Question

From the following heats of combustion,

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

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

H2(g) + ½O2(g) → H2O(l) ΔHorxn = –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

I got this from a previous session but they didn't explain where the Carbon wen to in the one section.

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

Now here is the question I have what happened to the graphite in multiplying equations 3 and 2.