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​

Answer 1

Answer:

The given chemical reactions are:

CCl

4

(l)→CCl

4

(g)Δ

vap

H

⊖

=30.5 kJ mol

−1

.

C(g)+2Cl

2

(g)→CCl

4

(g)Δ

f

H

⊖

=−135.5 kJ mol

−1

.

C(s)→C(g)Δ

a

H

⊖

=715.0 kJ mol

−1

.

Cl

2

(g)→2Cl(g)Δ

a

H

⊖

=242 kJ mol

−1

.

Now for calculating enthalpy change for the process CCl

4

(g)→C(g)+4Cl(g), we have to follow following algebric process

Equation(2)+2× Equation(3)−Equation(1)−Equation(4)

ΔH=Δ

a

H

⊖

+2Δ

a

H

⊖

Cl

2

−Δ

vap

H

⊖

−Δ

f

H

⊖

=715+2×242−30.5−(−135.5)

ΔH=1304 kJ mol

−1

.

Now,

The bond enthalpy of C−Cl in CCl

4

=

4

1304

=326 kJ mol

−1

.

Explanation:

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