Calculating the Equilibrium Constant
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AP Chemistry › Calculating the Equilibrium Constant
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 $K_c$?
0.040
0.10
0.25
2.5
25
Explanation
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.
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 $K_c$?
0.063
0.25
4.0
8.0
16
Explanation
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 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 $K_c$?
0.10
0.25
0.40
2.0
4.0
Explanation
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.
At equilibrium, the reaction $\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}$ has $\mathrm{NO}=0.20,\mathrm{M}$, $\mathrm{O_2}=0.10,\mathrm{M}$, and $\mathrm{NO_2}=0.20,\mathrm{M}$. What is $K_c$?
0.10
0.25
1.0
2.5
10
Explanation
This question involves calculating the equilibrium constant. For the balanced equation 2NO(g) + O₂(g) ⇌ 2NO₂(g), the equilibrium expression is written as products over reactants, with stoichiometric coefficients as exponents. Thus, Kc = [NO₂]² / ([NO]² [O₂]). Substituting the equilibrium concentrations [NO] = 0.20 M, [O₂] = 0.10 M, and [NO₂] = 0.20 M gives Kc = (0.20)² / ((0.20)² × 0.10) = 0.04 / (0.04 × 0.10) = 0.04 / 0.004 = 10. A tempting distractor is 0.10, which results from inverting the expression, but this is incorrect because Kc is products over reactants. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.
A closed container at constant temperature contains the equilibrium system
$$\mathrm{2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)}$$
At equilibrium, $\mathrm{HBr}=0.20,\text{M}$, $\mathrm{H_2}=0.10,\text{M}$, and $\mathrm{Br_2}=0.10,\text{M}$. What is the value of $K_c$?
0.040
0.25
0.50
2.0
4.0
Explanation
This question tests the skill of calculating the equilibrium constant. To calculate $K_c$, first write the equilibrium expression for the reaction $$\mathrm{2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)}$$, which is $$K_c = \frac{[\mathrm{H_2}][\mathrm{Br_2}]}{[\mathrm{HBr}]^2}$$. The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: $[\mathrm{HBr}] = 0.20,\text{M}$, $[\mathrm{H_2}] = 0.10,\text{M}$, and $[\mathrm{Br_2}] = 0.10,\text{M}$, so $$K_c = \frac{(0.10 \times 0.10)}{(0.20)^2} = \frac{0.01}{0.04} = 0.25$$. A tempting distractor is 4.0, 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 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?
$1.0 \times 10^{2}$
$1.0 \times 10^{-2}$
$1.0 \times 10^{-4}$
$1.0 \times 10^{-1}$
$1.0 \times 10^{-3}$
Explanation
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.
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?
0.10
1.0
4.0
10
100
Explanation
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.
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 $K_c$?
0.080
0.20
0.40
1.25
2.5
Explanation
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 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?
0.25
2.0
4.0
16
64
Explanation
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.
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 $K_c$?
0.10
0.25
2.0
4.0
10
Explanation
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.