For vaporization of water at 1 atmospheric pressure the values of `Delta H and Delta S` are `50.63 kJ mol^9-1)` and `118.8 JK ^(-1) mol^(-1)` respectively. The temperature when Gibbs energy change `(Delta G)` for this transformation will be zero, is

Options

A. `273.4 K`
B. `393.4 K`
C. `426.1 K`
D. `293.4 K`

Answer

To find the temperature at which the Gibbs energy change (ΔG) for the vaporization of water is zero, we can use the relationship between ΔG, ΔH, and ΔS given by the equation:

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

At the temperature where ΔG = 0, we can rearrange the equation to find T:

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

This can be rearranged to:

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

### Step 1: Convert ΔH from kJ to J
Given that:
\[ \Delta H = 50.63 \, \text{kJ/mol} \]
We need to convert this to joules:
\[ \Delta H = 50.63 \times 1000 \, \text{J/mol} = 50630 \, \text{J/mol} \]

### Step 2: Use the given ΔS value
Given:
\[ \Delta S = 118.8 \, \text{J/K/mol} \]

### Step 3: Substitute the values into the equation for T
Now we can substitute the values of ΔH and ΔS into the equation for T:
\[ T = \frac{50630 \, \text{J/mol}}{118.8 \, \text{J/K/mol}} \]

### Step 4: Calculate T
Perform the division:
\[ T = \frac{50630}{118.8} \approx 426.17 \, \text{K} \]

### Final Answer
Thus, the temperature when the Gibbs energy change (ΔG) for the vaporization of water will be zero is approximately:
\[ T \approx 426.17 \, \text{K} \]