At `25^(@)C`, `DeltaG^(@)` for the process `H_(2)O(l)iffH_(2)O(g)` is 8.6 kJ. The vapour pressure of water at this temperature, is nearly :

Options

A. 24 torr
B. 285 torr
C. 23.72 torr
D. 100 torr

Explanation

To solve the problem of finding the vapor pressure of water at 25°C given that ΔG° for the process H₂O(l) ⇌ H₂O(g) is 8.6 kJ, we can follow these steps:

### Step 1: Understand the relationship between ΔG° and the equilibrium constant (K)
The Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation:
\[ \Delta G° = -RT \ln K \]
Where:
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T \) is the temperature in Kelvin

### Step 2: Convert ΔG° from kJ to J
Given that ΔG° is 8.6 kJ, we need to convert this to joules:
\[ \Delta G° = 8.6 \, \text{kJ} \times 1000 \, \text{J/kJ} = 8600 \, \text{J} \]

### Step 3: Convert the temperature from Celsius to Kelvin
The temperature given is 25°C. To convert this to Kelvin:
\[ T = 25°C + 273.15 = 298.15 \, \text{K} \]

### Step 4: Rearrange the equation to solve for K
We can rearrange the equation for K:
\[ K = e^{-\Delta G° / (RT)} \]

### Step 5: Substitute the values into the equation
Substituting the values of ΔG°, R, and T into the equation:
\[ K = e^{-8600 / (8.314 \times 298.15)} \]

### Step 6: Calculate the exponent
First, calculate the denominator:
\[ 8.314 \times 298.15 \approx 2478.55 \]
Now calculate the exponent:
\[ -\frac{8600}{2478.55} \approx -3.47 \]

### Step 7: Calculate K
Now calculate K:
\[ K = e^{-3.47} \approx 0.031 \]

### Step 8: Relate K to vapor pressure
For the reaction H₂O(l) ⇌ H₂O(g), the equilibrium constant K is equal to the vapor pressure (P) of water at this temperature, since the activity of pure liquid water is 1:
\[ K = \frac{P}{P^0} \]
Where \( P^0 \) (the standard pressure) is 101325 Pa. Therefore:
\[ P = K \times P^0 \]
Substituting the values:
\[ P = 0.031 \times 101325 \, \text{Pa} \approx 3148.5 \, \text{Pa} \]

### Step 9: Convert the vapor pressure from Pa to torr
To convert from Pascals to torr, we use the conversion factor \( 1 \, \text{torr} \approx 133.322 \, \text{Pa} \):
\[ P \, (\text{torr}) = \frac{3148.5}{133.322} \approx 23.6 \, \text{torr} \]

### Final Answer
The vapor pressure of water at 25°C is approximately **23.6 torr**.

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