Question

In a closed cylinder of capacity 24.6 L the following reaction occurs at 27$^\circ$C
$A_2(s) \rightleftharpoons B_2(s) + 2C(g)$. At equilibrium 1 g of $B_2(s)$ (molar mass = 50 g mol\textsuperscript{--1}) is present. The equilibrium constant $K_p$ for the equilibrium in atm\textsuperscript{2 unit is ($R = 0.082$ L atm K\textsuperscript{--1} mol\textsuperscript{--1})}

• A. $1.6 \times 10^{-2}$
• B. $1.6 \times 10^{-5}$
• C. $1.6 \times 10^{-3}$
• D. $1.6 \times 10^{-4}$
• E. $1.6 \times 10^{-1}$

Hint:

Heterogeneous equilibrium is a favorite for examiners! Always ignore solids ($s$) and liquids ($l$) when writing $K_c$ or $K_p$ expressions—only gases ($g$) and aqueous ($aq$) species matter.

Answer 1

The Correct Option is C

Understanding the Concept: For heterogeneous equilibria, only gaseous species are included in the equilibrium constant expression ($K_p$).
• $K_p$ Expression: For the reaction $A_2(s) \rightleftharpoons B_2(s) + 2C(g)$, $K_p = (P_C)^2$.
• Stoichiometry: For every mole of $B_2(s)$ produced, 2 moles of $C(g)$ are produced.
• Ideal Gas Law: $PV = nRT \implies P = \frac{nRT}{V}$.

Step 1: Find the moles of gas $C$ at equilibrium. Moles of $B_2$ produced = $\frac{1 \text{ g}}{50 \text{ g/mol}} = 0.02$ mol. From the balanced equation, $n_C = 2 \times n_{B_2} = 2 \times 0.02 = 0.04$ mol.

Step 2: Calculate the partial pressure of $C$ and $K_p$. Given $V = 24.6$ L, $T = 27^\circ C = 300$ K, $R = 0.082$: \[ P_C = \frac{n_C RT}{V} = \frac{0.04 \times 0.082 \times 300}{24.6} \] Note that $24.6 = 0.082 \times 300$. \[ P_C = \frac{0.04 \times 24.6}{24.6} = 0.04 \text{ atm} \] \[ K_p = (P_C)^2 = (0.04)^2 = 0.0016 = 1.6 \times 10^{-3} \text{ atm}^2 \] If $K_p$ is requested in atm$^2$, $0.0016$ is $1.6 \times 10^{-3}$, which is Option C