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Question

CH3OH(1)+32O2(g)CO2(g)+2H2O(I);Δ,H=726kJmol1
C(graphite)+O2(g)CO2(g);ΔcH=393kJmol1
H2(g)+12O2(g)H2O(I);ΔfH=286kJmol1
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

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Solution

The reaction takes place during the formation of CH3OH (1) can be written as
C(s)+2H2O(g)+12O2(g)CH3OH(l) ---------------- (1)
The reaction (1) can be obtained from the given reactions by following algebraic calculations.
equation(ii)+2×equation (iii) - equation(iii)

ΔfHθ[CH3OH(l)]=ΔcHθ+2ΔfHθ[H2O(l)]ΔrHθ

(393KJmol1)+2(286KJmol1)(726KJmol1)

=(393572+726)KJmol1

ΔfHθ[CH3OH(l)]=239KJmol1



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