Question

Assuming ideal behaviour, the magnitude of log K for the following reaction at 25 °C is \( x \times 10^{-1} \). The value of \( x \) is __________ . (Integer answer) \( 3\,\text{HC}\equiv\text{CH (g)} \rightleftharpoons \text{C}_6\text{H}_6 \text{ (l)} \) [Given : \( \Delta_f G^\circ (\text{HC}\equiv\text{CH}) = -2.04 \times 10^5 \,\text{J mol}^{-1} \); \( \Delta_f G^\circ (\text{C}_6\text{H}_6) = -1.24 \times 10^5 \,\text{J mol}^{-1} \); \( R = 8.314 \,\text{J K}^{-1} \text{ mol}^{-1} \)]

Hint:

$\Delta G^\circ$ is positive here, implying the reaction is non-spontaneous at 25°C under standard conditions, resulting in a very small $K$ (large negative $\log K$).

Answer 1

Correct Answer: 855

Step 1: $\Delta G^\circ_{rxn} = \Delta_f G^\circ(C_6H_6) - 3\Delta_f G^\circ(C_2H_2) = (-1.24 \times 10^5) - 3(-2.04 \times 10^5)$.
Step 2: $\Delta G^\circ_{rxn} = -1.24 \times 10^5 + 6.12 \times 10^5 = 4.88 \times 10^5 \text{ J/mol}$.
Step 3: $\Delta G^\circ = -2.303 RT \log K \Rightarrow 4.88 \times 10^5 = -2.303 \times 8.314 \times 298 \times \log K$.
Step 4: $\log K = \frac{-488000}{5705.8} \approx -85.52$.
Step 5: Magnitude $| \log K | \approx 85.5 = 855 \times 10^{-1}$. Thus, $x = 855$.