Question

In a 7.0 L evacuated chamber, 0.50 mol H$_2$ and 0.50 mol I$_2$ react at 427°C. H$_2$(g) + I$_2$(g) $\rightleftharpoons$ 2HI(g). At the given temperature, K$_C$ = 49 for the reaction. What is the partial pressure (atm) of HI in the equilibrium mixture?

Answer 1

6.4 atm

Answer 2

1. The change in the number of moles of gas ($\Delta n$) for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ is $2 - (1+1) = 0$. Therefore, $K_P = K_C = 49$.
2. Calculate the initial molar concentrations of $H_2$ and $I_2$: $[H_2]_0 = [I_2]_0 = \frac{0.50 \text{ mol}}{7.0 \text{ L}} = \frac{1}{14}$ M.
3. Let the equilibrium concentration of HI be $y$. Then, $[H_2]_{eq} = [I_2]_{eq} = \frac{1}{14} - \frac{y}{2}$.
4. Substitute these into the $K_C$ expression: $K_C = \frac{[HI]^2}{[H_2][I_2]} \implies 49 = \frac{y^2}{(\frac{1}{14} - \frac{y}{2})^2}$.
5. Taking the square root of both sides: $7 = \frac{y}{\frac{1}{14} - \frac{y}{2}}$.
6. Solve for $y$: $y = \frac{1}{9}$ M.
7. Calculate the partial pressure of HI using the ideal gas law: $T = 427^\circ\text{C} + 273 = 700$ K. $P_{HI} = [HI]_{eq}RT = \frac{1}{9} \times 0.0821 \times 700 \approx 6.3856$ atm.
8. Rounding to two significant figures, the partial pressure of HI is $6.4$ atm.