Calculating the Equilibrium Constant - AP Chemistry

A sealed 1.0 L flask contains the reaction $\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g)$ . The system is at equilibrium at a certain temperature. The equilibrium concentrations are: $[\text{H}_2]=0.20\,M$ , $[\text{I}_2]=0.20\,M$ , and $[\text{HI}]=0.80\,M$ . What is the value of $K_c$ for the reaction at this temperature?
Answer: 16 This question tests the skill of calculating the equilibrium constant. For the balanced equation H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium expression is $K_c$ = [HI]^2 / ([H₂][I₂]). To calculate $K_c$ , substitute the given equilibrium concentrations: [HI] = 0.80 M, [H₂] = 0.20 M, and [I₂] = 0.20 M. This yields $K_c$ = $(0.80)^2$ / (0.20 × 0.20) = 0.64 / 0.04 = 16. A tempting distractor might be 4.0, which results from forgetting to square [HI] in the numerator. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
At a fixed temperature, the reaction $\mathrm{H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)}$ is at equilibrium. The equilibrium concentrations are $[\mathrm{H_2}]=0.30\,\mathrm{M}$ , $[\mathrm{Br_2}]=0.10\,\mathrm{M}$ , and $[\mathrm{HBr}]=0.30\,\mathrm{M}$ . What is the value of $K_c$ ?
Answer: 3.0 This problem involves calculating the equilibrium constant for H₂(g) + Br₂(g) ⇌ 2HBr(g). The equilibrium expression is $K_c$ = [HBr]²/([H₂][Br₂]), where [HBr] is squared because its stoichiometric coefficient is 2. Substituting the given concentrations: $K_c$ = (0.30)²/((0.30)(0.10)) = 0.09/0.03 = 3.0. Choice B (0.030) likely results from forgetting to square the HBr concentration, using [HBr]/([H₂][Br₂]) instead of [HBr]²/([H₂][Br₂]). Remember that stoichiometric coefficients become exponents in the equilibrium expression, so always check the balanced equation carefully.
The reaction $\mathrm{2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)}$ is at equilibrium. The equilibrium concentrations are $[\mathrm{NOCl}]=0.20\,\mathrm{M}$ , $[\mathrm{NO}]=0.20\,\mathrm{M}$ , and $[\mathrm{Cl_2}]=0.05\,\mathrm{M}$ . What is $K_c$ ?
Answer: 0.050 This question involves calculating the equilibrium constant. For the balanced equation 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), the equilibrium expression is written as products over reactants, with stoichiometric coefficients as exponents. Thus, Kc = ([NO]² [Cl₂]) / [NOCl]². Substituting the equilibrium concentrations [NOCl] = 0.20 M, [NO] = 0.20 M, and [Cl₂] = 0.05 M gives Kc = ((0.20)² × 0.05) / (0.20)² = 0.002 / 0.04 = 0.050. A tempting distractor is 0.25, which results from forgetting to square [NO], but this is incorrect because exponents are required for the coefficients. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
A reaction vessel at constant temperature contains the equilibrium system $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ The system is at equilibrium with $[\mathrm{CH_3COOH}]=0.10\,\text{M}$ , $[\mathrm{H^+}]=0.010\,\text{M}$ , and $[\mathrm{CH_3COO^-}]=0.010\,\text{M}$ . What is the value of $K_c$ for this reaction?
Answer: $1.0 \times 10^{-3}$ This question tests the skill of calculating the equilibrium constant. To calculate $K_c$ , first write the equilibrium expression for the reaction $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ , which is $K_c = \frac{[\mathrm{H^+}] [\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ . The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: $[\mathrm{CH_3COOH}] = 0.10\, \text{M}$ , $[\mathrm{H^+}] = 0.010\, \text{M}$ , and $[\mathrm{CH_3COO^-}] = 0.010\, \text{M}$ , so $K_c = (0.010 \times 0.010) / 0.10 = 0.0001 / 0.10 = 0.001 = 1.0 \times 10^{-3}$ . A tempting distractor is $1.0 \times 10^{-4}$ , which results from omitting the reactant concentration in the denominator. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.
In a sealed container at constant temperature, the reaction $\text{N}_2\text{O}_4(g)\rightleftharpoons 2\text{NO}_2(g)$ is at equilibrium. The equilibrium concentrations are $[\text{N}_2\text{O}_4]=0.50\,M$ and $[\text{NO}_2]=0.20\,M$ . What is the value of $K_c$ ?
Answer: 0.080 This question tests the skill of calculating the equilibrium constant. For the balanced equation N₂O₄(g) ⇌ 2NO₂(g), the equilibrium expression is $K_c$ = [NO₂]^2 / [N₂O₄]. To calculate $K_c$ , substitute the given equilibrium concentrations: [NO₂] = 0.20 M and [N₂O₄] = 0.50 M. This yields $K_c$ = $(0.20)^2$ / 0.50 = 0.04 / 0.50 = 0.080. A tempting distractor might be 1.25, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
A reaction mixture is at equilibrium for $\text{CO}_2(g)+\text{H}_2(g)\rightleftharpoons \text{CO}(g)+\text{H}_2\text{O}(g)$ . The equilibrium concentrations are $[\text{CO}_2]=0.25\,\text{M}$ , $[\text{H}_2]=0.50\,\text{M}$ , $[\text{CO}]=0.10\,\text{M}$ , and $[\text{H}_2\text{O}]=0.20\,\text{M}$ . What is the value of $K_c$ ?
Answer: 0.16 This problem involves calculating the equilibrium constant for CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g). The equilibrium expression is Kc = [CO][H₂O]/([CO₂][H₂]), with all species having exponents of 1 since all coefficients are 1. Substituting the given concentrations: Kc = (0.10)(0.20)/((0.25)(0.50)) = 0.02/0.125 = 0.16. A common mistake would be to invert the expression, calculating [CO₂][H₂]/([CO][H₂O]) = 0.125/0.02 = 6.25, which gives the reciprocal of the correct answer. When calculating equilibrium constants, always ensure products are in the numerator and reactants are in the denominator, then substitute values carefully.
For the equilibrium reaction $\text{COCl}_2(g)\rightleftharpoons \text{CO}(g)+\text{Cl}_2(g)$ , the system is at equilibrium with $[\text{COCl}_2]=0.50\,\text{M}$ , $[\text{CO}]=0.10\,\text{M}$ , and $[\text{Cl}_2]=0.20\,\text{M}$ . What is $K_c$ ?
Answer: 0.040 This problem involves calculating the equilibrium constant for COCl₂(g) ⇌ CO(g) + Cl₂(g). The equilibrium expression is Kc = [CO][Cl₂]/[COCl₂], with products in the numerator and reactant in the denominator. Substituting the equilibrium values: Kc = (0.10)(0.20)/(0.50) = 0.02/0.50 = 0.040. A student might mistakenly write the inverse expression Kc = [COCl₂]/([CO][Cl₂]) = 0.50/0.02 = 25, which gives the reciprocal of the correct answer. To avoid this error, always write the equilibrium expression for the forward reaction as given in the problem statement.
A system is at equilibrium for the reaction $\mathrm{A(g) + 2B(g) \rightleftharpoons C(g)}$ . The equilibrium concentrations are $[\mathrm{A}]=0.20\,\mathrm{M}$ , $[\mathrm{B}]=0.10\,\mathrm{M}$ , and $[\mathrm{C}]=0.20\,\mathrm{M}$ . What is the value of $K_c$ ?
Answer: 100 This problem requires calculating the equilibrium constant for A(g) + 2B(g) ⇌ C(g). The equilibrium expression is $K_c$ = [C]/([A][B]²), where [B] is squared because its stoichiometric coefficient is 2. Substituting the given concentrations: $K_c$ = 0.20/((0.20)(0.10)²) = 0.20/(0.20 × 0.01) = 0.20/0.002 = 100. Choice C (0.10) represents the error of forgetting to square the B concentration, using [C]/([A][B]) instead of the correct expression. When multiple molecules of a species appear in the balanced equation, always raise that species' concentration to the corresponding power in the equilibrium expression.
A sealed container is held at constant temperature. The reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is at equilibrium. The equilibrium concentrations are: $[\mathrm{H_2}]=0.20\,\mathrm{M}$ , $[\mathrm{I_2}]=0.20\,\mathrm{M}$ , and $[\mathrm{HI}]=0.80\,\mathrm{M}$ . What is the value of $K_c$ for the reaction?
Answer: 16 This problem requires calculating the equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is written as $K_c$ = [products]/[reactants], with each concentration raised to the power of its stoichiometric coefficient, giving $K_c$ = [HI]²/([H₂][I₂]). Substituting the given equilibrium concentrations: $K_c$ = (0.80)²/((0.20)(0.20)) = 0.64/0.04 = 16. Choice E (0.0625) represents the common error of inverting the equilibrium expression, calculating [H₂][I₂]/[HI]² instead of the correct products over reactants. To avoid errors, always write the equilibrium expression first with products in the numerator and reactants in the denominator, then substitute the equilibrium concentrations with proper exponents.
A container at constant temperature contains the equilibrium system: $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ At equilibrium, $[\mathrm{SO_2}]=0.20\,\text{M}$ , $[\mathrm{O_2}]=0.10\,\text{M}$ , and $[\mathrm{SO_3}]=0.20\,\text{M}$ . What is $K_c$ for the reaction?
Answer: 10 This question tests the skill of calculating the equilibrium constant. To calculate $K_c$ , first write the equilibrium expression for the reaction $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ , which is $K_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]}$ . The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: $[\mathrm{SO_2}] = 0.20 \, \mathrm{M}$ , $[\mathrm{O_2}] = 0.10 \, \mathrm{M}$ , and $[\mathrm{SO_3}] = 0.20 \, \mathrm{M}$ , so $K_c = (0.20)^2 / ((0.20)^2 \times 0.10) = 0.04 / (0.04 \times 0.10) = 0.04 / 0.004 = 10$ . A tempting distractor is 0.10, which results from inverting the expression by placing reactants over products. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.
In a rigid flask at constant temperature, the reaction $\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g)$ is at equilibrium. The equilibrium concentrations are $[\text{H}_2]=0.50\,\text{M}$ , $[\text{I}_2]=0.20\,\text{M}$ , and $[\text{HI}]=1.0\,\text{M}$ . What is $K_c$ for the reaction?
Answer: 1.0 × $10^1$ This problem involves calculating the equilibrium constant for H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is Kc = [HI]²/([H₂][I₂]), where the product HI is squared because its coefficient is 2 in the balanced equation. Substituting the equilibrium values: Kc = (1.0)²/((0.50)(0.20)) = 1.0/(0.10) = 10 = 1.0 × 10¹. A tempting error would be to forget to square the HI concentration, which would give Kc = 1.0/(0.50 × 0.20) = 10, but without properly accounting for the stoichiometry. When writing equilibrium expressions, always raise each concentration to the power of its stoichiometric coefficient from the balanced equation.
At a constant temperature, the following reaction mixture is at equilibrium: $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ Equilibrium concentrations are $[\mathrm{H_2}]=0.20\,\text{M}$ , $[\mathrm{I_2}]=0.20\,\text{M}$ , and $[\mathrm{HI}]=0.40\,\text{M}$ . What is $K_c$ for the reaction?
Answer: 4.0 This question tests the skill of calculating the equilibrium constant. To calculate $K_c$ , first write the equilibrium expression for the reaction H₂(g) + I₂(g) ⇌ 2HI(g), which is $K_c$ = [HI]^2 / ([H₂][I₂]). The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 0.40 M, so $K_c$ = $(0.40)^2$ / (0.20 × 0.20) = 0.16 / 0.04 = 4.0. A tempting distractor is 0.25, which results from inverting the expression by placing reactants over products. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.
A system is at equilibrium for the reaction $\text{SO}_2(g)+\text{NO}_2(g)\rightleftharpoons \text{SO}_3(g)+\text{NO}(g)$ . The equilibrium concentrations are $[\text{SO}_2]=0.50\,\text{M}$ , $[\text{NO}_2]=0.20\,\text{M}$ , $[\text{SO}_3]=0.10\,\text{M}$ , and $[\text{NO}]=0.40\,\text{M}$ . What is the value of $K_c$ ?
Answer: 0.40 This problem asks for calculating the equilibrium constant for SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g). The equilibrium expression is Kc = [SO₃][NO]/([SO₂][NO₂]), with products in the numerator and reactants in the denominator, all with exponents of 1. Substituting the concentrations: Kc = (0.10)(0.40)/((0.50)(0.20)) = 0.04/0.10 = 0.40. A typical error would be mixing up which species are products versus reactants, or perhaps calculating (0.50)(0.20)/(0.10)(0.40) = 2.5 by inverting the expression. When calculating equilibrium constants, carefully identify products and reactants from the reaction arrow, then construct your expression methodically before substituting values.
A reaction mixture is at equilibrium for the reaction $\text{CO}(g)+\text{Cl}_2(g)\rightleftharpoons \text{COCl}_2(g)$ . The equilibrium concentrations are $[\text{CO}]=0.40\,M$ , $[\text{Cl}_2]=0.20\,M$ , and $[\text{COCl}_2]=0.80\,M$ . What is the value of $K_c$ ?
Answer: 10 This question tests the skill of calculating the equilibrium constant. For the balanced equation CO(g) + Cl₂(g) ⇌ COCl₂(g), the equilibrium expression is $K_c$ = [COCl₂] / ([CO][Cl₂]). To calculate $K_c$ , substitute the given equilibrium concentrations: [COCl₂] = 0.80 M, [CO] = 0.40 M, and [Cl₂] = 0.20 M. This yields $K_c$ = 0.80 / (0.40 × 0.20) = 0.80 / 0.08 = 10. A tempting distractor might be 0.10, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
At equilibrium in a closed vessel at constant temperature, the reaction $\mathrm{A(g) + 2B(g) \rightleftharpoons AB_2(g)}$ has equilibrium concentrations $[\mathrm{A}]=0.20\,\text{M}$ , $[\mathrm{B}]=0.10\,\text{M}$ , and $[\mathrm{AB_2}]=0.20\,\text{M}$ . What is the value of $K_c$ for the reaction?
Answer: 100 This question tests the skill of calculating the equilibrium constant. To calculate $K_c$ , first write the equilibrium expression for the reaction $\mathrm{A(g) + 2B(g) \rightleftharpoons \mathrm{AB_2(g)}$ , which is $K_c = \frac{[\mathrm{AB_2}]}{[\mathrm{A}][\mathrm{B}]^2}$ . The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: $[\mathrm{A}] = 0.20\,\text{M}$ , $[\mathrm{B}] = 0.10\,\text{M}$ , and $[\mathrm{AB_2}] = 0.20\,\text{M}$ , so $K_c = 0.20 / (0.20 \times (0.10)^2) = 0.20 / (0.20 \times 0.01) = 0.20 / 0.002 = 100$ . A tempting distractor is 10, which results from forgetting to raise the concentration of B to the second power. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.
A reaction mixture is at equilibrium for $\text{CO}(g)+\text{H}_2\text{O}(g)\rightleftharpoons \text{CO}_2(g)+\text{H}_2(g)$ . The equilibrium concentrations are $[\text{CO}]=0.50\,\text{M}$ , $[\text{H}_2\text{O}]=0.25\,\text{M}$ , $[\text{CO}_2]=0.10\,\text{M}$ , and $[\text{H}_2]=0.20\,\text{M}$ . What is $K_c$ ?
Answer: 0.16 This problem requires calculating the equilibrium constant for CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). The equilibrium expression is Kc = [CO₂][H₂]/([CO][H₂O]), with all species having coefficients of 1, so no exponents are needed. Substituting the equilibrium concentrations: Kc = (0.10)(0.20)/((0.50)(0.25)) = 0.02/0.125 = 0.16. A student might incorrectly invert the expression to get Kc = [CO][H₂O]/([CO₂][H₂]) = 0.125/0.02 = 6.25, which is wrong. To avoid this error, consistently write products in the numerator and reactants in the denominator when setting up equilibrium expressions.
At 500 K, the reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is at equilibrium in a 1.0 L container. The equilibrium concentrations are: $[\mathrm{H_2}]=0.20\,\mathrm{M}$ , $[\mathrm{I_2}]=0.20\,\mathrm{M}$ , and $[\mathrm{HI}]=0.40\,\mathrm{M}$ . What is the value of $K_c$ for this reaction?
Answer: 4.0 This question involves calculating the equilibrium constant. For the balanced equation H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium expression is written as products over reactants, with stoichiometric coefficients as exponents. Thus, Kc = [HI]² / ([H₂][I₂]). Substituting the equilibrium concentrations [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 0.40 M gives Kc = (0.40)² / (0.20 × 0.20) = 0.16 / 0.04 = 4.0. A tempting distractor is 0.25, which results from inverting the expression to put reactants over products, but this is incorrect because the equilibrium constant is always products over reactants for Kc. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
A container holds the equilibrium system $\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)$ . At equilibrium, $[\text{PCl}_5]=0.40\,M$ , $[\text{PCl}_3]=0.20\,M$ , and $[\text{Cl}_2]=0.20\,M$ . What is the value of $K_c$ ?
Answer: 0.10 This question tests the skill of calculating the equilibrium constant. For the balanced equation PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), the equilibrium expression is $K_c$ = ([PCl₃][Cl₂]) / [PCl₅]. To calculate $K_c$ , substitute the given equilibrium concentrations: [PCl₃] = 0.20 M, [Cl₂] = 0.20 M, and [PCl₅] = 0.40 M. This yields $K_c$ = (0.20 × 0.20) / 0.40 = 0.04 / 0.40 = 0.10. A tempting distractor might be 4.0, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
In a sealed vessel, the reaction $\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g)$ is at equilibrium. The equilibrium concentrations are $[\text{H}_2]=0.20\,\text{M}$ , $[\text{I}_2]=0.20\,\text{M}$ , and $[\text{HI}]=0.80\,\text{M}$ . What is $K_c$ ?
Answer: 16 This question asks for calculating the equilibrium constant for H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is Kc = [HI]²/([H₂][I₂]), where the product HI is squared because its coefficient is 2 in the balanced equation. Substituting the given concentrations: Kc = (0.80)²/((0.20)(0.20)) = 0.64/0.04 = 16. A common mistake would be to write Kc = [HI]/([H₂][I₂]) without squaring [HI], which would give 0.80/0.04 = 20, an incorrect result. Remember to always include stoichiometric coefficients as exponents in the equilibrium expression before performing calculations.
A reaction mixture is at equilibrium for the reaction $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ . The equilibrium concentrations are $[\mathrm{SO_2}]=0.20\,\mathrm{M}$ , $[\mathrm{O_2}]=0.10\,\mathrm{M}$ , and $[\mathrm{SO_3}]=0.30\,\mathrm{M}$ . What is the value of $K_c$ ?
Answer: 22.5 This question asks for calculating the equilibrium constant for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium expression is $K_c$ = [SO₃]²/([SO₂]²[O₂]), with both SO₃ and SO₂ concentrations squared due to their coefficients of 2. Substituting the equilibrium concentrations: $K_c$ = (0.30)²/((0.20)²(0.10)) = 0.09/(0.04 × 0.10) = 0.09/0.004 = 22.5. Choice B (0.225) likely results from forgetting to square the concentrations, using [SO₃]/([SO₂][O₂]) instead of the correct expression with squared terms. Remember to always include stoichiometric coefficients as exponents in the equilibrium expression before performing calculations.

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