Question

Methanol (CH3OH) has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid methanol.

Standard Heats of Formation:

CH3OH(l) = –239 kJ/mol

O2(g) = 0 kJ/mol

CO2(g) = –393.5 kJ/mol

H2O(l) = –286 kJ/mol

ΔH =_______ kJ/g CH3OH

Answer 1

Methanol on combustion (on reaction with O₂) it forms Carbon di-oxide and Water.

The Complete equation is:

2CH₃OH (l) + 3O₂ (g)  = 2CO₂ (g) + 4H₂O (g)

H_combustion = H_products -  Hreactants

Now, H_products = [2*(-393.5) + 4*(-286)] kJ/mol = -1931 kJ/mol

Hreactants= 2*(-239)  kJ/mol = - 478  kJ/mol

So,

H_combustion = H_products -  Hreactants = (-1931 kJ/mol) - ( - 478  kJ/mol) = -1453 kJ/mol

Thus, the standard enthalpy of combustion 2 mole (i.e; 64g) of liquid methanol = -1453 kJ

Then, the standard enthalpy of combustion per gram of liquid methanol= -1453/64 kJ = -22.70kJ