Properties of the Equilibrium Constant - AP Chemistry

For the equilibrium reaction $\text{CO}(g)+\text{Cl}_2(g)\rightleftharpoons \text{COCl}_2(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $2\text{CO}(g)+2\text{Cl}_2(g)\rightleftharpoons 2\text{COCl}_2(g)$ ?
Answer: $K_{\text{new}}=K^2$ This question tests properties of the equilibrium constant for scaled reactions. The original is CO(g) + Cl₂(g) ⇌ COCl₂(g) with K = [COCl₂] / ([CO][Cl₂]), and the new has coefficients doubled, equivalent to multiplying by 2. Thus, K_new = K² as the expression becomes ([COCl₂]² / ([CO]²[Cl₂]²)) = K². This principle holds because scaling multiplies exponents in the equilibrium expression. A tempting distractor is D, 1/K², wrong if one mistakenly reverses the reaction, arising from the misconception of ignoring direction. Compare the new reaction to the original to find the multiplier n and set K_new = $K^n$ for consistent direction.
At a certain temperature, the equilibrium constant for $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ is $K$ . What is $K_{\text{new}}$ for the reaction $\mathrm{CaO(s) + CO_2(g) \rightleftharpoons CaCO_3(s)}$ ?
Answer: $K_{\text{new}} = \dfrac{1}{K}$ This question tests the understanding of the equilibrium constant for heterogeneous equilibria and its behavior upon reversal. The original reaction is CaCO₃(s) ⇌ CaO(s) + CO₂(g) with K = [CO₂], since pure solids are omitted from the expression. The new reaction is the reverse, CaO(s) + CO₂(g) ⇌ CaCO₃(s), so K_new = 1/K = 1/[CO₂]. The presence of solids does not change the reversal rule, as K still inverts for the opposite direction. A tempting distractor is C, K, which is incorrect because it assumes reversal has no effect, misunderstanding that directionality always requires taking the reciprocal. For heterogeneous equilibria, remember to exclude solids and liquids from K, but apply standard manipulation rules like reversal yielding 1/K regardless.
For the equilibrium reaction $2\text{H}_2\text{O}(g)\rightleftharpoons 2\text{H}_2(g)+\text{O}_2(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{H}_2(g)+\tfrac{1}{2}\text{O}_2(g)\rightleftharpoons \text{H}_2\text{O}(g)$ ?
Answer: $K_{\text{new}}=\dfrac{1}{\sqrt{K}}$ This question tests combined reversal and halving of a reaction. The original is 2H₂O(g) ⇌ 2H₂(g) + O₂(g) with K = ([H₂]²[O₂]) / [H₂O]², and the new is half the reverse, H₂(g) + ½O₂(g) ⇌ H₂O(g). Reversal gives 1/K, halving yields $(1/K)^{1/2}$ = 1/√K. The expression becomes [H₂O] / ([H₂][O₂]^{1/2}) = 1/√K. A tempting distractor is E, 1/K², wrong for using squared instead of square root, due to inverting the exponent. Apply reversal first (1/K) then the scaling power based on coefficient ratios.
For $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ with equilibrium constant $K$ , the reaction is reversed and multiplied by $\tfrac{1}{2}$ to form $\mathrm{\tfrac{1}{2}PCl_3(g) + \tfrac{1}{2}Cl_2(g) \rightleftharpoons \tfrac{1}{2}PCl_5(g)}$ . Which expression gives $K_{\text{new}}$ in terms of $K$ ?
Answer: $K_{\text{new}}=\dfrac{1}{\sqrt{K}}$ This question tests understanding of combined operations involving reversal and fractional multiplication. Starting with PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with equilibrium constant K, reversing gives K' = 1/K for PCl₃(g) + Cl₂(g) ⇌ PCl₅(g). Then multiplying all coefficients by ½ gives K_new = $(1/K)^{1/2}$ = 1/√K. This combines two rules: reversing inverts K, and multiplying coefficients by n raises K to the power n. Students who choose option B (√K) forget to account for the reversal and only consider the multiplication by ½. The key is to apply operations in order: reverse first (1/K), then raise to the power of the coefficient multiplier.
For the equilibrium reaction $\text{CH}_4(g)+\text{H}_2\text{O}(g)\rightleftharpoons \text{CO}(g)+3\text{H}_2(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $2\text{CH}_4(g)+2\text{H}_2\text{O}(g)\rightleftharpoons 2\text{CO}(g)+6\text{H}_2(g)$ ?
Answer: $K_{\text{new}}=K^2$ This question evaluates scaling effects on the equilibrium constant. The original is CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) with K = [CO][H₂]³ / ([CH₄][H₂O]), and the new is doubled, so multiplied by 2. Thus, K_new = K², as the expression raises to the second power. This principle applies regardless of coefficient complexity. A tempting distractor is C, 1/K², wrong if reversal is assumed, due to misreading direction. Determine the scaling factor by comparing coefficients and compute K_new = $K^n$ for the multiplier n.
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 $\text{NO}(g)+\tfrac{1}{2}\text{Cl}_2(g)\rightleftharpoons \text{NOCl}(g)$ ?
Answer: $K_{\text{new}}=\sqrt{K}$ 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.
At a given temperature, the equilibrium constant for the reaction $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ is $K$ . What is the relationship between $K_{\text{new}}$ and $K$ for the reaction $\mathrm{H^+(aq) + CH_3COO^-(aq) \rightleftharpoons CH_3COOH(aq)}$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K}$ This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction CH₃COOH ⇌ H⁺ + CH₃COO⁻ has K = [H⁺][CH₃COO⁻]/[CH₃COOH]. The new reaction H⁺ + CH₃COO⁻ ⇌ CH₃COOH reverses this, so K_new = [CH₃COOH]/([H⁺][CH₃COO⁻]) = 1/K. Therefore, K_new = 1/K, making choice C correct. A common error is confusing this with the relationship for conjugate acid-base pairs (Ka × Kb = Kw), but this question simply asks about reversing a reaction. For any reversed reaction, K_new = 1/K regardless of whether it involves acids and bases.
At a certain temperature, the equilibrium constant for $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ is $K$ . What is $K_{\text{new}}$ for the reaction $\mathrm{2CO(g) + 2H_2O(g) \rightleftharpoons 2CO_2(g) + 2H_2(g)}$ ?
Answer: $K_{\text{new}} = K^2$ This question tests the understanding of how the equilibrium constant changes when all coefficients in a reaction are multiplied by a factor. The original reaction is $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ with $K = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}$ . The new reaction is $\mathrm{2CO(g) + 2H_2O(g) \rightleftharpoons 2CO_2(g) + 2H_2(g)}$ , which is twice the original, so $K_{\text{new}} = K^2$ , as the exponent reflects the multiplication factor. This occurs because the equilibrium expression for the doubled reaction is the square of the original expression. A tempting distractor is B, $K$ , which is wrong due to the misconception that scaling coefficients does not affect K, ignoring the exponential relationship. Always calculate the scaling factor n by comparing coefficients and use $K_{\text{new}} = K^n$ for multiplied reactions.
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 $\mathrm{2PCl_5(g) \rightleftharpoons 2PCl_3(g) + 2Cl_2(g)}$ ?
Answer: $K_{\text{new}}=K^2$ 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.
For the equilibrium reaction $\text{CaCO}_3(s)\rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{CaO}(s)+\text{CO}_2(g)\rightleftharpoons \text{CaCO}_3(s)$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K}$ This question tests the effect of reversing a heterogeneous equilibrium on the constant. The original is CaCO₃(s) ⇌ CaO(s) + CO₂(g) with K = [CO₂], and the new is the reverse, CaO(s) + CO₂(g) ⇌ CaCO₃(s). Reversal gives K_new = 1/K, as solids do not affect the relationship. The new expression is 1/[CO₂] = 1/K. A tempting distractor is A, 1/K², incorrect if one wrongly scales by the coefficient of CO₂, stemming from misunderstanding heterogeneous equilibria. Check for reversal and apply 1/K, ignoring pure solids and liquids in the expression.
Given $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ with equilibrium constant $K$ . The reaction is reversed and then multiplied by 2 to obtain $\mathrm{2CO_2(g) + 2H_2(g) \rightleftharpoons 2CO(g) + 2H_2O(g)}$ . What is $K_{\text{new}}$ in terms of $K$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K^2}$ This question tests understanding of combined operations on equilibrium constants. Starting with CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) with equilibrium constant K, we first reverse it (giving K' = 1/K) and then multiply by 2 (giving K_new = (1/K)² = 1/K²). The reversal changes K to 1/K, and multiplying coefficients by 2 squares the equilibrium constant. Students who choose option A (1/K) forget to account for the multiplication by 2 after reversing. The strategy is to apply operations sequentially: reversing gives 1/K, then multiplying coefficients by n raises the result to the power n.
For the equilibrium reaction $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ , the equilibrium constant is $K$ . What is $K_\text{new}$ for the reaction $\mathrm{2PCl_5(g) \rightleftharpoons 2PCl_3(g) + 2Cl_2(g)}$ ?
Answer: $K_\text{new}=K^2$ 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.
At a certain temperature, the equilibrium constant for $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ is $K$ . What is $K_{\text{new}}$ for the reaction $\mathrm{2PCl_5(g) \rightleftharpoons 2PCl_3(g) + 2Cl_2(g)}$ ?
Answer: $K_{\text{new}} = K^2$ 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.
At a given temperature, the equilibrium constant for the reaction $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ is $K$ . What is the relationship between $K_{\text{new}}$ and $K$ for the reaction $\mathrm{2CaO(s) + 2CO_2(g) \rightleftharpoons 2CaCO_3(s)}$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K^2}$ This question tests understanding of how equilibrium constants change when reactions are reversed and coefficients are multiplied, considering that solids don't appear in equilibrium expressions. The original reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) has K = [CO₂] since solids are excluded. The new reaction reverses this AND doubles all coefficients, giving 2CaO(s) + 2CO₂(g) ⇌ 2CaCO₃(s), so K_new = 1/[CO₂]². Since K = [CO₂], we have K_new = 1/K², making choice A correct. A common error is forgetting that solids don't affect the equilibrium expression and thinking the answer is simply 1/K (choice C). When manipulating heterogeneous equilibria, remember that only gases and aqueous species appear in K expressions.
For the equilibrium $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ with equilibrium constant $K$ , the reaction is reversed to form $\mathrm{H^+(aq) + CH_3COO^-(aq) \rightleftharpoons CH_3COOH(aq)}$ . What is $K_{\text{new}}$ in terms of $K$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K}$ This question tests understanding of how reversing a reaction affects its equilibrium constant. For CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq) with equilibrium constant K = [H⁺][CH₃COO⁻]/[CH₃COOH], reversing to H⁺(aq) + CH₃COO⁻(aq) ⇌ CH₃COOH(aq) gives K_new = [CH₃COOH]/([H⁺][CH₃COO⁻]) = 1/K. This is the fundamental rule for reversed reactions: the equilibrium constant is inverted. Students who choose option A (K²) might confuse this with multiplying coefficients, but reversal only inverts K regardless of any coefficients. The key principle is that reversing any reaction always gives K_new = 1/K.
For the equilibrium reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ , the equilibrium constant is $K$ . What is the equilibrium constant, $K_\text{new}$ , for the reaction $\mathrm{2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)}$ ?
Answer: $K_\text{new}=\dfrac{1}{K}$ 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.
For the equilibrium reaction $2\text{SO}_2(g)+\text{O}_2(g)\rightleftharpoons 2\text{SO}_3(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{SO}_3(g)\rightleftharpoons \text{SO}_2(g)+\tfrac{1}{2}\text{O}_2(g)$ ?
Answer: $K_{\text{new}}=\dfrac{1}{\sqrt{K}}$ This question evaluates knowledge of equilibrium constant transformations for reversed and halved reactions. The original is 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) with K = [SO₃]² / ([SO₂]²[O₂]), and the new reaction is the reverse halved, SO₃(g) ⇌ SO₂(g) + ½O₂(g). Reversing gives 1/K, and halving raises to the ½ power, so K_new = $(1/K)^{1/2}$ = 1/√K. This combines the reciprocal for reversal and the root for scaling by 1/2. A tempting distractor is B, 1/K², incorrect because it over-applies the power to the reversal without considering the half factor, due to miscalculating the exponent. Always determine the reversal first (1/K) then apply the scaling exponent to find K_new.
At a given temperature, the equilibrium constant for the reaction $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ is $K$ . A student writes the reverse reaction: $\mathrm{CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g)}$ How is $K_{\text{new}}$ related to $K$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K}$ This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction CO + H₂O ⇌ CO₂ + H₂ has K = [CO₂][H₂]/([CO][H₂O]). The student's reversed reaction CO₂ + H₂ ⇌ CO + H₂O has K_new = [CO][H₂O]/([CO₂][H₂]), which is exactly 1/K. Therefore, K_new = 1/K, making choice B correct. Some students might think reversing has no effect on K (choosing K_new = K), but this is incorrect because products and reactants switch places in the equilibrium expression. The rule is simple: reversing a reaction always gives K_new = 1/K.
For the equilibrium reaction $\text{H}_2(g)+\text{Br}_2(g)\rightleftharpoons 2\text{HBr}(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{HBr}(g)\rightleftharpoons \tfrac{1}{2}\text{H}_2(g)+\tfrac{1}{2}\text{Br}_2(g)$ ?
Answer: $K_{\text{new}}=\dfrac{1}{\sqrt{K}}$ This question evaluates combined halving and reversal. The original is H₂(g) + Br₂(g) ⇌ 2HBr(g) with K = [HBr]² / ([H₂][Br₂]), and the new is half the reverse, HBr(g) ⇌ ½H₂(g) + ½Br₂(g). Reversal gives 1/K, halving yields $(1/K)^{1/2}$ = 1/√K. The expression is ([H₂]^{1/2}[Br₂]^{1/2}) / [HBr] = 1/√K. A tempting distractor is B, 1/K, incorrect for omitting the half factor, reflecting scaling oversight. Always apply reversal first, then the power for any fractional scaling.
For the equilibrium reaction $\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)$ , the equilibrium constant is $K$ . What is $K_{\text{new}}$ for the reaction $\text{PCl}_3(g)+\text{Cl}_2(g)\rightleftharpoons \text{PCl}_5(g)$ ?
Answer: $K_{\text{new}}=\dfrac{1}{K}$ 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.

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