Calculate the ∆H°rxn for the combustion of methane using the given ∆H°f.
CH4 (g) + 2 O2 (g) → CO2 (g) + 2H2O (g)
∆H°f, methane(g) = -74.60 kJ/mol
∆H°f, water(l) = -285.8 kJ/mol
∆H°f, carbon dioxide(g) = -393.5 kJ/mol
dHrxn = (n*dHo products) - (n*dHo reactants)
Substitute the numbers and calculate. Post your work if you get stuck.
To calculate the ∆H°rxn for the combustion of methane, we need to use the Hess's law and the standard enthalpies of formation (∆H°f) of the reactants and products.
The combustion reaction equation that you provided is:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2H2O (g)
First, let's determine the overall reaction by comparing the reactants and products. We have 1 mole of methane (CH4) reacting with 2 moles of oxygen gas (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).
The standard enthalpy change (∆H°rxn) for the reaction is given by the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
∆H°rxn = ∑∆H°f(products) - ∑∆H°f(reactants)
Now, let's calculate the ∆H°rxn:
∆H°rxn = [∆H°f(CO2) + 2 × ∆H°f(H2O)] - [∆H°f(CH4) + 2 × ∆H°f(O2)]
∆H°rxn = [(-393.5 kJ/mol) + 2 × (-285.8 kJ/mol)] - [(-74.60 kJ/mol) + 2 × 0 kJ/mol]
∆H°rxn = [-393.5 kJ/mol - 571.6 kJ/mol] - [-74.6 kJ/mol]
∆H°rxn = -965.1 kJ/mol + 74.6 kJ/mol
∆H°rxn = -890.5 kJ/mol
Therefore, the ∆H°rxn for the combustion of methane is -890.5 kJ/mol.