The standard free energy change for a reaction is -212.3 kJ `mol^(-1)`. If the enthalpy change of the reaction is `-216 kJ mol^(-1)` , calculate the entropy change for the reaction.

Answer

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

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

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

### Step 1: Identify the given values
From the problem, we have:
- ΔG = -212.3 kJ/mol
- ΔH = -216 kJ/mol
- Temperature (T) = 298 K (since it is not mentioned, we assume room temperature)

### Step 2: Rearrange the Gibbs free energy equation to solve for ΔS
We can rearrange the equation to isolate ΔS:

\[ \Delta S = \frac{\Delta H - \Delta G}{T} \]

### Step 3: Substitute the known values into the equation
Now, substitute the values we have into the equation:

\[ \Delta S = \frac{-216 \text{ kJ/mol} - (-212.3 \text{ kJ/mol})}{298 \text{ K}} \]

### Step 4: Simplify the equation
Calculating the numerator:

\[ -216 + 212.3 = -3.7 \text{ kJ/mol} \]

Now substituting this back into the equation for ΔS:

\[ \Delta S = \frac{-3.7 \text{ kJ/mol}}{298 \text{ K}} \]

### Step 5: Perform the division
Now, calculate ΔS:

\[ \Delta S = -0.0124 \text{ kJ/mol/K} \]

### Step 6: Convert ΔS to J/mol/K
Since entropy is often expressed in J/mol/K, we convert kJ to J by multiplying by 1000:

\[ \Delta S = -12.4 \text{ J/mol/K} \]

### Final Answer
The entropy change (ΔS) for the reaction is:

\[ \Delta S = -12.4 \text{ J/mol/K} \]

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