Question

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

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

So, 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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