To calculate the standard Gibbs free energy change (ΔG°) for the reaction:
\[
\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g)
\]
with the equilibrium constant \( K_p = 4.42 \times 10^4 \) at \( 25^\circ C \), we can use the following equation:
\[
\Delta G° = -2.303 \cdot R \cdot T \cdot \log K_p
\]
### Step-by-Step Solution:
**Step 1: Convert the temperature to Kelvin.**
- The temperature in Celsius is \( 25^\circ C \).
- To convert to Kelvin, use the formula:
\[
T(K) = T(°C) + 273.15
\]
- Thus, \( T = 25 + 273.15 = 298.15 \, K \).
**Step 2: Identify the value of the gas constant \( R \).**
- The value of \( R \) in kilojoules per mole per Kelvin is:
\[
R = 8.314 \, \text{J/(mol·K)} = 8.314 \times 10^{-3} \, \text{kJ/(mol·K)}
\]
**Step 3: Calculate the logarithm of the equilibrium constant \( K_p \).**
- Given \( K_p = 4.42 \times 10^4 \), we need to find \( \log K_p \):
\[
\log K_p = \log(4.42 \times 10^4) = \log(4.42) + \log(10^4) = \log(4.42) + 4
\]
- Using a calculator, \( \log(4.42) \approx 0.645 \).
- Therefore, \( \log K_p \approx 0.645 + 4 = 4.645 \).
**Step 4: Substitute the values into the ΔG° equation.**
- Now substitute \( R \), \( T \), and \( \log K_p \) into the equation:
\[
\Delta G° = -2.303 \cdot (8.314 \times 10^{-3} \, \text{kJ/(mol·K)}) \cdot (298.15 \, K) \cdot (4.645)
\]
**Step 5: Perform the calculation.**
- Calculate the product:
\[
\Delta G° = -2.303 \cdot 8.314 \times 10^{-3} \cdot 298.15 \cdot 4.645
\]
- First, calculate \( 2.303 \cdot 8.314 \times 10^{-3} \cdot 298.15 \):
\[
= 5.693 \, \text{kJ/mol}
\]
- Now, multiply by \( 4.645 \):
\[
\Delta G° = -5.693 \cdot 4.645 \approx -26.5 \, \text{kJ/mol}
\]
### Final Answer:
Thus, the standard Gibbs free energy change for the reaction is:
\[
\Delta G° \approx -26.5 \, \text{kJ/mol}
\]
