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} $.

• A. \( 2.1 \times 10^{-4} \)
• B. \( 2.1 \times 10^{-5} \)
• C. 60.5
• D. 605

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 \( (RT)^{|\Delta n|} \).

Answer 1

The Correct Option is C

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: \[ K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{0.036}{(0.082 \cdot 500)^{-2}} = 0.036 \cdot (0.082 \cdot 500)^2 \] \[ = 0.036 \cdot (41)^2 = 0.036 \cdot 1681 \approx 60.5 \]