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. Calculate enthalpy of formation of methane gas.

Answer 1

Given that

(i) CH₄ (g) + 2O₂ (g) → CO₂ (g) + CH₂O(g);∆_CH^Θ = −890.3 kJ mol⁻¹

(ii) C(s) + O₂ (g) → CO₂ (g);∆_CH^Θ = −393.5 kJ mol⁻¹

(iii) H₂ (g) + \(\frac{1}{2}\)O₂ (g) → H₂O(g); ∆_CH^Θ = −285.8 kJ mol⁻¹

We have to calculate ∆_fH^Θ of the equation:

C(s) + 2H₂ (g) → CH₄ (g); ∆_fH^Θ = ?

Appling inspection method,

−eq. (i) + eq. (ii) + 2 × eq. (iii)

−CH₄(g) − 2O₂(g) + C(s) + O₂(g) + 2H₂(g) + O₂ (g) → −CO₂ (g) − 2H₂O(g) + CO₂(g)

Or C(s) + 2H₂(g)

\(\therefore\) ∆_fH^Θ = −(−890.3) + (−393.5) + 2 × (−285.8)

= 890.3 − 393.5 − 571.6

= 890.3 − 965.1

= −75.8 kJ mol⁻¹