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

The Correct Option is C

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:
\[x = 10\]