Question

At $-20^\circ \text{C}$ and 1 atm pressure, a cylinder is filled with an equal number of $H_2$, $I_2$, and $HI$ molecules for the reaction:
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ The $K_P$ for the process is $x \times 10^{-1}$.
x = ___________) **Given:** \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}

• A. 2
• B. 1
• C. 10
• D. 0.01

Answer 1

Answer: (C)

10

Answer 2

The reaction given is:
$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$
At equilibrium, $\Delta n_g = 0$ (no change in the number of moles of gas). Therefore:
$K_p = \frac{(n_{\text{HI}})^2}{n_{\text{H}_2} \cdot n_{\text{I}_2}} \cdot \left( \frac{P_T}{P_T} \right)^{\Delta n_g}$
Since $\Delta n_g = 0$, the pressure term simplifies:
$K_p = \frac{(n_{\text{HI}})^2}{n_{\text{H}_2} \cdot n_{\text{I}_2}}$
Given that the number of moles of H$_2$, I$_2$, and HI are all initially equal:
$n_{\text{H}_2} = n_{\text{I}_2} = 1, \quad n_{\text{HI}} = 1$
Substituting into the formula:
$K_p = \frac{(1)^2}{1 \cdot 1} = 1 \times 10^1$
Thus:
$x = 10$