Question

The gas phase reaction

2A(g) $\rightleftharpoons$ A2(g)

at 400 K has $\Delta$G° = + 25.2 kJ mol$^{-1}$.

The equilibrium constant KC for this reaction is ......$\times$ 10$^{-2}$. (Round off to the Nearest integer)

[Use: R = 8.3 J mol$^{-1}$K$^{-1}$, ln 10 = 2.3

log$_{10}$ 2 = 0.30, 1 atm = 1 bar]

[antilog (-0.3) = 0.501]

Answer 1

166

Answer 2

1. Convert $\Delta G^\circ$ to Joules:
$\Delta G^\circ = 25.2 \times 10^3 \text{ J mol}^{-1}$.

2. Use $\Delta G^\circ = -RT \ln K_p$ to find $\ln K_p$:
$\ln K_p = -\frac{25.2 \times 10^3}{8.3 \times 400} = -\frac{25200}{3320} \approx -7.59036$.

3. Convert $\ln K_p$ to $\log_{10} K_p$:
$\log_{10} K_p = \frac{-7.59036}{2.3} \approx -3.30016$.

4. Calculate $K_p$:
$K_p = 10^{-3.30016} = 10^{0.69984} \times 10^{-4} \approx 5.010 \times 10^{-4}$.

5. Determine $\Delta n_g$ for $2A(g) \rightleftharpoons A_2(g)$:
$\Delta n_g = 1 - 2 = -1$.

6. Use $K_c = K_p (RT)^{-\Delta n_g} = K_p \times RT$:
$K_c = (5.010 \times 10^{-4}) \times (8.3 \times 400) = (5.010 \times 10^{-4}) \times 3320 \approx 1.66332$.

7. Express $K_c$ as $X \times 10^{-2}$:
$1.66332 = 166.332 \times 10^{-2}$.

8. Round $X$ to the nearest integer:
$166$.