Question

One mole each of He and $A(g)$ are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.
$$A(g) \rightleftharpoons B(g)$$
$K_c$ for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium)
(Assume He, $A(g)$ and $B(g)$ behave as ideal gases)
(Given : $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$)

• A. 3.28, 2.624
• B. 2.624, 3.28
• C. 3.28, 0.656
• D. 0.656, 6.56

Hint:

Calculate the partial pressure of Helium first using the ideal gas law. Then use the $K_c$ expression to find the moles of $B$ at equilibrium and convert that to partial pressure.

Answer 1

The Correct Option is A

We begin by recognizing that the total pressure in the flask is the sum of the partial pressures of all gases present (He, $A$, and $B$). Since He is an inert gas, its partial pressure depends only on its initial moles, volume, and temperature.
$P_{He} = \frac{n_{He}RT}{V} = \frac{1 \times 0.082 \times 400}{10} = 3.28$ atm.

For the reaction $A(g) \rightleftharpoons B(g)$, the change in moles $\Delta n_g = 1 - 1 = 0$. Thus, $K_p = K_c (RT)^{\Delta n_g} = K_c = 4.0$.
Let $P_A$ and $P_B$ be the partial pressures of $A$ and $B$ at equilibrium. Initially, only $A$ is present with a partial pressure $P_A^0 = \frac{1 \times 0.082 \times 400}{10} = 3.28$ atm.
At equilibrium, $P_A = 3.28 - p$ and $P_B = p$ (where $p$ is the change in pressure).
$$K_p = \frac{P_B}{P_A} = \frac{p}{3.28 - p} = 4$$
$$p = 4(3.28) - 4p \implies 5p = 13.12 \implies p = 2.624 \text{ atm}.$$
Thus, $P_B = 2.624$ atm. The equilibrium partial pressures of He and $B(g)$ are 3.28 atm and 2.624 atm.