Question

Calculate standard Gibbs energy for a certain reaction if \(\Delta \mathrm{H}^{\circ}\) and \(\Delta \mathrm{S}^{\circ}\) respectively are \(219 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(-20 \mathrm{JK}^{-1}\).

• A. \(224.96\text{ }\mathrm{kJ}\text{ }\mathrm{mol}^{-1}\)
• B. \(112.48\text{ }\mathrm{kJ}\text{ }\mathrm{mol}^{-1}\)
• C. \(312.10\text{ }\mathrm{kJ}\text{ }\mathrm{mol}^{-1}\)
• D. \(426.24\text{ }\mathrm{kJ}\text{ }\mathrm{mol}^{-1}\)

Hint:

Use the Gibbs-Helmholtz relation at 298 K with consistent units for entropy and enthalpy.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). Convert \(\Delta S^{\circ}\) to kJ so that units agree, using \(T=298\ \text{K}\).

Step 2: Substitution
\(\Delta H^{\circ}=219\ \text{kJ mol}^{-1}\) and \(\Delta S^{\circ}=-20\ \text{J K}^{-1}=-0.020\ \text{kJ K}^{-1}\).

Step 3: Calculation
\(\Delta G^{\circ}=219-298\times(-0.020)=219+5.96=224.96\ \text{kJ mol}^{-1}\).

Step 4: Compare the options
Option (A).

Final Answer:
The standard Gibbs energy is \(224.96\ \text{kJ mol}^{-1}\). \[ \boxed{\text{(A) }224.96\ \text{kJ mol}^{-1}} \]