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AP Chemistry

AP Chemistry Help: Properties Of The Equilibrium Constant

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

Question 1

For the equilibrium reaction H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}H2​(g)+I2​(g)⇌2HI(g), the equilibrium constant is KKK. What is KnewK_\text{new}Knew​ for the reaction 2HI(g)⇌H2(g)+I2(g)\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2(g)}2HI(g)⇌H2​(g)+I2​(g)?

  1. Knew=K2K_\text{new}=K^2Knew​=K2
  2. Knew=1K2K_\text{new}=\dfrac{1}{K^2}Knew​=K21​
  3. Knew=KK_\text{new}=\sqrt{K}Knew​=K​
  4. Knew=KK_\text{new}=KKnew​=K
  5. Knew=1KK_\text{new}=\dfrac{1}{K}Knew​=K1​
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 2NO(g)+Cl2(g)⇌2NOCl(g)2\text{NO}(g)+\text{Cl}_2(g)\rightleftharpoons 2\text{NOCl}(g)2NO(g)+Cl2​(g)⇌2NOCl(g), the equilibrium constant is KKK. What is KnewK_{\text{new}}Knew​ for the reaction NO(g)+12Cl2(g)⇌NOCl(g)\text{NO}(g)+\tfrac{1}{2}\text{Cl}_2(g)\rightleftharpoons \text{NOCl}(g)NO(g)+21​Cl2​(g)⇌NOCl(g)?

  1. Knew=KK_{\text{new}}=\sqrt{K}Knew​=K​
  2. Knew=1KK_{\text{new}}=\dfrac{1}{\sqrt{K}}Knew​=K​1​
  3. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
  4. Knew=K2K_{\text{new}}=K^2Knew​=K2
  5. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
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 CO2(g)+H2(g)⇌CO(g)+H2O(g)\text{CO}_2(g)+\text{H}_2(g)\rightleftharpoons \text{CO}(g)+\text{H}_2\text{O}(g)CO2​(g)+H2​(g)⇌CO(g)+H2​O(g), the equilibrium constant is KKK. What is KnewK_{\text{new}}Knew​ for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g)\text{CO}(g)+\text{H}_2\text{O}(g)\rightleftharpoons \text{CO}_2(g)+\text{H}_2(g)CO(g)+H2​O(g)⇌CO2​(g)+H2​(g)?

  1. Knew=K2K_{\text{new}}=K^2Knew​=K2
  2. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
  3. Knew=KK_{\text{new}}=KKnew​=K
  4. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
  5. Knew=KK_{\text{new}}=\sqrt{K}Knew​=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 CO(g)+H2O(g)⇌CO2(g)+H2(g)\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}CO(g)+H2​O(g)⇌CO2​(g)+H2​(g), the equilibrium constant is KKK. What is KnewK_\text{new}Knew​ for the reaction 2CO2(g)+2H2(g)⇌2CO(g)+2H2O(g)\mathrm{2CO_2(g) + 2H_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g)}2CO2​(g)+2H2​(g)⇌2CO(g)+2H2​O(g)?

  1. Knew=1KK_\text{new}=\dfrac{1}{K}Knew​=K1​
  2. Knew=K2K_\text{new}=K^2Knew​=K2
  3. Knew=1K2K_\text{new}=\dfrac{1}{K^2}Knew​=K21​
  4. Knew=KK_\text{new}=\sqrt{K}Knew​=K​
  5. Knew=KK_\text{new}=KKnew​=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 2NO2(g)⇌N2O4(g)\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}2NO2​(g)⇌N2​O4​(g), the equilibrium constant is KKK. What is KnewK_\text{new}Knew​ for the reaction N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}N2​O4​(g)⇌2NO2​(g)?

  1. Knew=1K2K_\text{new}=\dfrac{1}{K^2}Knew​=K21​
  2. Knew=K2K_\text{new}=K^2Knew​=K2
  3. Knew=1KK_\text{new}=\dfrac{1}{K}Knew​=K1​
  4. Knew=KK_\text{new}=\sqrt{K}Knew​=K​
  5. Knew=KK_\text{new}=KKnew​=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 C(s)+CO2(g)⇌2CO(g)\text{C}(s)+\text{CO}_2(g)\rightleftharpoons 2\text{CO}(g)C(s)+CO2​(g)⇌2CO(g), the equilibrium constant is KKK. What is KnewK_{\text{new}}Knew​ for the reaction 2CO(g)⇌C(s)+CO2(g)2\text{CO}(g)\rightleftharpoons \text{C}(s)+\text{CO}_2(g)2CO(g)⇌C(s)+CO2​(g)?

  1. Knew=KK_{\text{new}}=KKnew​=K
  2. Knew=K2K_{\text{new}}=K^2Knew​=K2
  3. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
  4. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
  5. Knew=KK_{\text{new}}=\sqrt{K}Knew​=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

PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}PCl5​(g)⇌PCl3​(g)+Cl2​(g)

is KKK. What is the relationship between KnewK_{\text{new}}Knew​ and KKK for the reaction

2PCl5(g)⇌2PCl3(g)+2Cl2(g)\mathrm{2PCl_5(g) \rightleftharpoons 2PCl_3(g) + 2Cl_2(g)}2PCl5​(g)⇌2PCl3​(g)+2Cl2​(g)?

  1. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
  2. Knew=KK_{\text{new}}=KKnew​=K
  3. Knew=K2K_{\text{new}}=K^2Knew​=K2
  4. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
  5. Knew=KK_{\text{new}}=\sqrt{K}Knew​=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

H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}H2​(g)+I2​(g)⇌2HI(g)

is KKK. A student doubles all coefficients to write

2H2(g)+2I2(g)⇌4HI(g)\mathrm{2H_2(g) + 2I_2(g) \rightleftharpoons 4HI(g)}2H2​(g)+2I2​(g)⇌4HI(g)

How is KnewK_{\text{new}}Knew​ related to KKK?

  1. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
  2. Knew=KK_{\text{new}}=KKnew​=K
  3. Knew=KK_{\text{new}}=\sqrt{K}Knew​=K​
  4. Knew=K2K_{\text{new}}=K^2Knew​=K2
  5. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
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 A(g)+B(g)⇌C(g)\mathrm{A(g) + B(g) \rightleftharpoons C(g)}A(g)+B(g)⇌C(g) has equilibrium constant KKK. A student adds this reaction to its reverse, C(g)⇌A(g)+B(g)\mathrm{C(g) \rightleftharpoons A(g) + B(g)}C(g)⇌A(g)+B(g), to obtain the net equation 0⇌0\mathrm{0 \rightleftharpoons 0}0⇌0. What is the equilibrium constant KnewK_{\text{new}}Knew​ for the net equation in terms of KKK?

  1. Knew=K2K_{\text{new}}=K^2Knew​=K2
  2. Knew=1KK_{\text{new}}=\dfrac{1}{K}Knew​=K1​
  3. Knew=1K2K_{\text{new}}=\dfrac{1}{K^2}Knew​=K21​
  4. Knew=1K_{\text{new}}=1Knew​=1
  5. Knew=KK_{\text{new}}=KKnew​=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 PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)PCl5​(g)⇌PCl3​(g)+Cl2​(g) is KKK. What is KnewK_\text{new}Knew​ for the reaction 2PCl5(g)⇌2PCl3(g)+2Cl2(g)2\text{PCl}_5(g)\rightleftharpoons 2\text{PCl}_3(g)+2\text{Cl}_2(g)2PCl5​(g)⇌2PCl3​(g)+2Cl2​(g)?

  1. Knew=1KK_\text{new}=\dfrac{1}{K}Knew​=K1​
  2. Knew=K2K_\text{new}=K^2Knew​=K2
  3. Knew=KK_\text{new}=\sqrt{K}Knew​=K​
  4. Knew=1K2K_\text{new}=\dfrac{1}{K^2}Knew​=K21​
  5. Knew=KK_\text{new}=KKnew​=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.