Calculate `Delta_(r)G^(@)` for `(NH_(4)Cl,s)` at 310K.
Given :`Delta_(r)H^(@)(NH_(4)Cl,s)` =-314 kj/mol,`Delta_(r)C_(p)=0`
`S_(N_(2)(g))^(@)=192 JK^(-1mol^(-1)),S_(H_(2)(g))^(@)=130.5JK^(-1)mol^(-1),`
`S_(Cl_(2)(g))^(@)=233JK mol^(-1), S_(NH_(4)Cl(s))^(@)=99.5JK^(-1)mol^(-1)`
All given data at 300K

Options

A. `-198.56kJ//mol`
B. `-426.7kJ//mol`
C. `-202.3kJ//mol`
D. None of these

Explanation

To calculate the standard Gibbs free energy change (Δ_rG°) for the reaction involving ammonium chloride (NH₄Cl) at 310 K, we will follow these steps:

### Step 1: Write the Reaction
The reaction for the formation of ammonium chloride from its elements is:
\[ \frac{1}{2} N_2(g) + H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow NH_4Cl(s) \]

### Step 2: Calculate the Standard Entropy Change (Δ_rS°)
Using the formula:
\[ \Delta_rS° = S_{products} - S_{reactants} \]

For the reaction:
- \( S_{NH_4Cl(s)} = 99.5 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_{N_2(g)} = 192 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_{H_2(g)} = 130.5 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_{Cl_2(g)} = 233 \, \text{J K}^{-1} \text{mol}^{-1} \)

Calculating the entropy of the reactants:
\[ S_{reactants} = \left( \frac{1}{2} \times S_{N_2} \right) + S_{H_2} + \left( \frac{1}{2} \times S_{Cl_2} \right) \]
\[ S_{reactants} = \left( \frac{1}{2} \times 192 \right) + 130.5 + \left( \frac{1}{2} \times 233 \right) \]
\[ S_{reactants} = 96 + 130.5 + 116.5 = 343 \, \text{J K}^{-1} \text{mol}^{-1} \]

Now, calculate Δ_rS°:
\[ \Delta_rS° = S_{NH_4Cl} - S_{reactants} \]
\[ \Delta_rS° = 99.5 - 343 = -243.5 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Convert Δ_rS° to kJ
\[ \Delta_rS° = -243.5 \, \text{J K}^{-1} \text{mol}^{-1} = -0.2435 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Step 4: Use the Gibbs Free Energy Equation
The Gibbs free energy change is given by:
\[ \Delta_rG° = \Delta_rH° - T \Delta_rS° \]

Given:
- \( \Delta_rH° = -314 \, \text{kJ/mol} \) (at 300 K)
- \( T = 310 \, \text{K} \)

Substituting the values:
\[ \Delta_rG° = -314 - (310 \times -0.2435) \]
\[ \Delta_rG° = -314 + 75.485 = -238.515 \, \text{kJ/mol} \]

### Step 5: Final Answer
Thus, the standard Gibbs free energy change (Δ_rG°) for the reaction at 310 K is:
\[ \Delta_rG° \approx -238.52 \, \text{kJ/mol} \]

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