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Difficulty: HardEquilibrium Constant Expression and Calculations
A 4.0 mol4.0\text{ mol} sample of hydrogen iodide, HI(g)\text{HI}(g), is placed in a 2.0 dm32.0\text{ dm}^3 rigid vessel and allowed to dissociate according to the equation:
2HI(g)⇌H2(g)+I2(g)2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)
If 20%20\% of the HI(g)\text{HI}(g) decomposes at equilibrium, what is the value of the equilibrium constant, KcK_c?
  1. 0.01560.0156Answer
  2. B
    0.06250.0625
  3. C
    0.02500.0250
  4. D
    0.25000.2500

Answer

0.01560.0156
To calculate KcK_c, first find the equilibrium amounts of all species. Starting with 4.0 mol4.0\text{ mol} of HI\text{HI}, a 20%20\% decomposition means 0.8 mol0.8\text{ mol} decomposes, leaving 3.2 mol3.2\text{ mol} of HI\text{HI} at equilibrium. Based on the balanced reaction stoichiometry 2HI(g)⇌H2(g)+I2(g)2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g), decomposing 0.8 mol0.8\text{ mol} of HI\text{HI} forms 0.4 mol0.4\text{ mol} of H2\text{H}_2 and 0.4 mol0.4\text{ mol} of I2\text{I}_2. Converting to concentrations in a 2.0 dm32.0\text{ dm}^3 vessel yields [HI]=1.6 mol dm−3[\text{HI}] = 1.6\text{ mol dm}^{-3}, [H2]=0.2 mol dm−3[\text{H}_2] = 0.2\text{ mol dm}^{-3}, and [I2]=0.2 mol dm−3[\text{I}_2] = 0.2\text{ mol dm}^{-3}. Substituting into Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} gives (0.2)(0.2)(1.6)2=0.0156\frac{(0.2)(0.2)}{(1.6)^2} = 0.0156.

Step-by-Step Solution

1
Determine initial moles and amount decomposed at equilibrium
Initial HI=4.0 mol\text{HI} = 4.0\text{ mol}. Amount decomposed =0.20×4.0 mol=0.8 mol= 0.20 \times 4.0\text{ mol} = 0.8\text{ mol}.
The reaction states that 20% of the initial HI dissociates at equilibrium.
2
Calculate equilibrium mole quantities
Equilibrium HI=4.0−0.8=3.2 mol\text{HI} = 4.0 - 0.8 = 3.2\text{ mol}. According to 2HI→H2+I22\text{HI} \rightarrow \text{H}_2 + \text{I}_2, equilibrium H2=0.4 mol\text{H}_2 = 0.4\text{ mol} and I2=0.4 mol\text{I}_2 = 0.4\text{ mol}.
Every 2 moles of HI decomposed produces 1 mole of H2 and 1 mole of I2.
3
Calculate equilibrium concentrations and solve for KcK_c
[HI]=3.22.0=1.6 mol dm−3[\text{HI}] = \frac{3.2}{2.0} = 1.6\text{ mol dm}^{-3}, [H2]=0.42.0=0.2 mol dm−3[\text{H}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}, [I2]=0.42.0=0.2 mol dm−3[\text{I}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}. Thus, Kc=[H2][I2][HI]2=(0.2)(0.2)(1.6)2=0.042.56=0.015625≈0.0156K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac{(0.2)(0.2)}{(1.6)^2} = \frac{0.04}{2.56} = 0.015625 \approx 0.0156.
Substitute equilibrium concentrations into the equilibrium constant expression.

Key Concept

Equilibrium Constant (KcK_c) Calculation from Percent Dissociation
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Topics

Equilibrium Constant Expression and Calculations