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: Understanding the Concept:
The standard Gibbs energy change of a reaction is linked to its equilibrium constant. A large \(K_p\) means a strongly negative \(\Delta G^{\circ}\).

Step 2: Key Formula:
\[ \Delta G^{\circ} = -2.303\, RT \log_{10} K_p \]
The temperature is not stated, so the standard temperature of 298 K is used.

Step 3: Detailed Explanation:
\(\log(2 \times 10^{17}) = \log 2 + 17 = 0.301 + 17 = 17.301\).
\[ \Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times 17.301 \]
\[ 2.303 \times 8.314 = 19.147,\quad 19.147 \times 298 = 5705.8 \]
\[ \Delta G^{\circ} = -5705.8 \times 17.301 \approx -98717 \text{ J mol}^{-1} \approx -98.7 \text{ kJ mol}^{-1} \]

Step 4: Why the other options are wrong.
The other values (-64.695, -44.08, -58.78 kJ) correspond to smaller \(\log K\) values, such as \(K\) near \(10^{11}\) to \(10^{8}\). They do not match \(K_p = 2\times10^{17}\).

Final Answer:
\(\Delta G^{\circ} \approx -98.716\) kJ mol\(^{-1}\), option (B). \[ \boxed{-98.716 \text{ kJ mol}^{-1}} \]