10^5 = 3.3 \times 10^4$.

Step Solution:

1.  Identify $K_a$ for Acetic Acid: $K_1 = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = 1.5 \times 10^{-5}$.

2.  Identify $K_a$ for HCN: $K_2 = \frac{[CN^-][H^+]}{[HCN]} = 4.5 \times 10^{-10}$.

3.  Express target $K$: $K_{net} = \frac{[HCN][CH_3COO^-]}{[CN^-][CH_3COOH]}$.

4.  Relate target to constants: $K_{net} = \frac{K_1}{K_2}$.

5.  Calculate: $\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} = \frac{1}{3} \times 10^5 \approx 3.33 \times 10^4 \approx 3 \times 10^4$.

Difficulty Level: Medium

Concept Name: Combining Equilibrium Constants

Short cut solution: $K_{net} = \frac{K_{acid}}{K_{conjugate\_acid}} = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}$.

Question 73

Question: The values of $K_p$ and $K_{p'}$ for the reactions (1) $X \rightleftharpoons Y + Z$ and (2) $A \rightleftharpoons 2B$ are in the ratio 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio:.

Options:

A. 36 : 1

B. 1 : 1

C. 3 : 1

D. 1 : 9

Correct Answer: A

Year: 2008

Solution: For reaction (1), total moles $= 1+\alpha$. $K_{p1} = \frac{\alpha^2 P_1}{1-\alpha^2}$. For reaction (2), $K_{p2} = \frac{4\alpha^2 P_2}{1-\alpha^2}$. Given $\frac{K_{p1}}{K_{p2}} = \frac{9}{1}$, substituting values gives $\frac{P_1}{4P_2} = \frac{9}{1}$, so $\frac{P_1}{P_2} = 36:1$.

Step Solution:

1.  Find total moles at equilibrium: For reaction 1, $n_{total} = (1-\alpha) + \alpha + \alpha = 1+\alpha$. For reaction 2, $n_{total} = (1-\alpha) + 2\alpha = 1+\alpha$.

2.  Express $K_{p1}$: $K_{p1} = \frac{(\frac{\alpha}{1+\alpha}P_1)(\frac{\alpha}{1+\alpha}P_1)}{(\frac{1-\alpha}{1+\alpha}P_1)} = \frac{\alpha^2 P_1}{1-\alpha^2}$.

3.  Express $K_{p2}$: $K_{p2} = \frac{(\frac{2\alpha}{1+\alpha}P_2)^2}{(\frac{1-\alpha}{1+\alpha}P_2)} = \frac{4\alpha^2 P_2}{1-\alpha^2}$.

4.  Set up the ratio: $\frac{K_{p1}}{K_{p2}} = \frac{(\frac{\alpha^2 P_1}{1-\alpha^2})}{(\frac{4\alpha^2 P_2}{1-\alpha^2})} = \frac{P_1}{4P_2}$.

5.  Calculate $P_1/P_2$: $\frac{9}{1} = \frac{P_1}{4P_2} \implies P_1 = 36P_2$.

Difficulty Level: Hard

Concept Name: $K_p$ and Degree of Dissociation ($\alpha$)

Short cut solution: Recognize that $K_{p1} \propto P_1$ and $K_{p2} \propto 4P_2$ for these specific reaction types when $\alpha$ is equal; thus $\frac{9}{1} = \frac{P_1}{4P_2}$.

Question 74

Question: The value of equilibrium constant of the reaction $HI(g) \rightleftharpoons \frac{1}{2}H_2(g) + \frac{1}{2}I_2(g)$ is 8.0. The equilibrium constant of the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:.

Options:

A. 16

B. 1/8

C. 1/16

D. 1/64

Correct Answer: D

Year: 2008

Solution: For $HI \rightleftharpoons \frac{1}{2}H_2 + \frac{1}{2}I_2$, $K = 8$. For the reverse reaction $\frac{1}{2}H_2 + \frac{1}{2}I_2 \rightleftharpoons HI$, $K_{rev} = \frac{1}{8}$. Multiplying this reversed reaction by 2 gives $H_2 + I_2 \rightleftharpoons 2HI$, so $K' = (\frac{1}{8})^2 = \frac{1}{64}$.

Step Solution:

1.  Analyze given constant: $K_1 = 8.0$ for the dissociation of 1 mole of HI into half moles of $H_2$ and $I_2$.

2.  Reverse the reaction: For $\frac{1}{2}H_2 + \frac{1}{2}I_2 \rightleftharpoons HI$, the constant is $K_2 = \frac{1}{K_1} = \frac{1}{8}$.

3.  Multiply reaction coefficients: The target reaction $H_2 + I_2 \rightleftharpoons 2HI$ is $2 \times$ the reversed reaction.

4.  Apply power rule: When a reaction is multiplied by $n$, the new constant is $K^n$. Here, $n = 2$.

5.  Calculate: $K_{final} = (\frac{1}{8})^2 = \frac{1}{64}$.

Difficulty Level: Easy

Concept Name: Properties of Equilibrium Constant

Short cut solution: The target reaction is the reverse and doubled version of the original, so $K_{new} = (\frac{1}{K})^2 = \frac{1}{8^2} = \frac{1}{64}$.

Question 77

Question: The dissociation equilibrium of a gas $AB_2$ can be represented as $2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$. The degree of dissociation is $x$ and is small compared [to 1]. The expression relating the degree of dissociation $x$, equilibrium constant $K_p$ and total pressure $P$ is:.

Options:

A. $(K_p/P)^{1/2}$

B. $(K_p/P)^{1/3}$

C. $(2K_p/P)^{1/2}$

D. $(2K_p/P)^{1/3}$

Correct Answer: D

Year: 2008

Solution: Moles at equilibrium are $2(1-x)$, $2x$, and $x$. Total moles $= 2-2x+2x+x = 2+x \approx 2$ (as $x$ is small). $K_p = \frac{P_{AB}^2 \cdot P_{B_2}}{P_{AB_2}^2}$. Substituting partial pressures: $K_p = \frac{(\frac{2x}{2}P)^2 (\frac{x}{2}P)}{(\frac{2}{2}P)^2} = \frac{x^3 P}{2}$. Solving for $x$ gives $x = (\frac{2K_p}{P})^{1/3}$.

Step Solution:

1.  Set up moles: Initial $AB_2 = 2$. Equilibrium: $AB_2 = 2(1-x)$, $AB = 2x$, $B_2 = x$.

2.  Total moles: $n_{total} = 2(1-x) + 2x + x = 2+x \approx 2$ since $x$ is small.

3.  Find partial pressures: $P_{AB2} \approx P$; $P_{AB} \approx \frac{2x}{2}P = xP$; $P_{B2} = \frac{x}{2}P$.

4.  Write $K_p$ expression: $K_p = \frac{(xP)^2 \cdot (\frac{x}{2}P)}{P^2} = \frac{x^3 P}{2}$.

5.  Solve for $x$: $x^3 = \frac{2K_p}{P} \implies x = (\frac{2K_p}{P})^{1/3}$.

Difficulty Level: Hard

Concept Name: Degree of Dissociation and $K_p$

Short cut solution: For the dissociation of $2AB_2$, $K_p \approx \frac{x^3 P}{2}$ when $x \ll 1$. Rearrange for $x$ to get the cube root of $2K_p/P$.

Question 79

Question: The following equilibrium constants are given:  

$N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$  

$N_2 + O_2 \rightleftharpoons 2NO; K_2$  

$H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$  

The equilibrium constant for the oxidation of $NH_3$ by oxygen to give NO is.  

Options:  

A. $K_2 K_3^2 / K_1$  

B. $K_2^2 K_3^2 / K_1$  

C. $K_1 K_2 K_3$  

D. $K_2 K_3^3 / K_1$  

Correct Answer: D  

Year: 2007  

Solution: $K = \frac{[NO]^2 [H_2O]^3}{[NH_3]^2 [O_2]^{5/2}}$. By manipulation of the given equations, $K = \frac{K_2 \times K_3^3}{K_1}$.  

Step Solution:  

1.  Identify target reaction: $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$.  

2.  Reverse Eq (1): $2NH_3 \rightleftharpoons N_2 + 3H_2$ with constant $K' = 1/K_1$.  

3.  Use Eq (2) as is: $N_2 + O_2 \rightleftharpoons 2NO$ with constant $K'' = K_2$.  

4.  Multiply Eq (3) by 3: $3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O$ with constant $K''' = K_3^3$.  

5.  Combine constants: Multiply $K', K'', \text{ and } K'''$ to get $K_{net} = \frac{K_2 K_3^3}{K_1}$.  

Difficulty Level: Medium  

Concept Name: Combining Equilibrium Constants  

Short cut solution: The target uses $2NO$ (from $K_2$), $3H_2O$ (requires $K_3^3$), and starts with $2NH_3$ (reverse of $K_1$), thus $K = \frac{K_2 \cdot K_3^3}{K_1}$.  

Question 83  

Question: For the reaction: $CH_4(g) + 2O_2(g) \rightleftharpoons CO_2(g) + 2H_2O(l), \Delta H_r = -170.8 kJ mol^{-1}$. Which of the following statements is not true?  

Options:  

A. The reaction is exothermic.  

B. At equilibrium, the concentrations of $CO_2(g)$ and $H_2O(l)$ are not equal.  

C. The equilibrium constant for the reaction is given by $K_p = \frac{[CO_2]}{[CH_4][O_2]}$  

D. Addition of $CH_4(g)$ or $O_2(g)$ at equilibrium will cause a shift to the right.  

Correct Answer: C  

Year: 2006  

Solution: $K_p$ should be expressed in terms of partial pressures: $K_p = \frac{P_{CO_2}}{P_{CH_4} \cdot P_{O_2}^2}$.  

Step Solution:  

1.  Evaluate A: $\Delta H$ is negative ($-170.8$), so it is exothermic (True).  

2.  Evaluate B: Concentrations depend on stoichiometry and initial amounts; they aren't inherently equal (True).  

3.  Evaluate D: Adding reactants ($CH_4, O_2$) shifts equilibrium forward (Right) per Le Chatelier's principle (True).  

4.  Evaluate C: $K_p$ requires partial pressures, not concentrations $[ ]$.  

5.  Identify error in C: The coefficient of $O_2$ is 2, so it must be squared in the expression (False).  

Difficulty Level: Easy  

Concept Name: Equilibrium Constant Expression  

Short cut solution: $K_p$ must use partial pressures and coefficients as powers; the expression in "C" uses concentrations and misses the exponent for $O_2$.  

Question 86  

Question: Equilibrium constants $K_1$ and $K_2$ for the following equilibria: $NO(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO_2(g)$ and $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$ are related as.  

Options:  

A. $K_2 = 1 / K_1^2$  

B. $K_2 = K_1^2$  

C. $K_2 = 1/K_1$  

D. $K_2 = K_1/2$  

Correct Answer: A  

Year: 2005  

Solution: $K_1 = \frac{p_{NO_2}}{p_{NO} \cdot (p_{O_2})^{1/2}}$ and $K_2 = \frac{(p_{NO})^2 \cdot p_{O_2}}{(p_{NO_2})^2}$. Thus, $\sqrt{K_2} = 1/K_1$, so $K_2 = 1/K_1^2$.  

Step Solution:  

1.  Write $K_1$ expression: $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}}$.  

2.  Write $K_2$ expression: $K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$.  

3.  Reverse Eq (1): $NO_2 \rightleftharpoons NO + \frac{1}{2}O_2$ has constant $1/K_1$.  

4.  Square the reversed Eq: $(1/K_1)^2$ corresponds to $2NO_2 \rightleftharpoons 2NO + O_2$.  

5.  Conclude relationship: $K_2 = (1/K_1)^2 = 1/K_1^2$.  

Difficulty Level: Easy  

Concept Name: Properties of Equilibrium Constant  

Short cut solution: Reaction 2 is the reverse of "double reaction 1," so $K_2 = (1/K_1)^2$.  

Question 104  

Question: Equilibrium constant $K_p$ for following reaction: $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$.  

Options:  

A. $K_p = P_{CO_2}$  

B. $K_p = P_{CO_2} \times \frac{P_{MgO}}{P_{MgCO_3}}$  

C. $K_p = \frac{P_{CO_2} + P_{MgO}}{P_{MgCO_3}}$  

D. $K_p = \frac{P_{MgCO_3}}{P_{CO_2} \times P_{MgO}}$  

Correct Answer: A  

Year: 2000  

Solution: $K_p = P_{CO_2}$. Solids do not exert pressure, so their partial pressure is taken as unity.  

Step Solution:  

1.  Identify species: $MgCO_3$ and $MgO$ are solids; $CO_2$ is a gas.  

2.  Recall the rule for heterogeneous equilibrium: Pure solids and liquids have an activity/effective pressure of 1.  

3.  Set up the $K_p$ expression: $K_p = \frac{P_{products}}{P_{reactants}}$.  

4.  Substitute values: $K_p = \frac{1 \cdot P_{CO_2}}{1}$.  

5.  Simplify: $K_p = P_{CO_2}$.  

Difficulty Level: Easy  

Concept Name: Heterogeneous Equilibrium  

Short cut solution: Always omit pure solids $(s)$ and liquids $(l)$ from equilibrium constant expressions.  

Question 108  

Question: If $K_1$ and $K_2$ are the respective equilibrium constants for the two reactions,  

$XeF_6 + H_2O_{(g)} \rightarrow XeOF_{4(g)} + 2HF_{(g)}$  

$XeO_{4(g)} + XeF_{6(g)} \rightarrow XeOF_{4(g)} + XeO_3F_{2(g)}$  

The equilibrium constant of the reaction $XeO_{4(g)} + 2HF_{(g)} \rightarrow XeO_3F_{2(g)} + H_2O_{(g)}$ will be.  

Options:  

A. $K_1/K_2$  

B. $K_1 \cdot K_2$  

C. $K_1/(K_2)^2$  

D. $K_2/K_1$  

Correct Answer: D  

Year: Late 1990s (Sequence context)  

Solution: The target reaction can be obtained by adding the second reaction to the reverse of the first reaction. The new constant is obtained by multiplying their individual constants: $K = K_2 \times (1/K_1) = K_2/K_1$.  

Step Solution:  

1.  Target reaction: $XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$.  

2.  Observe Reactants: $XeO_4$ is a reactant in reaction (2) ($K_2$).  

3.  Observe Products: $H_2O$ is a reactant in reaction (1), so reverse reaction (1) to make it a product ($1/K_1$).  

4.  Check intermediates: Adding these two cancels $XeF_6$ and $XeOF_4$.  

5.  Calculate final K: $K_{net} = K_2 \times \frac{1}{K_1} = \frac{K_2}{K_1}$.  

Difficulty Level: Medium  

Concept Name: Combining Equilibrium Constants  

Short cut solution: Subtract reaction 1 from reaction 2 to get the target; in constants, subtraction is division: $K = K_2 / K_1$.  

Question 112  

Question: The equilibrium constant for the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is K, then the equilibrium constant for the equilibrium $2NH_3 \rightleftharpoons N_2 + 3H_2$ is.  

Options:  

A. $\sqrt{K}$  

B. $\sqrt{1/K}$  

C. $1/K$  

D. $1/K^2$  

Correct Answer: C  

Year: 1996  

Solution: The equilibrium constant for the reverse reaction will be $1/K$.  

Step Solution:  

1.  Analyze given reaction: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ has constant $K$.  

2.  Analyze target reaction: $2NH_3 \rightleftharpoons N_2 + 3H_2$.  

3.  Compare reactions: The target is the exact reverse of the given reaction.  

4.  Apply rule: If a reaction is reversed, the new equilibrium constant is the reciprocal ($1/K_{old}$).  

5.  Conclusion: $K_{new} = 1/K$.  

Difficulty Level: Easy  

Concept Name: Properties of Equilibrium Constant  

Short cut solution: Reversing a reaction flips its equilibrium constant ($K \rightarrow 1/K$).

Question 120

Question: In liquid-gas equilibrium, the pressure of vapours above the liquid is constant at.

Options: 

A. constant temperature

B. low temperature

C. high temperature

D. none of these.

Correct Answer: A.

Year: 1995.

Solution: Vapour pressure is directly related to temperature. Greater is the temperature, greater will be the vapour pressure. So to keep it constant, temperature should be constant.

Step Solution:

1.  Identify the physical state: The system is in liquid-gas equilibrium.

2.  Define Vapour Pressure: It is the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phase at a given temperature.

3.  Analyze the dependency: Vapour pressure is an intensive property that depends solely on the nature of the liquid and the temperature.

4.  Apply the condition: If the temperature changes, the kinetic energy of the molecules changes, thereby changing the pressure.

5.  Conclusion: For the pressure to remain constant, the temperature must remain constant.

The difficulty level: Easy

The Concept Name: Vapour Pressure and Temperature Dependence

Short cut solution: Since vapour pressure is a function of temperature ($P \propto T$), a constant pressure necessitates a constant temperature.

Question 127

Question: $K_1$ and $K_2$ are equilibrium constant for reactions (i) and (ii):

(i) $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

(ii) $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$.

Options: 

A. $K_1 = \left( \frac{1}{K_2} \right)^2$

B. $K_1 = K_2^2$

C. $K_1 = K$

D. $K_1 = (K_2)^0$.

Correct Answer: A.

Year: 1989.

Solution: Reaction (ii) is the reversible reaction of (i) and is half of the reaction (i). Thus, rate constant can be given as: $K_2 = \sqrt{\frac{1}{K_1}}$ or $K_1 = \left[ \frac{1}{K_2} \right]^2$.

Step Solution:

1.  Define the first equilibrium constant: $K_1 = \frac{[NO]^2}{[N_2][O_2]}$.

2.  Define the second equilibrium constant: $K_2 = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.

3.  Observe the relationship: Reaction (ii) is the reverse of reaction (i) and its coefficients are multiplied by $1/2$.

4.  Apply the rules of equilibrium constants: Reversing a reaction gives $1/K$ and multiplying by $n$ raises the constant to the power of $n$. Thus, $K_2 = \left( \frac{1}{K_1} \right)^{1/2}$.

5.  Perform algebraic rearrangement: Square both sides: $K_2^2 = \frac{1}{K_1} \implies K_1 = \frac{1}{K_2^2} = \left( \frac{1}{K_2} \right)^2$.

The difficulty level: Easy

The Concept Name: Properties of Equilibrium Constant

Short cut solution: Reaction (ii) is the reverse and half of (i). Therefore, $K_2 = \sqrt{1/K_1}$. Squaring and inverting gives $K_1 = 1/K_2^2$.