calculate the enthalpy change for: (enthalpy of formation of aqueous sodium hydroxide is -469.60 kJ/mol. enthalpy of formation of liquid water is -285.8 kJ/mol) 2Na + 2H2O = 2 NaOH + H2
dHf = delta H formation
dHf rxn = (n*dHf products) - (n*dHf reactants)
To calculate the enthalpy change for the given reaction, you need to use the enthalpy of formation values for each compound involved. The enthalpy change (ΔH) can be calculated using the following formula:
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.