calculate delta H for N2O+4H2>N2+4H2O
You look these up in tables in your text.
delta Hrxn = [delta H products]-[delta H reactants]
To calculate the change in enthalpy (ΔH) for the reaction N2O + 4H2 → N2 + 4H2O, you need to know the enthalpy of formation (ΔHf) for each compound involved in the reaction. The ΔHf values represent the change in enthalpy that occurs when one mole of a compound is formed from its elements in their standard states at a given temperature and pressure.
1. Start by writing down the balanced chemical equation for the reaction: N2O + 4H2 → N2 + 4H2O
2. Look up the ΔHf values for each compound involved. You can find these values in reference books or online databases. As an example, here are the standard ΔHf values at 25°C (298 K):
ΔHf(N2O) = 82.05 kJ/mol
ΔHf(H2) = 0 kJ/mol
ΔHf(N2) = 0 kJ/mol
ΔHf(H2O) = -285.83 kJ/mol
3. Determine the change in enthalpy for the reaction by using the ΔHf values:
ΔH = Σ(nΔHf(products)) - Σ(nΔHf(reactants))
where n is the stoichiometric coefficient of each compound. Let's calculate:
ΔH = (1 mol × ΔHf(N2)) + (4 mol × ΔHf(H2O)) - (1 mol × ΔHf(N2O)) - (4 mol × ΔHf(H2))
ΔH = (1 mol × 0 kJ/mol) + (4 mol × -285.83 kJ/mol) - (1 mol × 82.05 kJ/mol) - (4 mol × 0 kJ/mol)
ΔH = -1143.32 kJ/mol - 82.05 kJ/mol
ΔH ≈ -1225.37 kJ/mol
Therefore, the change in enthalpy (ΔH) for the reaction N2O + 4H2 → N2 + 4H2O is approximately -1225.37 kJ/mol. This value indicates that the reaction is exothermic, as it releases energy in the form of heat.