At -20 °C and 1 atm pressure, a cylinder is filled with equal number of H2, I2 and HI molecules for the reaction H2(g)…

Step 1 — Identify the reaction:

$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$

$\Delta n_g = 2 - 2 = 0$, so $K_p = K_c$ (dimensionless).

Step 2 — Calculate $Q_p$ with equal moles:

Equal number of $\mathrm{H_2}$, $\mathrm{I_2}$, and HI molecules at 1 atm total pressure. Let each have $n$ moles. Total = $3n$.

$P_{H_2} = P_{I_2} = P_{HI} = \frac{1}{3}$ atm

$Q_p = \frac{P_{HI}^2}{P_{H_2} \cdot P_{I_2}} = \frac{(1/3)^2}{(1/3)(1/3)} = 1$

Step 3 — Determine $K_p$:

Since the system is at equilibrium at $-20°C$, $K_p = Q_p = 1$.

The question asks for $K_p$ in the form $x \times 10^{-1}$, where $x = 1$ means $K_p = 1 \times 10^{0} = 1$.

$K_p = 1$

Answer: Option (4) — $K_p = 1$