Question 14

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

CH₃OH (l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l) ; Δ_rH⁰= –726 kJ mol^–1

C(g) + O₂(g) → CO₂(g) ; Δ_cH⁰ = –393 kJ mol^–1

H₂(g) + 1/2 O₂(g) → H₂O(l) ; Δ_fH⁰ = –286 kJ mol^–1.

Answer

The reaction that takes place during the formation of CH₃OH(l) can be written as:

C_(s)+ 2H₂O_(g)+ ½O_2(g) →CH₃OH_(l) (1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) - equation (i)

Δ_fH^θ[CH₃OH_(l)] = Δ_cH^θ + 2Δ_fH^θ[H₂O_(l)] - Δ_rH^θ

= (-393 kJ mol^-1) + 2 (-286 kJ mol^-1) - (-726 kJ mol^-1)

= (-393 - 572 + 726) kJ mol^-1

$\therefore$ Δ_fH^θ[CH₃OH_(l)]

= -239 kJ mol^-1

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