Question

Calculate standard Gibbs energy for any gaseous reaction if values of \(K_p\) and \(R\) respectively are \(3.5\times 10^{17}\) ; \(8.314\,\text{JK}^{-1}\text{mol}^{-1}\)

• A. \(-100.1\,\text{kJ mol}^{-1}\)
• B. \(-79.5\,\text{kJ mol}^{-1}\)
• C. \(-71.4\,\text{kJ mol}^{-1}\)
• D. \(-89.5\,\text{kJ mol}^{-1}\)

Hint:

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

Answer 1

The Correct Option is A

Step 1: Understand the concept
The standard Gibbs energy change is linked to the equilibrium constant by \(\Delta G^\circ = -RT\ln K_p = -2.303\,RT\log K_p\). The temperature is taken as 298 K, the standard value.

Step 2: Find \(\log K_p\)
\[ \log(3.5 \times 10^{17}) = \log 3.5 + 17 = 0.544 + 17 = 17.544 \]

Step 3: Substitute
\[ \Delta G^\circ = -2.303 \times 8.314 \times 298 \times 17.544 \ \text{J mol}^{-1} \]
First \(2.303 \times 8.314 \times 298 = 5705.8\) J/mol. Then \(5705.8 \times 17.544 \approx 1.001 \times 10^{5}\) J/mol.

Step 4: Result and check
\[ \Delta G^\circ \approx -100.1\ \text{kJ mol}^{-1} \]
The sign is negative because \(K_p\) is very large, which means the reaction is strongly spontaneous. The other options (-79.5, -71.4, -89.5) would need a smaller \(\log K_p\), about 13.9, 12.5 or 15.7.

Final Answer:
Delta G is about -100.1 kJ per mole. This is option (A). \[ \boxed{\text{(A) }-100.1\ \text{kJ mol}^{-1}} \]