The standard free energy change for a reaction is `-213.3 KJ mol^(-1) "at" 25^(@)C`. If the enthalpy change of the reaction is `-217.77 KJ "mole"^(-1)` . Calculate the magnitude of entropy change for the reaction in Joule `"mole"^(-1)`

Explanation

To calculate the magnitude of the entropy change (ΔS) for the reaction, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature in Kelvin
- ΔS = change in entropy

### Step-by-Step Solution:

1. **Identify the given values:**
- ΔG = -213.3 kJ/mol
- ΔH = -217.77 kJ/mol
- Temperature (T) = 25°C = 298 K (since 25 + 273 = 298)

2. **Convert ΔG and ΔH from kJ to J:**
- ΔG = -213.3 kJ/mol × 1000 J/kJ = -213300 J/mol
- ΔH = -217.77 kJ/mol × 1000 J/kJ = -217770 J/mol

3. **Substitute the values into the Gibbs free energy equation:**
\[ -213300 = -217770 - 298 \Delta S \]

4. **Rearrange the equation to solve for ΔS:**
\[ 298 \Delta S = -217770 + 213300 \]
\[ 298 \Delta S = -4470 \]

5. **Calculate ΔS:**
\[ \Delta S = \frac{-4470}{298} \]
\[ \Delta S \approx -15.03 \text{ J/mol·K} \]

6. **Determine the magnitude of ΔS:**
The magnitude of ΔS is:
\[ |\Delta S| \approx 15.03 \text{ J/mol·K} \]

### Final Answer:
The magnitude of the entropy change for the reaction is approximately **15.03 J/mol·K**.

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