Question

The equilibrium constant for a reaction is \(100\). What will be the value of standard Gibbs energy change at \(298\) K ? (\(R = 8.314\) J \(\text{K}^{-1}\text{mol}^{-1}\))

• A. \(-11.411\) KJ/mol
• B. \(-5.744\) KJ/mol
• C. \(-570.584\) KJ/mol
• D. \(-57.058\) KJ/mol

Hint:

Use delta G standard = -RT ln K and convert the answer from J to kJ.

Answer 1

The Correct Option is A

Step 1: Use base-10 logarithm:
$\Delta G^{\circ} = -2.303\,RT\log_{10}K$. Here $\log_{10} 100 = 2$.

Step 2: Compute the product:
$8.314 \times 298 = 2477.6$ and $2477.6 \times 2.303 = 5705.8$ J/mol.

Step 3: Multiply by log K:
$\Delta G^{\circ} = -5705.8 \times 2 = -11411.6$ J/mol, which is $-11.411$ kJ/mol.

Step 4: Check the sign and units:
$K > 1$, so a negative value is expected. The answer in kJ/mol matches option A.

Final Answer:
Using base-10 logs, delta G standard works out to -11.411 kJ/mol. \[ \boxed{\text{(A) }-11.411\ \text{kJ/mol}} \]