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6.15 Calculate the enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ and calculate bond enthalpy of $C-Cl$ in $CCl_4(g)$
$\Delta _{vap} H ^{\ominus }(CCl_4) = 30.5 kJ mol^{-1}.\\\\ .\: \: \: \: \: \Delta _f H ^\ominus (CCl_4) = -135.5 kJ mol^{-1}.\\\\.\: \: \: \: \Delta _ a H ^ \ominus (C) = 715.0 kJ mol ^{-1} , where \: \: \Delta_ a H ^\ominus \: \: is\: \: enthalpy \: \: of \: \: atomisation\\\\.\: \: \: \: \Delta _ a H ^\ominus (Cl_2) = 242 kJ mol ^{-1}$

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We have the following chemical reactions equations-

$(1)\ CCl_{4}(l)\rightarrow CCl_{4}(g)$ ................ $\Delta _{vap}H^o = 30.5kJ/mol$
$(2)\ C(s)\rightarrow C(g)$ ........................ $\Delta _{a}H^o =715.0kJ/mol$
$(3)\ Cl2(g)\rightarrow 2Cl(g)$ .................. $\Delta _{a}H^o =242kJ/mol$
$(4)\ C(g)+4Cl\rightarrow CCl_4(g)$ ............ $\Delta _{f}H =-135.5kJ/mol$
The enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ by the above reaction is calculated as;

$\Delta H =\Delta _aH^o(C)+2(\Delta _aH^o(Cl_{2}))-\Delta _{vap}H^o-\Delta _fH$
$= [(715)+2(242)-(30.5)-(-135.5)] kJ/mol$
$=1304 kJmol^{-1}$

And the bond enthalpy of C-Cl bond in $CCl_4$ (g) = 1304/4 = 326 kJ/mol

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