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: Understanding the Concept:
The standard Gibbs energy change of a reaction is linked to its equilibrium constant. A large \(K\) (more than 1) means products are favoured, so \(\Delta G^{\circ}\) must be negative.

Step 2: Key Formula:
\[ \Delta G^{\circ} = -RT\ln K = -2.303\,RT\log K \]

Step 3: Substitute the values:
\(R = 8.314\) J K\(^{-1}\) mol\(^{-1}\), \(T = 298\) K, \(K = 100\), \(\ln 100 = 4.605\).
\[ \Delta G^{\circ} = -(8.314)(298)(4.605) = -11409 \text{ J/mol} \]
\[ \Delta G^{\circ} \approx -11.41 \text{ kJ/mol} \]

Step 4: Why the other options are wrong.
-5.744 kJ/mol comes from a slip in the logarithm step. -57.058 and -570.584 kJ/mol are off by factors of 5 and 50, from unit or arithmetic slips.

Final Answer:
The standard Gibbs energy change is \(-11.411\) kJ/mol. \[ \boxed{\text{(A) }-11.411\ \text{kJ/mol}} \]