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If for the reaction $\mathrm{CCl}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$ the following data is given
$\begin{aligned} & \Delta_{\mathrm{vap}} \mathrm{H}^\theta \mathrm{CCl}_4(\mathrm{l})=30 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {, vap = vaporization } \\ & \Delta_{\mathrm{f}} \mathrm{H}^\theta{ }_{\mathrm{CCl}_4}=-136.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{f}=\text { formation } \\ & \Delta_{\mathrm{a}} \mathrm{H}_{\mathrm{C}}^{\mathrm{q}}=714.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{a}=\text { atomization } \\ & \Delta_{\mathrm{a}} \mathrm{H}^\theta \mathrm{Cl}_2=242.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{a}=\text { atomization }\end{aligned}$
the bond mean enthalpy of $\mathrm{C}-\mathrm{Cl}$ in $\mathrm{CCl}_4(\mathrm{l})$ is
ChemistryThermodynamics (C)AP EAMCETAP EAMCET 2023 (17 May Shift 2)
Options:
  • A -319
  • B 326
  • C -326
  • D 292
Solution:
1187 Upvotes Verified Answer
The correct answer is: 326
$\mathrm{CCl}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$
or
$\begin{aligned} & \mathrm{CCl}_4(\mathrm{~g}) \mathrm{C}(\mathrm{g})+2 \mathrm{Cl}_2(\mathrm{~g}) \\ & \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\Delta_{\mathrm{a}} \mathrm{H}^{\circ}(\mathrm{c})+2 \Delta_{\mathrm{a}} \mathrm{H}^{\circ}\left(\mathrm{Cl}_2\right) \\ & -\Delta_{\text {vap }} \mathrm{H}^{\circ}-\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CCl}_4\right) \\ & =(714)+[2 \times 242.0]-(30.0)-(-136.0) \\ & =1304 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Rightarrow \text { Average bond enthalpy of } \mathrm{C}-\mathrm{Cl} \text { bond } \\ & =\frac{1304}{4}=326 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

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