6.15 Calculate the enthalpy change for the processCCl4(g) → C(g) + 4 Cl(g)and calculate bond enthalpy of C ? Cl in CCl4(g).ΔvapH0(CCl4) = 30.5 kJ \\mol?.ΔfH0 (CCl4) = ?135.5 kJ \\mol?.ΔaH0 (C) = 715.0 kJ \\mol? , where ΔaH0 is enthalpy of atomisationΔaH0 (Cl2) = 242 kJ \\mol?

Dear Student,

 vapHθ = 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)H = 1304 kJ mol–1
Bond enthalpy of C–Cl bond in CCl4 (g)
= 326 kJ mol–1

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