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) ; ΔrH0 = –726 kJ mol–1
C(g) + O2(g) → CO2(g) ; ΔcH0 = –393 kJ mol–1
H2(g) + 1/2 O2(g) → H2O(l) ; ΔfH0 = –286 kJ mol–1.
The reaction that takes place during the formation of CH3OH(l) can be written as:
C(s) + 2H2O(g) + ½O2(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)
ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] - ΔrHθ
= (-393 kJ mol-1) + 2 (-286 kJ mol-1) - (-726 kJ mol-1)
= (-393 - 572 + 726) kJ mol-1
= -239 kJ mol-1
