AP Chemistry · Learn by Concept

AP Chemistry Help: Properties Of The Equilibrium Constant

Review real example questions for Properties Of The Equilibrium Constant in AP Chemistry.

Question 1 / 10

0 of 10 answered

For the equilibrium reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ , the equilibrium constant is $K$ . What is $K_\text{new}$ for the reaction $\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2(g)}$ ?

All questions

Question 1

$K_\text{new}=K^2$
$K_\text{new}=\dfrac{1}{K^2}$
$K_\text{new}=\sqrt{K}$
$K_\text{new}=K$
$K_\text{new}=\dfrac{1}{K}$ (correct answer)

Explanation: This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction H₂(g) + I₂(g) ⇌ 2HI(g) has equilibrium constant K = [HI]²/([H₂][I₂]). The new reaction 2HI(g) ⇌ H₂(g) + I₂(g) is the exact reverse of the original. When a reaction is reversed, the new equilibrium constant is always the reciprocal of the original: K_new = 1/K. Students might incorrectly choose 1/K² (choice E), thinking the coefficient 2 in front of HI affects the relationship. Remember: reversing a reaction always gives K_new = 1/K, regardless of any coefficients in the balanced equation.

Question 2

For the equilibrium reaction $2\text{NO}(g)+\text{Cl}_2(g)\rightleftharpoons 2\text{NOCl}(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{NO}(g)+\tfrac{1}{2}\text{Cl}_2(g)\rightleftharpoons \text{NOCl}(g)$ ?

$K_{\text{new}}=\sqrt{K}$ (correct answer)
$K_{\text{new}}=\dfrac{1}{\sqrt{K}}$
$K_{\text{new}}=\dfrac{1}{K}$
$K_{\text{new}}=K^2$
$K_{\text{new}}=\dfrac{1}{K^2}$

Explanation: This question assesses halving a reaction's equilibrium constant. The original is 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) with K = [NOCl]² / ([NO]²[Cl₂]), and the new is halved, NO(g) + ½Cl₂(g) ⇌ NOCl(g). K_new = K^{1/2} = √K, as the expression takes the square root. This follows scaling by 1/2. A tempting distractor is D, K², incorrect for using the full power instead of half, stemming from inverting the factor. Calculate the scaling n and use K_new = K^n for same-direction reactions.

Question 3

For the equilibrium reaction $\text{CO}_2(g)+\text{H}_2(g)\rightleftharpoons \text{CO}(g)+\text{H}_2\text{O}(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{CO}(g)+\text{H}_2\text{O}(g)\rightleftharpoons \text{CO}_2(g)+\text{H}_2(g)$ ?

$K_{\text{new}}=K^2$
$K_{\text{new}}=\dfrac{1}{K}$ (correct answer)
$K_{\text{new}}=K$
$K_{\text{new}}=\dfrac{1}{K^2}$
$K_{\text{new}}=\sqrt{K}$

Explanation: This question assesses the equilibrium constant for a reversed reaction. The original is CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g) with K = [CO][H₂O] / ([CO₂][H₂]), and the new is the reverse. Thus, K_new = 1/K, as the expression inverts. This holds for balanced reactions with equal terms. A tempting distractor is D, 1/K², incorrect if scaling is mistakenly added, stemming from overcomplicating simple reversal. Identify pure reversal by swapped sides and directly use K_new = 1/K.

Question 4

For the equilibrium reaction $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ , the equilibrium constant is $K$ . What is $K_\text{new}$ for the reaction $\mathrm{2CO_2(g) + 2H_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g)}$ ?

$K_\text{new}=\dfrac{1}{K}$
$K_\text{new}=K^2$
$K_\text{new}=\dfrac{1}{K^2}$ (correct answer)
$K_\text{new}=\sqrt{K}$
$K_\text{new}=K$

Explanation: This question tests understanding of how equilibrium constants change when a reaction is both reversed and coefficients are multiplied. The original reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has equilibrium constant K. The new reaction 2CO₂(g) + 2H₂(g) ⇌ 2CO(g) + 2H₂O(g) involves two changes: reversing the reaction (which gives 1/K) and doubling all coefficients (which squares the result). Therefore, K_new = (1/K)² = 1/K². Students often incorrectly choose 1/K (choice A), forgetting to account for the coefficient change. Remember: when both reversing and changing coefficients, apply both transformations: reverse first, then adjust for coefficient changes.

Question 5

For the equilibrium reaction $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ , the equilibrium constant is $K$ . What is $K_\text{new}$ for the reaction $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ ?

$K_\text{new}=\dfrac{1}{K^2}$
$K_\text{new}=K^2$
$K_\text{new}=\dfrac{1}{K}$ (correct answer)
$K_\text{new}=\sqrt{K}$
$K_\text{new}=K$

Explanation: This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction 2NO₂(g) ⇌ N₂O₄(g) has equilibrium constant K = [N₂O₄]/[NO₂]². The new reaction N₂O₄(g) ⇌ 2NO₂(g) is the exact reverse of the original. When a reaction is reversed, the new equilibrium constant is always the reciprocal: K_new = [NO₂]²/[N₂O₄] = 1/K. Students might incorrectly choose 1/K² (choice A), thinking the coefficient 2 affects the reciprocal relationship. Remember: reversing a reaction always gives K_new = 1/K, regardless of coefficients in the balanced equation.

Question 6

For the equilibrium reaction $\text{C}(s)+\text{CO}_2(g)\rightleftharpoons 2\text{CO}(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $2\text{CO}(g)\rightleftharpoons \text{C}(s)+\text{CO}_2(g)$ ?

$K_{\text{new}}=K$
$K_{\text{new}}=K^2$
$K_{\text{new}}=\dfrac{1}{K}$ (correct answer)
$K_{\text{new}}=\dfrac{1}{K^2}$
$K_{\text{new}}=\sqrt{K}$

Explanation: This question tests reversal in a heterogeneous system. The original is C(s) + CO₂(g) ⇌ 2CO(g) with K = [CO]² / [CO₂], and the new is the reverse, 2CO(g) ⇌ C(s) + CO₂(g). K_new = 1/K, inverting the expression. Solids are omitted, preserving the relationship. A tempting distractor is D, 1/K², wrong for extra powering, due to misapplying the coefficient of CO. Confirm exact reversal and use K_new = 1/K, checking coefficients match.

Question 7

At a given temperature, the equilibrium constant for the reaction

$\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$

is $K$ . What is the relationship between $K_{\text{new}}$ and $K$ for the reaction

$\mathrm{2PCl_5(g) \rightleftharpoons 2PCl_3(g) + 2Cl_2(g)}$ ?

$K_{\text{new}}=\dfrac{1}{K^2}$
$K_{\text{new}}=K$
$K_{\text{new}}=K^2$ (correct answer)
$K_{\text{new}}=\dfrac{1}{K}$
$K_{\text{new}}=\sqrt{K}$

Explanation: This question tests understanding of how equilibrium constants change when all coefficients are multiplied by the same factor. The original reaction PCl₅ ⇌ PCl₃ + Cl₂ has K = [PCl₃][Cl₂]/[PCl₅]. When all coefficients are doubled, K_new = [PCl₃]²[Cl₂]²/[PCl₅]² = ([PCl₃][Cl₂]/[PCl₅])² = K². Therefore, K_new = K², making choice C correct. A common misconception is thinking that doubling coefficients has no effect on K (choosing K_new = K), but each concentration term is raised to the power of its coefficient. When coefficients are multiplied by n, raise the original K to the nth power.

Question 8

At a given temperature, the equilibrium constant for the reaction

$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$

is $K$ . A student doubles all coefficients to write

$\mathrm{2H_2(g) + 2I_2(g) \rightleftharpoons 4HI(g)}$

How is $K_{\text{new}}$ related to $K$ ?

$K_{\text{new}}=\dfrac{1}{K^2}$
$K_{\text{new}}=K$
$K_{\text{new}}=\sqrt{K}$
$K_{\text{new}}=K^2$ (correct answer)
$K_{\text{new}}=\dfrac{1}{K}$

Explanation: This question tests understanding of how equilibrium constants change when all coefficients in a reaction are multiplied by the same factor. The original reaction H₂ + I₂ ⇌ 2HI has K = [HI]²/([H₂][I₂]). When all coefficients are doubled, the new equilibrium expression becomes K_new = [HI]⁴/([H₂]²[I₂]²) = ([HI]²/([H₂][I₂]))² = K². Therefore, K_new = K², making choice D correct. Students might incorrectly think doubling coefficients doubles K (which would give K_new = 2K, not an option), but the relationship is exponential, not linear. When all coefficients are multiplied by n, the new equilibrium constant equals K^n.

Question 9

The equilibrium reaction $\mathrm{A(g) + B(g) \rightleftharpoons C(g)}$ has equilibrium constant $K$ . A student adds this reaction to its reverse, $\mathrm{C(g) \rightleftharpoons A(g) + B(g)}$ , to obtain the net equation $\mathrm{0 \rightleftharpoons 0}$ . What is the equilibrium constant $K_{\text{new}}$ for the net equation in terms of $K$ ?

$K_{\text{new}}=K^2$
$K_{\text{new}}=\dfrac{1}{K}$
$K_{\text{new}}=\dfrac{1}{K^2}$
$K_{\text{new}}=1$ (correct answer)
$K_{\text{new}}=K$

Explanation: This question tests understanding of equilibrium constants for combined reactions. When adding A(g) + B(g) ⇌ C(g) with K₁ = K to its reverse C(g) ⇌ A(g) + B(g) with K₂ = 1/K, the net equation is 0 ⇌ 0. For combined reactions, equilibrium constants multiply: K_new = K₁ × K₂ = K × (1/K) = 1. This makes physical sense because the net equation represents no net change, and at equilibrium, there's no driving force in either direction. Students who choose option B (1/K) might think only one reaction's K matters. The key principle is that when reactions are added, their equilibrium constants multiply, and K × (1/K) always equals 1.

Question 10

At a given temperature, the equilibrium constant for $\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)$ is $K$ . What is $K_\text{new}$ for the reaction $2\text{PCl}_5(g)\rightleftharpoons 2\text{PCl}_3(g)+2\text{Cl}_2(g)$ ?

$K_\text{new}=\dfrac{1}{K}$
$K_\text{new}=K^2$ (correct answer)
$K_\text{new}=\sqrt{K}$
$K_\text{new}=\dfrac{1}{K^2}$
$K_\text{new}=K$

Explanation: This question tests understanding of how equilibrium constants change when reaction coefficients are multiplied. The original reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has equilibrium constant K = [PCl₃][Cl₂]/[PCl₅]. When all coefficients are doubled to get 2PCl₅(g) ⇌ 2PCl₃(g) + 2Cl₂(g), the new equilibrium expression becomes K_new = [PCl₃]²[Cl₂]²/[PCl₅]² = ([PCl₃][Cl₂]/[PCl₅])² = K². This follows the rule that when coefficients are multiplied by n, the equilibrium constant is raised to the nth power. A common error is thinking the equilibrium constant remains unchanged (choice E), failing to recognize that the exponents in the equilibrium expression change with the coefficients. To determine the new K value, identify the coefficient multiplier and raise the original K to that power.