Question

At 300 K, for the reaction, $$\text{A}_2 \text{B}_2(g) \rightleftharpoons \text{A}_2(g) + \text{B}_2(g)$$ 
$\text{is 100 mol L}^{-1}$. $\text{What is its } K_p \text{(in atm)} \text{at the same temperature?}$
$\text{(R = 0.082 L atm mol}^{-1}$ $\text{K}^{-1})$

• A. 100
• B. 2460
• C. 4.06
• D. 246

Hint:

For reactions where the change in the number of moles of gas is zero, $K_p = K_c$, which simplifies calculations.

Answer 1

The Correct Option is B

To determine $K_p$ for the given reaction at 300 K, we use the relationship between $K_c$ and $K_p$: $$K_p = K_c (RT)^{\Delta n}$$ Given: - $K_c = 100 \, \text{mol L}^{-1}$ 
- Temperature, $T = 300 \, \text{K}$ 
- Gas constant, $R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1}$ 
- Change in the number of moles of gas, $\Delta n$ 
Step 1: Determine $\Delta n$ For the reaction: $$\text{A}_2 \text{B}_2(g) \rightleftharpoons \text{A}_2(g) + \text{B}_2(g)$$ 
- Reactants: 1 mole of $\text{A}_2 \text{B}_2$ 
- Products: 1 mole of $\text{A}_2$ + 1 mole of $\text{B}_2$ $$\Delta n = (1 + 1) - 1 = 1$$ 
Step 2: Calculate $K_p$ Using the formula: $$K_p = K_c (RT)^{\Delta n}$$ Substitute the values: $$K_p = 100 \times (0.082 \times 300)^1$$ $$K_p = 100 \times 24.6$$ $$K_p = 2460 \, \text{atm}$$ Final Answer: $$\boxed{2460}$$ This corresponds to option (2).