Question

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–1C(g) + O2(g) → CO2(g) ;    ΔcH0 = –393 kJ mol–1H2(g) + 1/2 O2(g) → H2O(l) ;    ΔfH0 = –286 kJ mol–1

Answer 1

Hey dear,

● 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

Hope this is helpful...

Answer 2

Hey dear,

● 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

Hope this is helpful...