When NH3 is treated with oxygen gas, the products obtained are N2(g) and H2O(l). If standard enthalpies of formation at 298 K for NH3(g) and H2O(l) are –46.00 kJ/mol and –286.0 kJ/mol respectively, calculate the enthalpy change of the reaction.
To calculate the enthalpy change of the reaction, we can use the concept of Hess's law.
The balanced chemical equation for the reaction is:
4NH3(g) + 5O2(g) -> 4N2(g) + 6H2O(l)
According to Hess's law, the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of individual reactions that lead to the same overall reaction. We can break down the given reaction into two steps:
Step 1: Formation of N2(g) from nitrogen atoms:
N2(g) -> 2N(g) ΔH1
Step 2: Formation of H2O(l) from hydrogen and oxygen atoms:
2H2(g) + O2(g) -> 2H2O(l) ΔH2
Since we know the standard enthalpies of formation of NH3(g) and H2O(l), we can use them to calculate the enthalpy change for each step.
Step 1: We need to double the enthalpy change of formation for N2 since the balanced equation has four moles of N2. Therefore, the enthalpy change for Step 1 is:
2 × ΔHf(N2) = 2 × 0 kJ/mol = 0 kJ/mol
Step 2: We need to use the coefficients from the balanced equation to determine the molar enthalpy change for Step 2. The molar enthalpy change for Step 2 is:
2 × ΔHf(H2O) - [2 × ΔHf(H2) + ΔHf(O2)] = 2 × (-286.0 kJ/mol) - [2 × 0 kJ/mol + 0 kJ/mol] = -572.0 kJ/mol
Now, we can calculate the overall enthalpy change by summing up the enthalpy changes for both steps:
ΔH = ΔH1 + ΔH2 = 0 kJ/mol + (-572.0 kJ/mol) = -572.0 kJ/mol
Therefore, the enthalpy change of the reaction is -572.0 kJ/mol.