1. Tardigrade
  2. Question
  3. Chemistry
  4. he enthalpy change for the reaction, CCl4(g) → C(g)+4Cl(g) is Given: Δ textvapH°(CCl4)=30.5 KJ mol-1, ΔfH°(CCl4)=-135.5 KJ mol-1, ΔaH°(C)=715.0 kJ mol-1, ΔaH° (Cl2)=242 KJ mol-1 where ΔaH° is enthalpy of atomisation

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Q. he enthalpy change for the reaction, $CCl_{4(g)} \to C_{(g)}+4Cl_{(g)}$ is
Given:
$\Delta_{\text{vap}}H^{\circ}(CCl_{4})=30.5\,KJ\,mol^{-1}, \Delta_{f}H^{\circ}(CCl_{4})=-135.5\,KJ\,mol^{-1}$,
$\Delta_{a}H^{\circ}(C)=715.0\,kJ\,mol^{-1}, \Delta_{a}H^{\circ} (Cl_{2})=242\,KJ\,mol^{-1}$
where $\Delta_{a}H^{\circ}$ is enthalpy of atomisation

Thermodynamics

Solution:

Given :
(i) $CCl_{4\left(l\right)} \rightarrow CCl_{4\left(g\right)}, \Delta_{\text{vap}} H=30.5\,kJ \,mol^{-1}$
(ii) $C_{\left(s\right)}+2Cl_{2\left(g\right)} \rightarrow CCl_{4\left(l\right)}, \Delta_{f} H^{\circ}=-135.5 \,KJ \,mol^{-1}$
(iii) $C_{\left(s\right)}\rightarrow C_{\left(g\right)}, \Delta_{a}H^{\circ}=715.0\, KJ\,mol^{-1}$
(iv) $Cl_{2\left(g\right)}\rightarrow2Cl_{\left(g\right)}, \Delta_{a}H^{\circ}=242\,KJ \,mol^{-1}$
Required equation is : $CCl_{4(g)} \to C_{(g)}+4Cl_{(g)}; \Delta H=$ ?
From Hess’s law,
eqn.(iii) $+ 2 \times$ eqn.(iv) - eqn.(i) - eqn.(ii) gives required equation:
$\therefore \Delta H = 715.0+2(242)-30.5-(-135.5)$
$=1304\,kJ\,mol^{-1}$