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

To solve this problem, we first calculate the partial pressure of Helium (He), which is an inert gas and does not participate in the chemical reaction. Using the Ideal Gas Equation $PV = nRT$:
For He:
Moles ($n$) = 1 mol
Volume ($V$) = 10 L
Temperature ($T$) = 400 K
Gas constant ($R$) = 0.082 L atm K$^{-1}$ mol$^{-1}$
Partial pressure of He ($P_{He}$) = $\frac{nRT}{V} = \frac{1 \times 0.082 \times 400}{10} = \frac{32.8}{10} = 3.28$ atm.

Now, let's analyze the equilibrium for the reaction $A(g) \rightleftharpoons B(g)$. Let $x$ be the moles of $B$ formed at equilibrium.
Initially: Moles of $A = 1$, Moles of $B = 0$.
At equilibrium: Moles of $A = 1 - x$, Moles of $B = x$.
The equilibrium constant $K_c$ is given by:
$$K_c = \frac{[B]}{[A]} = \frac{x/V}{(1-x)/V} = \frac{x}{1-x}$$
Given $K_c = 4.0$:
$$4 = \frac{x}{1-x}$$
$$4 - 4x = x \implies 5x = 4 \implies x = 0.8 \text{ moles}$$
So, at equilibrium, the number of moles of $B$ is 0.8 moles. The partial pressure of $B$ ($P_B$) is:
$$P_B = \frac{n_B RT}{V} = \frac{0.8 \times 0.082 \times 400}{10} = 0.8 \times 3.28 = 2.624 \text{ atm}.$$
Therefore, the partial pressures of He and $B(g)$ are 3.28 atm and 2.624 atm respectively.