If the enthalpy change for the transition of liquid water to steam is 300kJ `mol^(-1)" at "27^(@)C`, the entropy change for the proces would be

Options

A. `1000 JK^(-1)mol^(-1)`
B. `10 JK^(-1)mol^(-1)`
C. `1.0 JK^(-1)mol^(-1)`
D. `0.1JK^(-1)mol^(-1)`

Answer

To find the entropy change (ΔS) for the transition of liquid water to steam, we can use the relationship between enthalpy change (ΔH), temperature (T), and entropy change (ΔS) at equilibrium. The formula we will use is derived from the Gibbs free energy equation:

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

At equilibrium, the change in Gibbs free energy (ΔG) is zero. Therefore, we can rearrange the equation to solve for ΔS:

\[ 0 = \Delta H - T \Delta S \implies \Delta H = T \Delta S \implies \Delta S = \frac{\Delta H}{T} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Enthalpy change (ΔH) for the transition of liquid water to steam = 300 kJ/mol
- Temperature (T) = 27°C = 27 + 273 = 300 K

2. **Convert ΔH from kJ to Joules:**
\[ \Delta H = 300 \text{ kJ/mol} = 300 \times 1000 \text{ J/mol} = 300000 \text{ J/mol} \]

3. **Substitute the values into the ΔS formula:**
\[ \Delta S = \frac{\Delta H}{T} = \frac{300000 \text{ J/mol}}{300 \text{ K}} \]

4. **Calculate ΔS:**
\[ \Delta S = \frac{300000}{300} = 1000 \text{ J/mol·K} \]

### Final Answer:
The entropy change (ΔS) for the process is **1000 J/mol·K**.

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