For vaporization of water at 1 atmospheric pressure the values of `Delta H and Delta S` are `50.63 kJ mol^-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 free energy change (ΔG) for the vaporization of water is zero, we can use the Gibbs free energy equation:

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

### Step 1: Set ΔG to zero
Since we want to find the temperature when ΔG is zero, we can set the equation to:

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

### Step 2: Rearrange the equation
Rearranging the equation gives us:

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

### Step 3: Solve for T
Now, we can solve for T:

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

### Step 4: Substitute the values
We know that:
- ΔH = 50.63 kJ/mol = 50.63 × 10³ J/mol (convert kJ to J)
- ΔS = 118.8 J/(K·mol)

Substituting these values into the equation:

\[ T = \frac{50.63 \times 10^3 \text{ J/mol}}{118.8 \text{ J/(K·mol)}} \]

### Step 5: Calculate T
Now, we perform the calculation:

\[ T = \frac{50630 \text{ J/mol}}{118.8 \text{ J/(K·mol)}} \approx 426.1 \text{ K} \]

### Final Answer
Thus, the temperature at which ΔG for the vaporization of water is zero is approximately:

\[ T \approx 426.1 \text{ K} \]

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