The Total solution for NCERT class 6-12
Calculate the enthalpychange for the process
CCl4(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C–Cl in CCl4(g).
ΔvapHθ (CCl4) = 30.5 kJ mol–1.
ΔfHθ (CCl4) = –135.5 kJ mol–1.
ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy ofatomisation
ΔaHθ (Cl2) = 242 kJ mol–1
The chemical equationsimplying to the given values of enthalpies are:
ΔvapHθ = 30.5 kJ mol–1
ΔaHθ = 715.0 kJ mol–1
ΔaHθ = 242 kJ mol–1
ΔfH = –135.5 kJ mol–1
Equation (ii) + 2 ×Equation (iii) – Equation (i) – Equation (iv)
ΔH = ΔaHθ(C) + 2ΔaHθ (Cl2) – ΔvapHθ – ΔfH
= (715.0 kJ mol–1) + 2(242 kJ mol–1) – (30.5 kJ mol–1) – (–135.5 kJ mol–1)
ΔH = 1304 kJ mol–1
Bond enthalpy of C–Cl bond in CCl4(g)
=326 kJ mol–1