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



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