Chemistry: Equilibrium Constants from Concentration Tables Worksheet

Question 1

For the reaction A(g) ⇌ B(g), the equilibrium concentrations are [A] = 0.25 M and [B] = 0.75 M. Write the Kc expression and calculate Kc.

Accepted Answer

The expression is Kc = [B]/[A]. Substituting the values gives Kc = 0.75/0.25 = 3.0.

Question 2

For the reaction H2(g) + I2(g) ⇌ 2HI(g), the equilibrium concentration table is: [H2] = 0.020 M, [I2] = 0.020 M, and [HI] = 0.160 M. Calculate Kc.

Accepted Answer

The expression is Kc = [HI]^2/([H2][I2]). Substituting the values gives Kc = (0.160)^2/(0.020 × 0.020) = 0.0256/0.000400 = 64.

Question 3

For the reaction N2O4(g) ⇌ 2NO2(g), the equilibrium concentrations are [N2O4] = 0.12 M and [NO2] = 0.30 M. Calculate Kc.

Accepted Answer

The expression is Kc = [NO2]^2/[N2O4]. Substituting the values gives Kc = (0.30)^2/0.12 = 0.090/0.12 = 0.75.

Question 4

For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), the equilibrium concentrations are [SO2] = 0.10 M, [O2] = 0.20 M, and [SO3] = 0.40 M. Calculate Kc.

Accepted Answer

The expression is Kc = [SO3]^2/([SO2]^2[O2]). Substituting the values gives Kc = (0.40)^2/((0.10)^2 × 0.20) = 0.16/0.0020 = 80.

Question 5

For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), the initial concentrations are [PCl5] = 0.500 M, [PCl3] = 0 M, and [Cl2] = 0 M. At equilibrium, [PCl5] = 0.200 M. Use the concentration changes to find the equilibrium concentrations of PCl3 and Cl2, then calculate Kc.

Accepted Answer

PCl5 decreases by 0.300 M, so PCl3 and Cl2 each increase by 0.300 M. At equilibrium, [PCl3] = 0.300 M and [Cl2] = 0.300 M. The expression is Kc = ([PCl3][Cl2])/[PCl5], so Kc = (0.300 × 0.300)/0.200 = 0.45.

Question 6

For the reaction A(g) + 2B(g) ⇌ C(g), the equilibrium concentrations are [A] = 0.10 M, [B] = 0.20 M, and [C] = 0.80 M. Calculate Kc.

Accepted Answer

The expression is Kc = [C]/([A][B]^2). Substituting the values gives Kc = 0.80/(0.10 × (0.20)^2) = 0.80/0.0040 = 200.

Question 7

For the reaction CO(g) + Cl2(g) ⇌ COCl2(g), the equilibrium concentrations are [CO] = 0.050 M, [Cl2] = 0.100 M, and [COCl2] = 0.350 M. Calculate Kc.

Accepted Answer

The expression is Kc = [COCl2]/([CO][Cl2]). Substituting the values gives Kc = 0.350/(0.050 × 0.100) = 0.350/0.00500 = 70.

Question 8

For the reaction 2NOBr(g) ⇌ 2NO(g) + Br2(g), the initial concentrations are [NOBr] = 0.60 M, [NO] = 0 M, and [Br2] = 0 M. At equilibrium, [NOBr] = 0.20 M. Use an ICE table to calculate Kc.

Accepted Answer

NOBr decreases by 0.40 M. Since 2NOBr produces 2NO and 1Br2, [NO] increases by 0.40 M and [Br2] increases by 0.20 M. The equilibrium concentrations are [NOBr] = 0.20 M, [NO] = 0.40 M, and [Br2] = 0.20 M. The expression is Kc = ([NO]^2[Br2])/[NOBr]^2, so Kc = ((0.40)^2 × 0.20)/(0.20)^2 = 0.032/0.040 = 0.80.

Question 9

For the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g), the equilibrium concentrations are [H2] = 0.15 M, [CO2] = 0.25 M, [H2O] = 0.35 M, and [CO] = 0.45 M. Calculate Kc.

Accepted Answer

The expression is Kc = ([H2O][CO])/([H2][CO2]). Substituting the values gives Kc = (0.35 × 0.45)/(0.15 × 0.25) = 0.1575/0.0375 = 4.2.

Question 10

For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium concentration of CO2 is 0.020 M. Write the Kc expression and calculate Kc.

Accepted Answer

Pure solids are not included in the Kc expression. The expression is Kc = [CO2], so Kc = 0.020.

Question 11

For the reaction A(g) ⇌ 2B(g), the equilibrium concentrations are [A] = 0.40 M and [B] = 0.20 M. Calculate Kc for the forward reaction. Then calculate Kc for the reverse reaction, 2B(g) ⇌ A(g).

Accepted Answer

For the forward reaction, Kc = [B]^2/[A] = (0.20)^2/0.40 = 0.040/0.40 = 0.10. For the reverse reaction, Kc is the reciprocal, so Kc = 1/0.10 = 10.

Question 12

For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), the equilibrium concentrations are [N2] = 0.50 M, [H2] = 0.20 M, and [NH3] = 0.30 M. Calculate Kc.

Accepted Answer

The expression is Kc = [NH3]^2/([N2][H2]^3). Substituting the values gives Kc = (0.30)^2/(0.50 × (0.20)^3) = 0.090/(0.50 × 0.0080) = 0.090/0.0040 = 22.5.

Question 13

For the reaction A(g) + 2B(g) ⇌ 3C(g), the equilibrium concentrations are [A] = 0.250 M, [B] = 0.400 M, and [C] = 0.500 M. Calculate Kc.

Accepted Answer

The expression is Kc = [C]^3/([A][B]^2). Substituting the values gives Kc = (0.500)^3/(0.250 × (0.400)^2) = 0.125/(0.250 × 0.160) = 0.125/0.0400 = 3.125.

Question 14

For the reaction N2O4(g) ⇌ 2NO2(g), the initial concentration of N2O4 is 0.100 M and the initial concentration of NO2 is 0 M. At equilibrium, [NO2] = 0.080 M. Find [N2O4] at equilibrium, then calculate Kc.

Accepted Answer

Because 2NO2 forms for every 1N2O4 used, forming 0.080 M NO2 means 0.040 M N2O4 was consumed. The equilibrium concentration of N2O4 is 0.100 M - 0.040 M = 0.060 M. The expression is Kc = [NO2]^2/[N2O4], so Kc = (0.080)^2/0.060 = 0.0064/0.060 = 0.107.

Question 15

For the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g), the initial concentrations are [PCl3] = 0.40 M, [Cl2] = 0.40 M, and [PCl5] = 0 M. At equilibrium, [PCl5] = 0.25 M. Use an ICE table to calculate Kc.

Accepted Answer

If 0.25 M PCl5 forms, then 0.25 M PCl3 and 0.25 M Cl2 are consumed. The equilibrium concentrations are [PCl3] = 0.15 M, [Cl2] = 0.15 M, and [PCl5] = 0.25 M. The expression is Kc = [PCl5]/([PCl3][Cl2]), so Kc = 0.25/(0.15 × 0.15) = 0.25/0.0225 = 11.1.

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