Question

Calculate C-Cl bond enthalpy from the following data :

CH₃Cl_(g) + Cl_2(g) → CH₂Cl_2(g) + HCl_(g) ΔH⁰ = – 104 kJ.

Bond	C–H	Cl–Cl	H–Cl
ΔH0/KJ mol-1	414	243	431

(330 kJ mol^-1).

Answer 1

Given :

Bond	C-H	Cl-Cl	H-Cl
ΔH0/KJ mol-1	414	243	431

 

For the given reaction,

 ΔH⁰_r = -104 kJ

Bond enthalpy of C-Cl = ΔH⁰C–Cl] = ?

In this reaction, 1 C–H, 1 Cl–Cl bonds of the reactants are broken while 1C–Cl and 1H–Cl bonds of the products are formed.

Sum of bond enthalpies of bonds formed of the products

∴ Bond enthalpy of C–Cl = ΔH⁰C–Cl 

= 330 kJ mol^-1