Question

The bond enthalpies of H_2(g),Br_2(g) and HBr_(g) are 436 kJ mol^-1,193 kJ mol^-1 and 366 kJ mol^-1 respectively. 

Calculate the enthalpy change for the following reaction :

H_2(g)+ Br_2(g) → 2HBr_(g).

Answer 1

Given : Bond enthalpy of H_2(g) = ΔH⁰H_2(g)

= 436 kJ mol^-1

Bond enthalpy of Br_2(g) = ΔH⁰Br_2(g) = 193 kJ mol^-1

Bond enthalpy of HBr_(g) =  ΔH⁰Br_(g) = 366 kJ mol^-1

Given reaction,

H_2(g) + Br_2(g) → 2HBr_(g)

OR

H-H_(g) + Br-Br_(g) → 2H-Br_(g)

The enthalpy change of the reaction is,

= [436 + 193] – 2[366] 

= 629 – 732 

= -103 kJ

∴ Enthalpy change for the reaction = Δ_rH⁰ 

= -103 kJ