Calculate the magnitude of standard free energy of formation of ammonium chloride at `25^(@)C` (approximate integer in Kcal`mol^(-1)`), the equation showing the formation of `NH_(4)CI` from its elements is
`1/2NH_(4)(g) + 2H_(2)(g) + 1/2CI_(2)(g) rarr NH_(4)CI(s)`
For `NH_(4)CI, DeltaH_(f)^(@)` is `-313 KJ mol^(01)`, Also given that
`" "S_(N_(2))^(@) = 191.5 JK^(-1) mol^(-1)" "S_(N_(2))^(@) = 130.6 JK^(-1) mol^(-1)`
`" "S_(CI_(2))^(@) = 223.0 JK^(-1) mol^(-1) " "S_(NH_(4)CI)^(@) = 94.6 JK^(-1)mol^(-1)`

Explanation

To calculate the magnitude of the standard free energy of formation of ammonium chloride (NH₄Cl) at 25°C, we will follow these steps:

### Step 1: Write the Gibbs Free Energy Equation
The Gibbs free energy change (ΔG) can be calculated using the formula:
\[ \Delta G = \Delta H - T \Delta S \]

### Step 2: Identify Given Values
From the problem statement, we have:
- ΔH (enthalpy of formation of NH₄Cl) = -313 kJ/mol
- Temperature (T) = 25°C = 298 K (since T in Kelvin = °C + 273)
- Entropy values:
- S(N₂) = 191.5 J/K·mol
- S(H₂) = 130.6 J/K·mol
- S(Cl₂) = 223.0 J/K·mol
- S(NH₄Cl) = 94.6 J/K·mol

### Step 3: Calculate ΔS (Change in Entropy)
The change in entropy (ΔS) can be calculated using:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]
For the reaction:
\[ \frac{1}{2} \text{N}_2(g) + 2 \text{H}_2(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NH}_4\text{Cl}(s) \]
The entropy of the reactants is:
\[ S_{\text{reactants}} = \frac{1}{2} S(\text{N}_2) + 2 S(\text{H}_2) + \frac{1}{2} S(\text{Cl}_2) \]
Substituting the values:
\[ S_{\text{reactants}} = \frac{1}{2} \times 191.5 + 2 \times 130.6 + \frac{1}{2} \times 223.0 \]
Calculating this:
\[ S_{\text{reactants}} = 95.75 + 261.2 + 111.5 = 468.45 \text{ J/K·mol} \]
Now, calculate ΔS:
\[ \Delta S = S(\text{NH}_4\text{Cl}) - S_{\text{reactants}} = 94.6 - 468.45 = -373.85 \text{ J/K·mol} \]

### Step 4: Convert ΔS to kJ
To convert ΔS from J to kJ:
\[ \Delta S = -373.85 \text{ J/K·mol} \div 1000 = -0.37385 \text{ kJ/K·mol} \]

### Step 5: Calculate TΔS
Now calculate TΔS:
\[ T \Delta S = 298 \text{ K} \times (-0.37385 \text{ kJ/K·mol}) = -111.56 \text{ kJ/mol} \]

### Step 6: Calculate ΔG
Now substitute ΔH and TΔS into the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
\[ \Delta G = -313 \text{ kJ/mol} - (-111.56 \text{ kJ/mol}) = -313 + 111.56 = -201.44 \text{ kJ/mol} \]

### Step 7: Calculate Magnitude
The magnitude of ΔG is:
\[ \text{Magnitude of } \Delta G \approx 201 \text{ kJ/mol} \]

### Final Answer
The magnitude of the standard free energy of formation of ammonium chloride at 25°C is approximately **201 kJ/mol**.

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