AP Chemistry Quiz: Properties Of The Equilibrium Constant

What this quiz covers

This quiz focuses on Properties Of The Equilibrium Constant, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $K_\text{new}=K^2$
2. $K_\text{new}=\dfrac{1}{K^2}$
3. $K_\text{new}=\sqrt{K}$
4. $K_\text{new}=K$
5. $K_\text{new}=\dfrac{1}{K}$ (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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Question 11

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

Explanation: This question tests understanding of how equilibrium constants change when a reaction is reversed. For the original reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant is K = [NH₃]²/([N₂][H₂]³). When the reaction is reversed to 2NH₃(g) ⇌ N₂(g) + 3H₂(g), the products and reactants switch places, so K_new = [N₂][H₂]³/[NH₃]² = 1/K. This follows the general rule that reversing a reaction inverts the equilibrium constant. A common misconception is thinking that reversing doubles the K value (choice A), but this confuses reversal with doubling coefficients. When manipulating equilibrium expressions, remember: reversing inverts K, halving coefficients takes the square root, and doubling coefficients squares K.

Question 12

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

Explanation: This question tests understanding of how equilibrium constants change when a reaction is reversed. For the original reaction H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium constant is K = [HI]²/([H₂][I₂]). When reversed to 2HI(g) ⇌ H₂(g) + I₂(g), the products and reactants exchange positions, giving K_new = [H₂][I₂]/[HI]² = 1/K. This demonstrates the fundamental principle that reversing any chemical equation inverts its equilibrium constant. A common misconception is thinking that the coefficient 2 on HI affects the transformation (leading to choice E, 1/K²), but the reversal operation always gives 1/K regardless of the coefficients. To solve equilibrium constant transformations, first identify whether the reaction is reversed, then check if coefficients are multiplied or divided.

Question 13

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

Explanation: This question tests understanding of how equilibrium constants change when all coefficients are divided by the same factor. The original reaction 2NO + O₂ ⇌ 2NO₂ has K = [NO₂]²/([NO]²[O₂]). When all coefficients are divided by 2, K_new = [NO₂]/([NO][O₂]^(1/2)) = ([NO₂]²/([NO]²[O₂]))^(1/2) = K^(1/2) = √K. Therefore, K_new = √K, making choice A correct. Students might incorrectly think halving coefficients means K_new = K/2, but the relationship involves taking the square root, not dividing by 2. When all coefficients are divided by n, take the nth root of K.

Question 14

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

Explanation: This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with equilibrium constant K = [NH₃]²/([N₂][H₂]³). When a reaction is reversed, the new equilibrium constant is the reciprocal of the original: K_new = 1/K. For the reversed reaction 2NH₃(g) ⇌ N₂(g) + 3H₂(g), K_new = [N₂][H₂]³/[NH₃]² = 1/K. Students often confuse this with squaring K (choice C), thinking that the coefficient 2 in front of NH₃ means squaring the equilibrium constant. Remember: reversing a reaction always gives K_new = 1/K, regardless of coefficients.

Question 15

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

Explanation: This question tests understanding of how equilibrium constants change when a reaction is both reversed and coefficients are divided. The original reaction 2H₂S(g) + 3O₂(g) ⇌ 2SO₂(g) + 2H₂O(g) has equilibrium constant K. The new reaction SO₂(g) + H₂O(g) ⇌ H₂S(g) + (3/2)O₂(g) involves reversing (giving 1/K) and dividing all coefficients by 2 (giving the square root). Therefore, K_new = (1/K)^(1/2) = 1/√K. Students often incorrectly choose 1/K (choice E), forgetting to account for the coefficient change. Remember: when both reversing and changing coefficients, apply both transformations: reverse first, then adjust for the coefficient change.

Question 16

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

Explanation: This question tests the understanding of scaling the coefficients of a dissociation reaction and its impact on the equilibrium constant. The original reaction is PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with K = [PCl₃][Cl₂] / [PCl₅]. The new reaction is 2PCl₅(g) ⇌ 2PCl₃(g) + 2Cl₂(g), which is twice the original, so K_new = K², following the rule for multiplying coefficients by 2. The equilibrium expression for the doubled reaction is ([PCl₃]²[Cl₂]² / [PCl₅]²) = (K)². A tempting distractor is D, K, which is incorrect because it ignores the scaling effect, mistakenly treating the constant as invariant to coefficient changes. When scaling reactions, identify the multiplication factor and raise the original K to that power for the new constant.

Question 17

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

Explanation: This question tests understanding of how equilibrium constants change when all reaction coefficients are multiplied by a factor. The original reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has equilibrium constant K = [PCl₃][Cl₂]/[PCl₅]. When all coefficients are doubled to give 2PCl₅(g) ⇌ 2PCl₃(g) + 2Cl₂(g), the new equilibrium constant 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, K_new = K^n. Students often incorrectly choose √K (choice C), confusing multiplication with division of coefficients. Remember: when multiplying all coefficients by n, raise K to the nth power.

Question 18

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

Explanation: This question tests understanding of how equilibrium constants change when reactions are reversed and coefficients are divided. The original reaction 2SO₂ + O₂ ⇌ 2SO₃ has K = [SO₃]²/([SO₂]²[O₂]). The new reaction reverses this AND divides all coefficients by 2, so K_new = ([SO₂][O₂]^(1/2))/[SO₃]. For reversal alone, we'd get 1/K, but dividing coefficients by 2 means taking the square root, so K_new = (1/K)^(1/2) = 1/√K, making choice C correct. A common error is forgetting to account for both operations and choosing 1/K (choice A). When a reaction is both reversed and scaled, apply both transformations: reciprocal for reversal, then the appropriate power for scaling.

Question 19

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

Explanation: This question tests the understanding of how the equilibrium constant changes when a reaction is reversed. The original reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with K = [NH₃]² / ([N₂][H₂]³), and the new reaction is the reverse, 2NH₃(g) ⇌ N₂(g) + 3H₂(g). Reversing the reaction means the new equilibrium constant is the reciprocal of the original, so K_new = 1/K. This follows the principle that for a reverse reaction, products become reactants and vice versa, inverting the expression. A tempting distractor is C, 1/K², which is incorrect because it assumes the reaction is also scaled by a factor, reflecting the misconception of confusing reversal with coefficient changes. To solve similar problems, always identify if the reaction is reversed (invert K) and if scaled (raise to the power of the factor), applying these steps sequentially.

Question 20

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

Explanation: This question examines how the equilibrium constant changes upon reaction reversal. The original is PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with K = [PCl₃][Cl₂] / [PCl₅], and the new is the exact reverse, PCl₃(g) + Cl₂(g) ⇌ PCl₅(g). Reversal inverts the constant, so K_new = 1/K. The expression becomes [PCl₅] / ([PCl₃][Cl₂]), confirming the reciprocal. A tempting distractor is D, 1/K², incorrect as it assumes additional scaling, reflecting the misconception of altering coefficients unnecessarily. Identify if the reaction is reversed and simply take 1/K, verifying with the equilibrium expression.