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Question 15

Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl4(g).

ΔvapH0(CCl4) = 30.5 kJ mol–1.

ΔfH0 (CCl4) = –135.5 kJ mol–1.

ΔaH0 (C) = 715.0 kJ mol–1 , where ΔaH0 is enthalpy of atomisation

ΔaH0 (Cl2) = 242 kJ mol–1

Answer

The chemical equations implying to the given values of enthalpies are:

(i) CCl4(l)    →    CCL4(g)      ΔvapHθ = 30.5 kJ mol-1

(ii) C(s)    →   C(g)               ΔaHθ = 715.0 kJ mol-1

(iii) Cl2(g)  →   2Cl(g)          ΔaHθ = 242 kJ mol-1

(iv) C(g)  + 4Cl(g)  →  CCl4(g)  ΔfH = -135.5 kJ mol-1

Enthalpy change for the given process  C(g)  + 4Cl(g)  →  CCl4(g)   can be calculated using the following algebraic calculations as:

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

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