Answer:
a) Heat of combustion of 1 g of methanol = -22.6 kJ = (-2.26 × 10) kJ
b) Heat of combustion of 1 g of ethanol = -29.7 kJ = (-2.97 × 10) kJ
c) Heat of combustion of 1 g of propanol = -33.5 kJ = (-3.35 × 10) kJ
Explanation:
a) The equation for the combustion of methanol is given as
CH₃OH + (3/2)O₂ → CO₂ + 2H₂O
The standard heat of combustion of methanol is given as -726 kJ/mol from literature.
But, 1 g of methanol will have the heat of combustion of the number of moles of methanol contained in 1 g of methanol.
Number of moles = (mass)/(molar mass)
Molar mass of (CH₃OH) = 32.04 g/mol
Number of moles = (1/32.04) = 0.03121 moles
1 mole of methanol has a heat of combustion of -726 kJ
0.03121 mole of methanol will have a heat of combustion of (0.03121 × -726) = -22.6 kJ
b) The equation for the combustion of ethanol is given as
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
The standard heat of combustion of ethanol is given as -1367.6 kJ/mol from literature.
But, 1 g of ethanol will have the heat of combustion of the number of moles of ethanol contained in 1 g of ethanol.
Number of moles = (mass)/(molar mass)
Molar mass of (C₂H₅OH) = 46.07 g/mol
Number of moles = (1/46.07) = 0.0217 moles
1 mole of ethanol has a heat of combustion of -1367.6 kJ
0.0217 mole of ethanol will have a heat of combustion of (0.03121 × -1367.6) = -29.7 kJ
c) The equation for the combustion of propanol is given as
C₃H₇OH + (9/2)O₂ → 3CO₂ + 4H₂O
The standard heat of combustion of propanol is given as -2020 kJ/mol from literature.
But, 1 g of propanol will have the heat of combustion of the number of moles of propanol contained in 1 g of propanol.
Number of moles = (mass)/(molar mass)
Molar mass of (C₃H₇OH) = 60.09 g/mol
Number of moles = (1/60.09) = 0.0166 moles
1 mole of propanol has a heat of combustion of -2020 kJ
0.0166 mole of propanol will have a heat of combustion of (0.0166 × -2020) = -33.5 kJ
Hope this Helps!!!
