Calculate the heat of reaction (ΔHrxn) for the following reactions, using heats of formation (ΔHf). Show your work. Refer to a table of heats of formation in your textbook as needed.
5O2(g) + 4NH3(g) → 6H2O(g) + 4NO(g)
dHrxn = (n*dHf products) - (n*dHf reactants)
To find the heat of reaction (ΔHrxn) for the given reaction using the heats of formation (ΔHf), you need to follow these steps:
1. Write down the balanced chemical equation for the reaction:
5O2(g) + 4NH3(g) → 6H2O(g) + 4NO(g)
2. Look up the heats of formation (ΔHf) values for each compound involved in the reaction. The heats of formation are typically given in kilojoules per mole (kJ/mol). You can find a table of heats of formation in your textbook or online.
ΔHf(O2) = 0 kJ/mol (oxygen is in its standard state)
ΔHf(NH3) = -46.2 kJ/mol (given)
ΔHf(H2O) = -241.8 kJ/mol (given)
ΔHf(NO) = +90.4 kJ/mol (given)
3. Calculate the ΔHrxn by using the formula:
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
ΣΔHf(products) = (6 mol H2O × ΔHf(H2O)) + (4 mol NO × ΔHf(NO))
ΣΔHf(reactants) = (5 mol O2 × ΔHf(O2)) + (4 mol NH3 × ΔHf(NH3))
4. Plug in the values and calculate:
ΣΔHf(products) = (6 mol H2O × -241.8 kJ/mol) + (4 mol NO × 90.4 kJ/mol)
= -1450.8 kJ + 361.6 kJ
= -1089.2 kJ
ΣΔHf(reactants) = (5 mol O2 × 0 kJ/mol) + (4 mol NH3 × -46.2 kJ/mol)
= 0 kJ + (-184.8 kJ)
= -184.8 kJ
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
= -1089.2 kJ - (-184.8 kJ)
= -1089.2 kJ + 184.8 kJ
= -904.4 kJ
Therefore, the heat of reaction (ΔHrxn) for the given reaction is -904.4 kJ.