Partial Pressures at Equilibrium for Ideal Gases Reaction

To find the partial pressures of $\text{He}$ and $\text{B(g)}$ at equilibrium, we apply the ideal gas equation, $P = \frac{nRT}{V}$, and the principles of chemical equilibrium.

1. Partial pressure of Helium ($\text{He}$): Since $\text{He}$ is an inert gas and does not participate in the reaction, its number of moles remains constant at $n_{\text{He}} = 1\text{ mol}$.

Using the ideal gas equation: $P_{\text{He}} = \frac{n_{\text{He}} R T}{V}$

Given:

• $n_{\text{He}} = 1\text{ mol}$
• $R = 0.082\text{ L atm K}^{-1}\text{ mol}^{-1}$
• $T = 400\text{ K}$
• $V = 10\text{ L}$

Substituting these values: $P_{\text{He}} = \frac{1 \times 0.082 \times 400}{10} = \frac{32.8}{10} = 3.28\text{ atm}$

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2. Partial pressure of $\text{B(g)}$ at equilibrium: Consider the equilibrium reaction: $\text{A(g)} \rightleftharpoons \text{B(g)}$

Let $x$ be the moles of $\text{A(g)}$ converted to $\text{B(g)}$ at equilibrium.

• Initial moles: $n_{\text{A}} = 1\text{ mol}, \quad n_{\text{B}} = 0\text{ mol}$
• Equilibrium moles: $n_{\text{A}} = (1 - x)\text{ mol}, \quad n_{\text{B}} = x\text{ mol}$

The equilibrium concentrations are: $[\text{A}] = \frac{1 - x}{10}\text{ M}$ $[\text{B}] = \frac{x}{10}\text{ M}$

The equilibrium constant $K_c$ is given by: $K_c = \frac{[\text{B}]}{[\text{A}]} = \frac{\frac{x}{10}}{\frac{1 - x}{10}} = \frac{x}{1 - x}$

Given $K_c = 4.0$: $4.0 = \frac{x}{1 - x}$ $4.0(1 - x) = x$ $4 - 4x = x \implies 5x = 4 \implies x = 0.8\text{ mol}$

Therefore, the number of moles of $\text{B(g)}$ at equilibrium is $n_{\text{B}} = 0.8\text{ mol}$.

The partial pressure of $\text{B(g)}$ at equilibrium is: $P_{\text{B}} = \frac{n_{\text{B}} R T}{V} = \frac{0.8 \times 0.082 \times 400}{10} = 0.8 \times 3.28 = 2.624\text{ atm}$

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Thus, the partial pressures of $\text{He}$ and $\text{B(g)}$ at equilibrium are $3.28\text{ atm}$ and $2.624\text{ atm}$, respectively.

Correct Option: A