Question

ch3oh(l)+3/2o2 =co2+2h2o(l)​

Answer 1

Answer:

239 kJ/mol

Explanation:

Let's number our eqns-

CH3OH + 3/2 O2 → CO2 + 2H2O ...(1)

ΔH1 = –726 kJ/mol

C + O2 → CO2 ...(2)

ΔH2 = –393 kJ/mol

H2 + 1/2 O2 → H2O ...(3)

ΔH3 = –286 kJ/mol

Eqn(2) + 2×eqn(3) - eqn(1) :-

C + O2 + 2H2 + O2 + CO2 + 2H20→ CO2 + 2H20 + CH3OH + 3/2 O2

C + 1/2 O2 + 2H2 → CH3OH

Thus enthalpy of formation-

∆Hf(CH3OH) = ∆H2 + 2∆H3 - ∆H1

∆Hf(CH3OH) = -393 + 2(-286) + 726

∆Hf(CH3OH) = -239 kJ/mol

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