At 300 K, for the reaction, \[ \text{A}_2 \text{B}_2(g) \rightleftharpoons \text{A}_2(g) + \text{B}_2(g) \]
\(\text{is 100 mol L}^{-1}\). \(\text{What is its } K_p \text{(in atm)} \text{at the same temperature?} \)
\(\text{(R = 0.082 L atm mol}^{-1}\) \(\text{K}^{-1}) \)
\(\text{is 100 mol L}^{-1}\). \(\text{What is its } K_p \text{(in atm)} \text{at the same temperature?} \)
\(\text{(R = 0.082 L atm mol}^{-1}\) \(\text{K}^{-1}) \)
Show Hint
- 100
- 2460
- 4.06
- 246
The Correct Option is B
Solution and Explanation
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).
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