Question

Calculate Δ_rH⁰ of the reaction CH_4(g) + O_2(g) → CH₂O_(g) + H₂O_(g) 

From the following data :

Bond	C - H	O = O	C = O	O - H
ΔH0/kJ mol-1	414	499	745	464

Answer 1

Given:

 

Bond	C - H	O = O	C = O	O - H
ΔH0/kJ mol-1	414	499	745	464

Standard enthalpy change for the reaction = Δ_rH⁰ = ?

 

 

= [2ΔH⁰_{C - H} + ΔH⁰_{o = o}] – [ΔH⁰_{C = 0} + 2ΔH⁰_o-H]

= [2 × 414 + 499] – [745 + 2 × 464]

= [828 + 499] – [745 + 928] 

= -346 kJ

∴ Standard enthalpy change for the reaction = Δ_rH⁰ = -346 kJ