Calculate the ΔHrxn for the following reactions using the heats of formation given in the table at the end of this assignment:
a)The complete combustion of methane, CH4, producing water as a liquid.
b)The complete combustion of glucose, C6H12O6, producing water as a liquid.
c)The reaction of iron(III) oxide with carbon monoxide to produce solid iron and carbon dioxide.
d)The reaction of aluminum with Fe3O4
e)The combustion of carbon monoxide, producing carbon dioxide gas.
Write and balance the equation.
Then dHrxn = (n*dHf products) - (n*dHf reactants)
To calculate the ΔHrxn (change in enthalpy of reaction) for each of these reactions using the heats of formation, you need to follow these steps:
Step 1: Write the balanced chemical equation for the given reaction.
Step 2: Determine the heats of formation (ΔHf) for each compound involved in the reaction.
Step 3: Calculate the sum of the heats of formation for the products (ΔHf products).
Step 4: Calculate the sum of the heats of formation for the reactants (ΔHf reactants).
Step 5: Calculate the ΔHrxn by subtracting the sum of ΔHf reactants from the sum of ΔHf products.
Now, let's solve each of the given reactions:
a) The complete combustion of methane, CH4, producing water as a liquid.
Step 1: The balanced chemical equation is:
CH4 + 2O2 → CO2 + 2H2O
Step 2: The heats of formation for the compounds involved are as follows:
ΔHf(CH4) = -74.8 kJ/mol
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
Step 3: ΔHf products = 1(-393.5 kJ/mol) + 2(-285.8 kJ/mol) = -965.1 kJ/mol
Step 4: ΔHf reactants = 1(-74.8 kJ/mol) + 2(0 kJ/mol) = -74.8 kJ/mol
Step 5: ΔHrxn = ΔHf products - ΔHf reactants = -965.1 kJ/mol - (-74.8 kJ/mol) = -890.3 kJ/mol
Therefore, the ΔHrxn for the complete combustion of methane is -890.3 kJ/mol.
You can follow the same steps to solve the remaining reactions (b, c, d, and e) using the heats of formation provided in the table at the end of your assignment.