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: Plan
Convert the equilibrium constant to a free energy through $\Delta G^\circ = -RT\ln K_p$.

Step 2: Natural log
$\ln K_p = \ln 3.5 + 17\ln 10 = 1.2528 + 39.1439 = 40.397$.

Step 3: Evaluate
$RT = 8.314 \times 298 = 2477.6$ J/mol. So $\Delta G^\circ = -2477.6 \times 40.397 = -1.0008 \times 10^{5}$ J/mol.

Step 4: Interpret
This is -100.1 kJ/mol. A very negative value matches a very large $K_p$. The smaller magnitudes in the other options would correspond to smaller equilibrium constants.

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