dHf = delta H formation
dHf rxn = (n*dHf products) - (n*dHf reactants)
dHf rxn = (n*dHf products) - (n*dHf reactants)
ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))
Where:
ΔHf is the enthalpy of formation
Σ signifies a sum (you need to calculate the sum of the enthalpies of formation for all the reactants and products)
Given the enthalpy of formation values:
ΔHf(NaOH) = -469.60 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
For the reaction: 2Na + 2H2O → 2NaOH + H2
The enthalpy change can be calculated as follows:
ΔH = [2ΔHf(NaOH) + ΔHf(H2)] - [2ΔHf(Na) + 2ΔHf(H2O)]
Substituting the values:
ΔH = [2(-469.60 kJ/mol) + 0 kJ/mol] - [2(0 kJ/mol) + 2(-285.8 kJ/mol)]
Simplifying:
ΔH = -939.20 kJ/mol - (-571.6 kJ/mol)
ΔH = -939.20 kJ/mol + 571.6 kJ/mol
ΔH = -367.60 kJ/mol
Therefore, the enthalpy change for the given reaction is -367.60 kJ/mol.