Given `triangle_(r)H^(@) ("kJ/mol") S_(m)^(@) ("J/mol-K")`
`C Cl_(4)(l)-135.4" "214.4`
`C Cl_(4)(g) -103.0" "308.7`
The boiling point of carbon tetrachloride is

Options

A. `8.25^(@)C`
B. `70.5^(@)C`
C. `92.3^(@)C`
D. `45.8^(@)C`

Answer

To find the boiling point of carbon tetrachloride (CCl4) using the given enthalpy and entropy values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Enthalpy of CCl4 (liquid): \( \Delta H_{CCl_4(l)} = -135.4 \, \text{kJ/mol} \)
- Enthalpy of CCl4 (gas): \( \Delta H_{CCl_4(g)} = -103.0 \, \text{kJ/mol} \)
- Entropy of CCl4 (liquid): \( S_{CCl_4(l)} = 214.4 \, \text{J/mol-K} \)
- Entropy of CCl4 (gas): \( S_{CCl_4(g)} = 308.7 \, \text{J/mol-K} \)

2. **Calculate the Change in Enthalpy (\( \Delta H \))**:
\[ \Delta H = \Delta H_{CCl_4(g)} - \Delta H_{CCl_4(l)} = (-103.0) - (-135.4) = 32.4 \, \text{kJ/mol} \]

3. **Calculate the Change in Entropy (\( \Delta S \))**:
\[ \Delta S = S_{CCl_4(g)} - S_{CCl_4(l)} = 308.7 - 214.4 = 94.3 \, \text{J/mol-K} \]

4. **Convert \( \Delta H \) to Joules**:
\[ \Delta H = 32.4 \, \text{kJ/mol} = 32.4 \times 10^3 \, \text{J/mol} = 32400 \, \text{J/mol} \]

5. **Use the Formula to Find the Boiling Point (T)**:
The boiling point can be calculated using the equation:
\[ T = \frac{\Delta H}{\Delta S} \]
Substituting the values:
\[ T = \frac{32400 \, \text{J/mol}}{94.3 \, \text{J/mol-K}} \approx 343.58 \, \text{K} \]

6. **Convert Kelvin to Celsius**:
To convert the temperature from Kelvin to Celsius:
\[ T(°C) = T(K) - 273 = 343.58 - 273 \approx 70.58 \, °C \]

### Final Answer:
The boiling point of carbon tetrachloride (CCl4) is approximately **70.58 °C**.