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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(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
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(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
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(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
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(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
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(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
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