Question

Calculate \(△G^{\circ}\) for reaction,
\(\text{CH}_{4(g)}+\text{H}_{2(g)}\rightarrow \text{C}_2\text{H}_{6(g)}\) (\(K_p = 2\times 10^{17}\), \(R = 8.314 \text{JK}^{-1}\text{mol}^{-1}\))

• A. \(-64.695\) kJ mol\(^{-1}\)
• B. \(-98.716\) kJ mol\(^{-1}\)
• C. \(-44.08\) kJ mol\(^{-1}\)
• D. \(-58.78\) kJ mol\(^{-1}\)

Hint:

Use delta G = -2.303 RT log Kp at 298 K.

Answer 1

The Correct Option is B

Step 1: Idea:
At equilibrium $\Delta G = 0$, which gives $\Delta G^{\circ} = -RT\ln K$. I will use natural logarithm form and then convert.

Step 2: Compute ln K:
$\ln(2\times10^{17}) = \ln 2 + 17\ln 10 = 0.693 + 39.144 = 39.837$.

Step 3: Compute RT:
$RT = 8.314 \times 298 = 2477.6$ J mol$^{-1}$.

Step 4: Multiply:
\[ \Delta G^{\circ} = -2477.6 \times 39.837 \approx -98700 \text{ J mol}^{-1} \]
This rounds to about -98.7 kJ mol$^{-1}$, matching the listed value of -98.716 kJ mol$^{-1}$ (the small difference is rounding of logs).

Final Answer:
The value is about -98.7 kJ per mole, option (B). \[ \boxed{-98.716 \text{ kJ mol}^{-1}} \]