What is `DeltaG^(ɵ)` for the following reaction? `1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g), K_(p)=4.42xx10^(4)` at `25^(@)C`

Options

A. `-26.5 kJ mol^(-1)`
B. `-11.5 kJ mol^(-1)`
C. `-2.2 kJ mol^(-1)`
D. `-0.97 kJ mol^(-1)`

Explanation

To find the standard Gibbs free energy change (ΔG°) for the reaction given, we can use the relationship between ΔG° and the equilibrium constant (Kp). The formula is:

\[ \Delta G^\circ = -2.303 \cdot R \cdot T \cdot \log(K_p) \]

### Step 1: Identify the values needed for the calculation
- **Kp** = \(4.42 \times 10^4\)
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