Question

For the following reaction,
\(\text{C}(s)+2\text{H}_{2}(g)\rightarrow \text{CH}_{4}(g);\ \Delta _{r}H^{0}=-75\ \text{kJ}\).
Calculate \(\Delta G^{0}\) if \(\Delta S^{0}\) is \(185\ \text{JK}^{-1}\).

• A. \(-130.13\ \text{kJ}\)
• B. \(130.13\ \text{kJ}\)
• C. \(19.87\ \text{kJ}\)
• D. \(-19.87\ \text{kJ}\)

Hint:

Use the Gibbs equation with the enthalpy in kJ and entropy converted to kJ per kelvin.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The Gibbs equation is \(\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}\), with temperature taken as \(298\ \text{K}\).

Step 2: Unit Conversion
\(\Delta S^{0}=185\ \text{J K}^{-1}=0.185\ \text{kJ K}^{-1}\)

Step 3: Calculation
\(T\Delta S^{0}=298\times 0.185=55.13\ \text{kJ}\)
\(\Delta G^{0}=-75-55.13=-130.13\ \text{kJ}\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The standard free energy change is minus 130.13 kJ. \[ \boxed{\text{(A) }-130.13\ \text{kJ}} \]