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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2A + B → C
ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol-1
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(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
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(iii) +74.8 kJ mol–1
(iv) +52.26 kJ mol–1
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(ii) whose value is independent of path
(iii) used to determine pressure volume work
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(b) NaHSO4
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