Question

The enthalpy of combustion of methane, graphite and dihydrogen at 298K are -890.3 kJ mol^-1, -393.5 kJ mol^-1 and 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄ will be …

(a) – 74.8 kJ mol^-1 

(b) – 52.27 k mol^-1 

(c) 74.8 kJ mol^-1

(d) + 52.26 kJ mol^-1

Answer 1

(a) – 74.8 kJ mol^-1 

CH₄ + 3O₂ → CO + 2H₂O ∆H₁ = – 890.3 kJ mol^-1 …. (1) 

C + O₂ → CO₂ = -393.5 kJ mol^-1  ….(2) 

H₅ + 1/2 O₂ → H₂O = – 285.8 kJ mol^-1 ……….(3) 

Equation (1) is reversed. 

Equation (3) x 2

∆H_f = +890.3 – 965.1 = -74.8 KJ mol^-1