Question

From the following heats of combustion,

CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l)        ΔH^o_rxn= –726.4 kJ/mol

C(graphite) + O₂(g) → CO₂(g)                               ΔH^o_rxn = –393.5 kJ/mol

H₂(g) + ½O₂(g) → H₂O(l)                                                  ΔH^o_rxn = –285.8 kJ/mol

Calculate the enthalpy of formation of methanol (CH₃OH) from its elements.

C(graphite) + 2H₂(g) + ½O₂(g) → CH₃OH(l)

Answer 1

Number equations 1,2,3

Now,

Invert EQN1

CO2(g) + 2H2O(l)   → CH3OH(l) + 3/2O2(g)     ΔHorxn = +726.4 kJ/mol

Multiply EQN3 by 2

2H2(g) + O2(g) → 2H2O(l) ΔHorxn = 2*(–285.8 kJ/mol) = -571.+6 kJ/mol

ADd equations

CO2(g) + 2H2O(l)   → CH3OH(l) + 3/2O2(g)     ΔHorxn = +726.4 kJ/mol

2H2(g) + O2(g) → 2H2O(l) ΔHorxn = 2*(–285.8 kJ/mol) = -571.+6 kJ/mol

CO2(g) + 2H2O(l) + 2H2(g) + O2(g) --> CH3OH(l) + 3/2O2(g) +   2H2O(l)   ΔHorxn = +154.8 kJ/mol

Cancel common terms

CO2(g) + + 2H2(g)    --> CH3OH(l) + 1/2O2(g)    ΔHorxn = +154.8 kJ/mol

Now, add EQN3

C(graphite) + O2(g) → CO2(g)                               ΔHorxn = –393.5 kJ/mol

add both equations

CO2(g) +  + 2H2(g)    --> CH3OH(l) + 1/2O2(g)    ΔHorxn = +154.8 kJ/mol

C(graphite) + O2(g) → CO2(g)                               ΔHorxn = –393.5 kJ/mol

CO2(g) +  + 2H2(g) + C(graphite) + O2(g) -->  CH3OH(l) + 1/2O2(g) +CO2(g)   ΔHorxn = -238.7 kJ/mol

cancel terms

2H2(g) + C(graphite) + 1/2O2(g) -->  CH3OH(l) ΔHorxn = -238.7 kJ/mol

NOTE that Hf of C = 0 so...

2H2(g) + 1/2O2(g)   -->  CH3OH(l)   ΔHorxn = -238.7 kJ/mol

Which is what we wanted :)