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# Calculate the Enthalpy Change for the Process and Calculate Bond Enthalpy of C–Cl in Ccl4(G) Where δAhθ Is Enthalpy of Atomisation - CBSE (Science) Class 11 - Chemistry

ConceptEnthalpies for Different Types of Reactions Bond Enthalpy

#### Question

Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g)

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

ΔvapHθ (CCl4) = 30.5 kJ mol–1.

ΔfHθ (CCl4) = –135.5 kJ mol–1.

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

ΔaHθ (Cl2) = 242 kJ mol–1

#### Solution

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

1)"CCl"_(4(l)) -> "CCl"_(4(g))  ΔvapHθ = 30.5 kJ mol–1

2) C_((s)) -> C_(g) aHθ = 715.0 kJ mol–1

3) Cl_(2(g))  -> 2Cl_(g) ΔaHθ = 242 kJ mol–1

4) C_(g) +  4Cl_(g) -> "CCl"_(4(g)) ΔfH = –135.5 kJ mol–1

Enthalpy change for the given process "CCl"_(4(g)) ->  C_(g) + 4Cl_(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)

= 1304/4 kJ mol^(-1)

= 326 kJ mol–1

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Solution Calculate the Enthalpy Change for the Process and Calculate Bond Enthalpy of C–Cl in Ccl4(G) Where δAhθ Is Enthalpy of Atomisation Concept: Enthalpies for Different Types of Reactions - Bond Enthalpy.
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