Question

Calculate \(\Delta \mathrm{H}^{\circ}\) for a reaction if \(\Delta \mathrm{S}^{\circ}=120 \mathrm{JK}^{-1}\) and \(\Delta \mathrm{G}^{\circ}=28000 \mathrm{~J}\)

• A. 15.94 kJ
• B. 31.83 kJ
• C. 94.12 kJ
• D. 63.76 kJ

Hint:

Use the Gibbs equation, taking the standard temperature of 298 K.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The Gibbs equation is \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), so \(\Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ}\).

Step 2: Substitution
Take the standard temperature \(T = 298\text{ K}\).
\(T\Delta S^{\circ} = 298\times 120 = 35760\text{ J}\).

Step 3: Calculation
\(\Delta H^{\circ} = 28000 + 35760 = 63760\text{ J} = 63.76\text{ kJ}\).

Final Answer:
The enthalpy change is 63.76 kJ. \[ \boxed{\text{(D) }63.76\text{ kJ}} \]