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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics5 Marks
Question
Calculate the enthalpy change for the process$\text{CCI}_4(\text{g})\xrightarrow{ \ \ \ \ \ }\text{C}(\text{g})+4\text{CI}(\text{g})$
and calculate bond enthalpy of $C – Cl$ in $CCl_4$(g).$\Delta_\text{vap}\text{H}^\ominus(\text{CCI}_4)=30.5\text{kJ} \ \text{mol}^{-1}.$
$\Delta_\text{f}\text{H}^\ominus(\text{CCI}_4)=-135.5\text{kJ} \ \text{mol}^{-1}.$
$\Delta_\text{a}\text{H}^\ominus(\text{C})=-715.0\text{kJ} \ \text{mol}^{-1},$ where $\Delta_\text{a}\text{H}^\ominus$ is enthalpy of atomisation
$\Delta_\text{a}\text{H}^\ominus(\text{CI}_2)=242\text{kJ} \ \text{mol}^{-1}$
✓

Answer

The chemical equations implying to the given values of enthalpies are:
  1. $\text{CCl}_{4(\text{l})}\xrightarrow{ \ \ \ \ \ }\text{CCl}_{4(\text{g})}\Delta_\text{vap}\text{H}^\ominus=30.5\text{kJ} \ \text{mol}^{-1}$
  2. $\text{C}_{(\text{s})}\xrightarrow[]{\ \ \ \ \ \ \ }\text{C}_{(\text{g})} \ \Delta_{\text{a}}\text{H}^\ominus=715.0\text{kJ} \ \text{mol}^{-1}$
  3. $\text{Cl}_{2(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ }2\text{Cl}_{(\text{g})} \ \Delta_{\text{a}}\text{H}^\ominus=242\text{kJ} \ \text{mol}^{-1}$
  4. $\text{C}_{(\text{g})}+4\text{Cl}_{(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ }\text{CCl}_{4(\text{g})} \ \Delta_{\text{f}}\text{H}=-135.5\text{kJ} \ \text{mol}^{-1}$
Enthalpy change for the given process $\text{CCI}_{4(\text{g})}\xrightarrow{ \ \ \ \ \ }\text{C}_{(\text{g})}+4\text{Cl}_{(\text{g})}$, can be calculated
using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)
$\Delta\text{H}=\Delta_\text{a}\text{H}^\ominus(\text{C})+2\Delta_\text{a}\text{H}^\ominus(\text{Cl}_2)-\Delta_\text{vap}\text{H}^\ominus-\Delta_\text{f}\text{H} $
$= (715.0\text{kJ} \ \text{mol}^{–1}) + 2(242\text{kJ} \ \text{mol}^{–1}) – (30.5\text{kJ} \ \text{mol}^{–1}) – (–135.5\text{kJ} \ \text{mol}^{–1})$
$\therefore\Delta\text{H}=1304\text{kJ} \ \text{mol}^{-1}$
Bond enthalpy of $C–Cl$ bond in $CCl_{4(g)}$
$=\frac{1304}{4}\text{kJ} \ \text{mol}^{-1}$
$=326\text{kJ} \ \text{mol}^{-1}$

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