Liquid methanol (CH
3
OH) is used as an alternative fuel in truck engines. When a 0.150 g sample
of methanol was burned in a calorimeter, open to
the atmosphere, the temperature increased 1.52 ºC.
The heat capacity of the calorimeter was 1.97 kJ
/ ºC. The products of the reaction were carbon
dioxide,
'
Hº
f
(CO
2
) = -393.5 kJ/mol and water vapour,
'
Hº
f
(H
2
O(g)) = -241.8 kJ/mol.
Calculate the enthalpy of formation,
'
Hº
f
, for CH
3
OH(l).
To calculate the enthalpy of formation, ΔHf, for CH3OH(l), we need to use the equation:
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
where ΔHrxn is the enthalpy change of the reaction.
In this case, the reaction is the combustion of methanol (CH3OH), and the reactants are CH3OH and O2, while the products are CO2 and H2O(g).
The enthalpy change of the reaction, ΔHrxn, can be calculated using the equation:
ΔHrxn = q / n
where q is the heat absorbed by the reaction and n is the number of moles of the substance undergoing the reaction.
First, we need to calculate the heat absorbed by the reaction, q. This can be done using the equation:
q = C∆T
where C is the heat capacity of the calorimeter and ∆T is the change in temperature.
Given:
- C = 1.97 kJ/ºC (heat capacity of the calorimeter)
- ∆T = 1.52 ºC (change in temperature)
q = C∆T = (1.97 kJ/ºC) * (1.52 ºC)
q = 2.99 kJ
Next, we need to calculate the number of moles of methanol burned in the reaction. To do this, we can use the molar mass of methanol (CH3OH):
Molar mass of CH3OH = (1*12.01 g/mol) + (4*1.01 g/mol) + (1*16.00 g/mol)
Molar mass of CH3OH = 32.04 g/mol
Given:
- Mass of methanol = 0.150 g
Number of moles of CH3OH = mass / molar mass
Number of moles of CH3OH = 0.150 g / 32.04 g/mol
Number of moles of CH3OH = 0.00468 mol
Now, we can calculate the enthalpy change of the reaction, ΔHrxn:
ΔHrxn = q / n
ΔHrxn = 2.99 kJ / 0.00468 mol
ΔHrxn = 640 kJ
Finally, we can calculate the enthalpy of formation, ΔHf, for CH3OH(l):
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
Given:
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(H2O(g)) = -241.8 kJ/mol
ΔHrxn = ΔHf(CO2) + ΔHf(H2O(g)) - ΔHf(CH3OH(l))
640 kJ = (-393.5 kJ/mol) + (-241.8 kJ/mol) - ΔHf(CH3OH(l))
ΔHf(CH3OH(l)) = (-393.5 kJ/mol) + (-241.8 kJ/mol) - 640 kJ
ΔHf(CH3OH(l)) = -875.3 kJ/mol
Therefore, the enthalpy of formation, ΔHf, for CH3OH(l) is -875.3 kJ/mol.