Question:
At 500 K, for the reaction: $$ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $$ Given: $ K_p = 0.036\, \text{atm}^{-2} $, $ R = 0.082\, \text{L atm mol}^{-1}\text{K}^{-1} $.
Find $ K_c $ in $ \text{L}^2 \text{mol}^{-2} $.
At 500 K, for the reaction: $$ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $$ Given: $ K_p = 0.036\, \text{atm}^{-2} $, $ R = 0.082\, \text{L atm mol}^{-1}\text{K}^{-1} $.
Find $ K_c $ in $ \text{L}^2 \text{mol}^{-2} $.
Show Hint
Use the formula \( K_c = K_p / (RT)^{\Delta n} \), where \( \Delta n = \text{products} - \text{reactants} \). For negative \( \Delta n \), multiply \( K_p \) by \( (RT)^{|\Delta n|} \).
Updated On: Mar 18, 2026
- \( 2.1 \times 10^{-4} \)
- \( 2.1 \times 10^{-5} \)
- 60.5
- 605
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The Correct Option is C
Solution and Explanation
The relation between \( K_p \) and \( K_c \) is:
\[
K_p = K_c (RT)^{\Delta n}
\Rightarrow K_c = \frac{K_p}{(RT)^{\Delta n}}
\]
Step 1: Find \( \Delta n \):
\[
\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}
= 2 - (1 + 3) = 2 - 4 = -2
\]
Step 2: Substitute values:
\[
K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{0.036}{(0.082 \cdot 500)^{-2}}
= 0.036 \cdot (0.082 \cdot 500)^2
\]
\[
= 0.036 \cdot (41)^2 = 0.036 \cdot 1681 \approx 60.5
\]
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