Given: `Delta_(f)H^(@) (kJ//mol) " " S_(m)^(@)(J//K mol)`
`{:( C Cl_(4)(l), -135 , 215.4 ),( C Cl_(4)(g), -103.0, 308.7):}:`
What is the boiling point of carbon tetrachloride?

Options

A. `8.25^(@)` C
B. `69.97^(@)` C
C. `92.3^(@)` C
D. `45.8^(@)` C

Explanation

To find the boiling point of carbon tetrachloride (CCl₄), we can use the relationship between the enthalpy change (ΔH) and the entropy change (ΔS) at the boiling point. The boiling point occurs when the change in enthalpy is equal to the change in entropy multiplied by the temperature.

### Step-by-Step Solution:

1. **Identify the Reaction:**
The reaction we are considering is the phase change from liquid CCl₄ to gaseous CCl₄:
\[ \text{CCl}_4(l) \rightarrow \text{CCl}_4(g) \]

2. **Calculate ΔH° (Standard Enthalpy Change of Reaction):**
The standard enthalpy change of the reaction (ΔH°) can be calculated using the standard enthalpy of formation values provided:
\[ ΔH° = ΔH_f°(\text{CCl}_4(g)) - ΔH_f°(\text{CCl}_4(l)) \]
Substituting the values:
\[ ΔH° = (-103.0 \, \text{kJ/mol}) - (-135.0 \, \text{kJ/mol}) = 32.0 \, \text{kJ/mol} \]

3. **Convert ΔH° to Joules:**
Since the entropy is given in Joules, we need to convert ΔH° from kJ to J:
\[ ΔH° = 32.0 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 32000 \, \text{J/mol} \]

4. **Calculate ΔS° (Standard Entropy Change of Reaction):**
The standard entropy change of the reaction (ΔS°) is calculated as follows:
\[ ΔS° = S_m°(\text{CCl}_4(g)) - S_m°(\text{CCl}_4(l)) \]
Substituting the values:
\[ ΔS° = 308.7 \, \text{J/K mol} - 215.4 \, \text{J/K mol} = 93.3 \, \text{J/K mol} \]

5. **Calculate the Boiling Point (T_boiling):**
The boiling point can be found using the formula:
\[ T_{\text{boiling}} = \frac{ΔH°}{ΔS°} \]
Substituting the values:
\[ T_{\text{boiling}} = \frac{32000 \, \text{J/mol}}{93.3 \, \text{J/K mol}} \approx 342.97 \, \text{K} \]

6. **Convert Temperature to Celsius:**
To convert the boiling point from Kelvin to Celsius:
\[ T_{\text{boiling}}(°C) = T_{\text{boiling}}(K) - 273.15 \]
\[ T_{\text{boiling}}(°C) \approx 342.97 \, \text{K} - 273.15 \approx 69.82 \, °C \]

7. **Final Answer:**
The boiling point of carbon tetrachloride is approximately **69.82 °C**.