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InterviewSolution
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Calculate the enthalpy change for the process `C Cl_(4)(g) rarrC(g) +4Cl(g)` and calculate bond enthalpyof`C-Cl` in `C Cl_(4)(g)` Given `: V_(vap) H^(@) (C C l_(4)) =30.5 kJ mol^(-1), Delta_(f)H^(@) (C C l_94))= -135.5 kJ mol^(-1)` `Delta_(a)H^(@) ( C ) = 715.0 kJ mol^(-1) ` where`Delta_(a)H^(@)`is enthalpy of atomisation`Delta_(a) H^(@)( Cl_(2)) = 242 kJ mol^(-1)` |
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Answer» The given data imply as under `: (i)C Cl_(4) (l) rarr C C l_(4)(g) , DeltaH= 30.5 kJ mol^(-1)` (ii)`C(s) + 2Cl_(2)(g) rarrC C l_(4)(l) , DeltaH =-135.5 kJ mol^(-1)` (iii) `C(s)rarr C(g) , DeltaH= 715 .0 kJ mol^(-1)` (iv) `Cl_(2)(g) rarr2Cl(g), DeltaH = 242 kJ mol^(-1)` Aim `:C C l_(4)(g) rarr C(g) + 4Cl(g) , DeltaH = ?` Eqn. (iii)`+2 xx` Eqn. (iv) - Eqn. (i). Eqn. (ii) gives the required equation with `DeltaH=715.0 + 2 ( 242)-30.5 - ( - 135.5) kJ mol^(-1) = 1304 kJ mol^(-1)` Bond enthalpy ofC-Cl in `C C l_(4)` ( average value)`= ( 1304)/( 4) = 326 kJ mol^(-1)` |
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