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
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)
Hcombustion = Hproducts - Hreactants
Now, Hproducts = [2*(-393.5) + 4*(-286)] kJ/mol = -1931 kJ/mol
Hreactants= 2*(-239) kJ/mol = - 478 kJ/mol
Hcombustion = Hproducts - 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
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