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    				| 1. | Calculate the enthalpy change for the processCCl4(g) → C(g) + 4Cl(g)and calculate bond enthalpy of C-Cl in CCl4(g),ΔvapH° (CCl4) = 30.5 kJ mol-1ΔfH° (CCl4) = -135.5 kJ mol-1ΔaH°(c) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation. ΔaH°(Cl2) = 242 kJ mol-1 | 
| Answer» CCl4(g) → C(g) + 4Cl(g) Bond enthalpy of C-Cl bond = 1/4 x Heat energy required to break four C-Cl bonds in CCl4(g) = 1/4 x ΔH The ΔH for the reaction CCl4(g) → C(g) + 4Cl(g) is the enthalpy of atomisation of CCl4 ΔrH° = Σ bond enthalpies reactants - Σ bond enthalpies products = 30.5 - (715 + 4 x 242) kJ = 30.5 - (715 + 968) kJ = (30.5 - 1683) kJ = -1652.5 kJ | |