For vaporization of water at 1 atmospheric pressure, the values of `DeltaH and DeltaS` are `40.63 "kJ mol"^(-1) and 108.8 "JK"^(-1) mol^(-1)` respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero , is :

Options

A. 293.4 K
B. 273.4 K
C. 393.4 K
D. 373.4 K

Answer

To solve the problem, we need to find the temperature at which the Gibbs free energy change (ΔG) for the vaporization of water is zero. We can use the Gibbs free energy equation:

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

Given that ΔG = 0 at the temperature we are looking for, we can rearrange the equation:

\[ 0 = \Delta H - T \Delta S \]

This simplifies to:

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

### Step 1: Convert ΔH to Joules
The value of ΔH is given as 40.63 kJ/mol. We need to convert this to Joules:

\[ \Delta H = 40.63 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40630 \, \text{J/mol} \]

### Step 2: Use the given ΔS value
The value of ΔS is given as 108.8 J/K·mol.

### Step 3: Calculate the temperature T
Now we can substitute the values of ΔH and ΔS into the equation for T:

\[ T = \frac{\Delta H}{\Delta S} = \frac{40630 \, \text{J/mol}}{108.8 \, \text{J/K·mol}} \]

### Step 4: Perform the calculation
Calculating the temperature:

\[ T = \frac{40630}{108.8} \approx 373.0 \, \text{K} \]

Thus, the temperature at which the Gibbs energy change (ΔG) for the vaporization of water is zero is approximately **373 K**.

### Final Answer:
The temperature when Gibbs energy change (ΔG) for the vaporization of water will be zero is **373 K**.

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