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Question -

Calculate the standard enthalpy of formationof CH3OH(l) from the followingdata:

CH3OH(l) + O2(g)  CO2(g) + 2H2O(l) ; ΔrHθ = –726 kJ mol–1

C(g) + O2(g)  CO2(g) ; ΔcHθ = –393 kJ mol–1

H2(g) +O2(g)  H2O(l) ; ΔfHθ = –286 kJ mol–1.



Answer -

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

C(s) + 2H2O(g) + O2(g) CH3OH(l) (1)

The reaction (1) can be obtained from the given reactions by following thealgebraic 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

ΔfHθ [CH3OH(l)] = –239 kJ mol–1

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