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
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 15: Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond....
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