Question 15

Calculate the enthalpy change for the process

CCl_{4}(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl_{4}(g).

Δ_{vap}H^{0}(CCl_{4}) = 30.5 kJ mol^{–1}.

Δ_{f}H^{0} (CCl_{4}) = –135.5 kJ mol^{–1}.

Δ_{a}H^{0} (C) = 715.0 kJ mol^{–1} , where Δ_{a}H^{0} is enthalpy of atomisation

Δ_{a}H^{0} (Cl_{2}) = 242 kJ mol^{–1}

Answer

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

(i) CCl_{4(l)} → CCL_{4(g)} Δ_{vap}H^{θ} = 30.5 kJ mol^{-1}

(ii) C_{(s) } → C_{(g)} Δ_{a}H^{θ} = 715.0 kJ mol^{-1}

(iii) Cl_{2(g)} → 2Cl_{(g) } Δ_{a}H^{θ} = 242 kJ mol^{-1}

(iv) C_{(g) } + 4Cl_{(g)} → CCl_{4(g)} Δ_{f}H = -135.5 kJ mol^{-1}

Enthalpy change for the given process C_{(g) } + 4Cl_{(g)} → CCl_{4(g)} can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)

ΔH = Δ_{a}H^{θ}(C) + 2Δ_{a}H^{θ} (Cl_{2}) - Δ_{vap}H^{θ} - Δ_{f}H

= (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 CCl_{4(g)} = 326 kJ mol^{-1}

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Elements
Δ
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(d) the least reactive non-metal.

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- NCERT Chapter

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