Question:
Consider the following reaction:
\[ \text{CCl}_4(g) \rightarrow \text{C}(g) + 4 \text{Cl}(g) \quad \Delta H = 1304\text{ kJ} \]
What is $\Delta_{\text{vap}} H^\ominus$ of $\text{CCl}_4(l)$ (in $\text{kJ}\cdot\text{mol}^{-1}$)?
Given: $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{Cl}_2) = 242\text{ kJ}\cdot\text{mol}^{-1}$
Consider the following reaction:
\[ \text{CCl}_4(g) \rightarrow \text{C}(g) + 4 \text{Cl}(g) \quad \Delta H = 1304\text{ kJ} \]
What is $\Delta_{\text{vap}} H^\ominus$ of $\text{CCl}_4(l)$ (in $\text{kJ}\cdot\text{mol}^{-1}$)?
Given: $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{Cl}_2) = 242\text{ kJ}\cdot\text{mol}^{-1}$
Given: $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{Cl}_2) = 242\text{ kJ}\cdot\text{mol}^{-1}$
Show Hint
Always pay close attention to whether the atomization value given is per mole of molecules (like $\text{Cl}_2 \rightarrow 2\text{Cl}$) or per mole of atoms formed.
Dividing by 2 is essential for $\text{Cl}_2$ to find the enthalpy of single Cl atoms.
Dividing by 2 is essential for $\text{Cl}_2$ to find the enthalpy of single Cl atoms.
Updated On: Jul 22, 2026
- $+272.5$
- $-30.5$
- $-272.5$
- $+30.5$
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to find the standard enthalpy of vaporization ($\Delta_{\text{vap}} H^\ominus$) of liquid carbon tetrachloride ($\text{CCl}_4$).
This represents the enthalpy change for the process:
\[ \text{CCl}_4(l) \rightarrow \text{CCl}_4(g) \]
Step 2: Key Formula or Approach:
By definition, the enthalpy of vaporization is:
\[ \Delta_{\text{vap}} H^\ominus = \Delta_f H^\ominus(\text{CCl}_4(g)) - \Delta_f H^\ominus(\text{CCl}_4(l)) \] We are given $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$.
We need to determine the standard enthalpy of formation of gaseous $\text{CCl}_4$, i.e., $\Delta_f H^\ominus(\text{CCl}_4(g))$, from the other given reactions.
Step 3: Detailed Explanation:
• Let us write the formation reaction of $\text{CCl}_4(g)$ from its elements:
\[ \text{C}(graphite) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \]
• The given atomization reaction of $\text{CCl}_4(g)$ is:
\[ \text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g) \quad \Delta H = 1304\text{ kJ}\cdot\text{mol}^{-1} \] Therefore, the enthalpy of formation of $\text{CCl}_4(g)$ is related to the atomization of elements as:
\[ \Delta_{\text{atom}} H^\ominus(\text{CCl}_4(g)) = \Delta_f H^\ominus(\text{C}(g)) + 4\Delta_f H^\ominus(\text{Cl}(g)) - \Delta_f H^\ominus(\text{CCl}_4(g)) \]
• From the given data:
$\Delta_f H^\ominus(\text{C}(g)) = \Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$
The enthalpy of atomization of $\text{Cl}_2$ is $242\text{ kJ}\cdot\text{mol}^{-1}$ for $\text{Cl}_2(g) \rightarrow 2\text{Cl}(g)$.
Therefore, the enthalpy of formation of chlorine atoms is:
\[ \Delta_f H^\ominus(\text{Cl}(g)) = \frac{1}{2} \Delta_a H^\ominus(\text{Cl}_2) = \frac{242}{2} = 121\text{ kJ}\cdot\text{mol}^{-1} \]
• Substitute these values into the atomization equation of $\text{CCl}_4(g)$:
\[ 1304 = 715 + 4(121) - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ 1304 = 715 + 484 - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ 1304 = 1199 - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ \Delta_f H^\ominus(\text{CCl}_4(g)) = 1199 - 1304 = -105\text{ kJ}\cdot\text{mol}^{-1} \]
• Now, we calculate the enthalpy of vaporization:
\[ \Delta_{\text{vap}} H^\ominus = \Delta_f H^\ominus(\text{CCl}_4(g)) - \Delta_f H^\ominus(\text{CCl}_4(l)) \] \[ \Delta_{\text{vap}} H^\ominus = -105 - (-135.5) \] \[ \Delta_{\text{vap}} H^\ominus = -105 + 135.5 = +30.5\text{ kJ}\cdot\text{mol}^{-1} \]
Step 4: Final Answer:
The standard enthalpy of vaporization of $\text{CCl}_4(l)$ is $+30.5\text{ kJ}\cdot\text{mol}^{-1}$.
The question asks us to find the standard enthalpy of vaporization ($\Delta_{\text{vap}} H^\ominus$) of liquid carbon tetrachloride ($\text{CCl}_4$).
This represents the enthalpy change for the process:
\[ \text{CCl}_4(l) \rightarrow \text{CCl}_4(g) \]
Step 2: Key Formula or Approach:
By definition, the enthalpy of vaporization is:
\[ \Delta_{\text{vap}} H^\ominus = \Delta_f H^\ominus(\text{CCl}_4(g)) - \Delta_f H^\ominus(\text{CCl}_4(l)) \] We are given $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$.
We need to determine the standard enthalpy of formation of gaseous $\text{CCl}_4$, i.e., $\Delta_f H^\ominus(\text{CCl}_4(g))$, from the other given reactions.
Step 3: Detailed Explanation:
• Let us write the formation reaction of $\text{CCl}_4(g)$ from its elements:
\[ \text{C}(graphite) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \]
• The given atomization reaction of $\text{CCl}_4(g)$ is:
\[ \text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g) \quad \Delta H = 1304\text{ kJ}\cdot\text{mol}^{-1} \] Therefore, the enthalpy of formation of $\text{CCl}_4(g)$ is related to the atomization of elements as:
\[ \Delta_{\text{atom}} H^\ominus(\text{CCl}_4(g)) = \Delta_f H^\ominus(\text{C}(g)) + 4\Delta_f H^\ominus(\text{Cl}(g)) - \Delta_f H^\ominus(\text{CCl}_4(g)) \]
• From the given data:
$\Delta_f H^\ominus(\text{C}(g)) = \Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$
The enthalpy of atomization of $\text{Cl}_2$ is $242\text{ kJ}\cdot\text{mol}^{-1}$ for $\text{Cl}_2(g) \rightarrow 2\text{Cl}(g)$.
Therefore, the enthalpy of formation of chlorine atoms is:
\[ \Delta_f H^\ominus(\text{Cl}(g)) = \frac{1}{2} \Delta_a H^\ominus(\text{Cl}_2) = \frac{242}{2} = 121\text{ kJ}\cdot\text{mol}^{-1} \]
• Substitute these values into the atomization equation of $\text{CCl}_4(g)$:
\[ 1304 = 715 + 4(121) - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ 1304 = 715 + 484 - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ 1304 = 1199 - \Delta_f H^\ominus(\text{CCl}_4(g)) \] \[ \Delta_f H^\ominus(\text{CCl}_4(g)) = 1199 - 1304 = -105\text{ kJ}\cdot\text{mol}^{-1} \]
• Now, we calculate the enthalpy of vaporization:
\[ \Delta_{\text{vap}} H^\ominus = \Delta_f H^\ominus(\text{CCl}_4(g)) - \Delta_f H^\ominus(\text{CCl}_4(l)) \] \[ \Delta_{\text{vap}} H^\ominus = -105 - (-135.5) \] \[ \Delta_{\text{vap}} H^\ominus = -105 + 135.5 = +30.5\text{ kJ}\cdot\text{mol}^{-1} \]
Step 4: Final Answer:
The standard enthalpy of vaporization of $\text{CCl}_4(l)$ is $+30.5\text{ kJ}\cdot\text{mol}^{-1}$.
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