Question

Value of equilibrium constant for following reaction, N$_2$(g) + 3H$_2$(g) $\rightarrow$ 2NH$_3$, (g) is 0.50 at 400°C. What is the value of K$_p$ at 400°C ?

• A. 1.64 × 10$^{-3}$
• B. 1.64 × 10$^4$
• C. 1.64 × 10$^{-4}$
• D. 164 × 10$^4$

Answer 1

Answer: (C)

1.64 × 10$^{-4}$

Answer 2

The equilibrium constant $K_p$ is related to $K_c$ by the equation $K_p = K_c (RT)^{\Delta n_g}$.

For the reaction N$_2$(g) + 3H$_2$(g) $\rightarrow$ 2NH$_3$(g), $\Delta n_g = 2 - (1+3) = -2$.

The temperature is $T = 400^\circ\text{C} = 400 + 273 = 673 \text{ K}$.

Using $R = 0.0821 \text{ L atm/mol K}$ and $K_c = 0.50$, we calculate $K_p$.

$K_p = 0.50 \times (0.0821 \times 673)^{-2} = 0.50 \times (55.2533)^{-2} = 0.50 \times \frac{1}{3053.02} \approx 0.50 \times 3.275 \times 10^{-4} \approx 1.6375 \times 10^{-4}$.

This value is approximately $1.64 \times 10^{-4}$.