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: $\Delta_f H^\ominus(\text{CCl}_4(l)) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{C}) = 715\text{ kJ}\cdot\text{mol}^{-1}$, $\Delta_a H^\ominus(\text{Cl}_2) = 242\text{ kJ}\cdot\text{mol}^{-1}$

• A. $+272.5$
• B. $-30.5$
• C. $-272.5$
• D. $+30.5$

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.

Answer 1

The Correct Option is D

Step 1: Build a Hess cycle for gaseous $\text{CCl}_4$.
Forming $\text{CCl}_4(g)$ from elements can be routed through atoms: $\text{C}(s) \to \text{C}(g)$ costs $\Delta_aH = 715$, $2\,\text{Cl}_2(g) \to 4\,\text{Cl}(g)$ costs $2\times 242 = 484$, and then $\text{C}(g)+4\,\text{Cl}(g)\to \text{CCl}_4(g)$ releases the reverse of the given 1304 kJ atomisation step.
Step 2: Add the cycle to get $\Delta_fH^\ominus(\text{CCl}_4, g)$.
\[ \Delta_fH^\ominus(\text{CCl}_4,g) = 715 + 484 - 1304 = -105 \text{ kJ/mol} \]
Step 3: Use the definition of vaporisation enthalpy.
\[ \Delta_{\text{vap}}H^\ominus = \Delta_fH^\ominus(\text{CCl}_4,g) - \Delta_fH^\ominus(\text{CCl}_4,l) = -105 - (-135.5) \]
\[ \boxed{\Delta_{\text{vap}}H^\ominus = +30.5 \text{ kJ/mol}} \]