Question 15.8: A reaction vessel containing 1.000 × 10^- 3 M H2 gas and 2.0...
A reaction vessel containing 1.000 × 10^{- 3} M H_{2} gas and 2.000 × 10^{- 3} M I_{2} gas is heated to 448 °C where the following reaction takes place
H_{2} (g) + I_{2} (g)⇌2 HI (g)What is the value of the equilibrium constant K_{c} if once the system comes to equilibrium at 448 °C the concentration of HI is 1.87 × 10^{-3}
Analyze We are given the initial concentrations of H_{2} and I_{2} and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant K_{c} for H_{2} (g) + I_{2} (g)⇌2 HI (g).
Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant.
Solve
(1) We tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table.
| Initial concentration (M) | 1.000 × 10^{-3} | 2.000 × 10^{-3} | 0 |
| Change in concentration (M) | |||
| Equilibrium concentration (M) | 1.87 × 10^{-3} |
(2) We calculate the change in HI concentration, which is the difference between the equilibrium and initial values:
Change in [HI] = 1.87 × 10^{-3} M – 0 = 1.87 × 10^{-3} M
(3) We use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H_{2}] and [I_{2}]:
\left(1.87 × 10^{-3} \frac{\cancel{ mol HI}}{L} \right)\left(\frac{1 mol H_{2}}{2 \cancel{mol HI}} \right)= 0.935 × 10^{-3} \frac{mol H_{2}}{L}\left(1.87 × 10^{-3} \frac{\cancel{ mol HI}}{L} \right)\left(\frac{1 mol I_{2}}{2 \cancel{mol HI}} \right)= 0.935 × 10^{-3} \frac{mol I_{2}}{L}
(4) We calculate the equilibrium concentrations of H_{2} and I_{2}, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed:
[H_{2}] = (1.000 × 10^{-3} M) – (0.935 × 10^{-3} M) = 0.065 × 10^{-3} M[I_{2}] = (2.000 × 10^{-3} M) – (0.935 × 10^{-3} M) = 1.065 × 10^{-3} M
(5) Our table now is complete (with equilibrium concentrations in blue for emphasis):
H_{2} (g) + I_{2} (g)⇌2 HI (g)| Initial concentration (M) | 1.000 × 10^{-3} | 2.000 × 10^{-3} | 0 |
| Change in concentration (M) | -0.935 × 10^{-3} | -0.935 × 10^{-3} | +1.87 × 10^{-3} |
| Equilibrium concentration (M) | 0.065 × 10^{-3} | 1.065 × 10^{-3} | 1.87 × 10^{-3} |
Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed.
Finally, we use the equilibrium-constant expression to calculate the equilibrium constant:
K_{c} =\frac{[HI]²}{[H_{2}][I_{2}]}=\frac{(1.87 ×10^{-3})²}{(0.065 ×10^{-3})(1.065 ×10^{-3})}= 51Comment The same method can be applied to gaseous equilibrium problems to calculate K_{p}, in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for \underline{I}nitial – \underline{C}hange – \underline{E}quilibrium.