The molar heat of vapourization of toluene is `DeltaH_v` if its vapour pressure at 315 K is 60 torr & that at 355 K is 300 torr then `DeltaH_v=?`(log 2=0.3)

Options

A. 37.5 kJ/mole
B. 3.75 kJ/mole
C. 37.5 J/mol
D. 3.75 J/mole

Explanation

To find the molar heat of vaporization (ΔH_v) of toluene using the given vapor pressures at two different temperatures, we can apply the Clausius-Clapeyron equation. Here’s a step-by-step solution:

### Step 1: Write the Clausius-Clapeyron Equation
The Clausius-Clapeyron equation relates the change in vapor pressure with temperature to the heat of vaporization:
\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_v}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
Where:
- \( P_1 \) = vapor pressure at temperature \( T_1 \)
- \( P_2 \) = vapor pressure at temperature \( T_2 \)
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( \Delta H_v \) = molar heat of vaporization

### Step 2: Identify the Given Values
From the problem, we have:
- \( P_1 = 60 \) torr at \( T_1 = 315 \) K
- \( P_2 = 300 \) torr at \( T_2 = 355 \) K

### Step 3: Convert the Equation to Logarithmic Form
We can convert the equation using the logarithm base 10:
\[ \log\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_v}{2.303R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 4: Substitute the Values into the Equation
Substituting the known values into the equation:
\[ \log\left(\frac{300}{60}\right) = -\frac{\Delta H_v}{2.303 \times 8.314} \left(\frac{1}{355} - \frac{1}{315}\right) \]

### Step 5: Calculate the Logarithm
Calculate the logarithm:
\[ \log\left(\frac{300}{60}\right) = \log(5) \approx 0.7 \quad (\text{since } \log(2) = 0.3 \text{ and } \log(5) = \log(2) + \log(2.5) \approx 0.3 + 0.4) \]

### Step 6: Calculate the Temperature Difference
Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):
\[ \frac{1}{355} - \frac{1}{315} = \frac{315 - 355}{355 \times 315} = \frac{-40}{111825} \approx -0.000357 \]

### Step 7: Substitute and Rearrange to Solve for ΔH_v
Now substituting back into the equation:
\[ 0.7 = -\frac{\Delta H_v}{2.303 \times 8.314} \times (-0.000357) \]
Rearranging gives:
\[ \Delta H_v = 0.7 \times 2.303 \times 8.314 \times \frac{1}{0.000357} \]

### Step 8: Calculate ΔH_v
Calculating the values:
- \( 2.303 \times 8.314 \approx 19.1 \)
- \( \Delta H_v \approx 0.7 \times 19.1 \times 2800 \approx 37.414 \text{ kJ/mol} \)

### Final Answer
\[ \Delta H_v \approx 37.5 \text{ kJ/mol} \]