Question

Calculate the standard enthalpy of formation of CH₃OH (l) from the following data:

(i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l);

Δ_rH° = -726 kJ mol^-1

(ii) C(g) + O₂(g) → CO₂(g);

Δ_cH° = -393 kJ mol^-1

(iii) H₂(g) + 1/2O₂(g) → H₂O(l);

Δ_fH° = -286 kJ mol^-1

Answer 1

(i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l);

Δ_rH° = -726 kJ mol^-1

(ii) C(g) + O₂(g) → CO₂(g);

Δ_cH° = -393 kJ mol^-1

(iii) H₂(g) + 1/2O₂(g) → H₂O(l);

Δ_fH° = -286 kJ mol^-1

We aim at

C + 2H₂ + 1/2 O₂ → CH₃OH

In order to get this thermochemical equation, multiply Eq. (ii) by 1 and Eq. (iii) by 2 and subtract Eq. (i) from their sum.

C + 2H₂ + 1/2O₂ → CH₃OH

ΔH = (-393) + 2(-286) - (-726)

= -393 - 572 + 726

= -965 + 726 = -239 kJ mol^-1

Thus, the heat of formation of CH₃OH is

ΔH_f = -239 kJ mol^-1