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Question

calculate sandard enthalpy of formation of CH3OH from:

CH3OH + 3/2O2 ---- CO2 + 2H2O delta r H = -726 Kj/Mol

C + O2 ---- CO2 dlta fH = -393.5 Kj/mol

H2 + 1/2O2 ----- H2O dlta fH = -286 Kj/Mol

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Solution

to calculate the standard heat of formation of methanol your equations should be arranged like this

CO2 + 2H2O → CH3OH + 3/2O2 =ΔHf, 726 Kj/Mol (just reversing the equation.)

C + O2→ CO2 ΔHf, =-393.5 Kj/mol

2H2 + O2 →2H2O = ΔHf,-286 x2 = -572 Kj/Mol(multiplying the equation by two)

Now by summing up all 3 equations we get

C + 1/2O2 + 2H2→CH3OH = ( 726-393.5-572 = -239)KJmol-1


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