The heats of combustion for methanol, ethanol, and n-propanol are -7.24 × 10 kJ/mol, -1.37 × 10 kJ/mol, and -2.01 × 10 kJ/mol respectively, calculated by multiplying the heat liberated per gram by their respective molecular weights.
To calculate the heats of combustion of methanol, ethanol, and n-propanol in kJ/mol, we need to first determine the molecular weights of these compounds. Once we have the molecular weights, we can multiply the heat released per gram by the molecular weight to find the heat of combustion per mole.
a) Methanol (CH3OH) has a molecular weight of 32.04 g/mol. So, the heat of combustion for methanol is -22.6 kJ/g × 32.04 g/mol = -724.104 kJ/mol, which is -7.24 × 10 kJ/mol when expressed in scientific notation.
b) Ethanol (C2H5OH) has a molecular weight of 46.07 g/mol. For ethanol, the heat of combustion is -29.7 kJ/g × 46.07 g/mol = -1367.079 kJ/mol, or -1.37 × 10 kJ/mol in scientific notation.
c) n-Propanol (C3H7OH) has a molecular weight of 60.10 g/mol. The heat of combustion for n-propanol is -33.4 kJ/g × 60.10 g/mol = -2007.34 kJ/mol, which translates to -2.01 × 10 kJ/mol in scientific notation.
Therefore, the heats of combustion for methanol, ethanol, and n-propanol are -7.24 × 10 kJ/mol, -1.37 × 10 kJ/mol, and -2.01 × 10 kJ/mol respectively.
Complete question is:
Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as indicated. Calculate the heats of combustion of these alcohols in kJ/mol.
(a) methanol (CH3OH), −22.6 kJ kJ/mol
(b) ethanol (C2H5OH), −29.7 kJ Enter your answer in scientific notation. × 10 kJ/mol
(c) n−propanol (C3H7OH), −33.4 kJ Enter your answer in scientific notation. × 10 kJ/mol