Question

given the following standard enthalpy of formation values, calculate the enthalpy change, δh°rxn, for the following reaction: n2o4(g) 4h2(g) → n2(g) 4h2o(g)

Answer 1

The standard enthalpy of formation, ΔH°f, is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states.

Here are the standard enthalpy of formation values needed for the calculation:

ΔH°f(N2O4) = +9.16 kJ/mol

ΔH°f(N2) = 0 kJ/mol

ΔH°f(H2) = 0 kJ/mol

ΔH°f(H2O) = -241.8 kJ/mol

To calculate the enthalpy change for the reaction, ΔH°rxn, we need to use Hess's law. Hess's law states that the enthalpy change for a reaction is the sum of the enthalpy changes for any series of reactions that add up to the original reaction, regardless of the number of steps or intermediates involved.

We can write the reaction as a sum of the formation reactions of the products minus the sum of the formation reactions of the reactants:

ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)

ΔH°rxn = [ΔH°f(N2) + 4ΔH°f(H2O)] - [ΔH°f(N2O4) + 4ΔH°f(H2)]

ΔH°rxn = [0 + 4(-241.8 kJ/mol)] - [9.16 kJ/mol + 4(0)]

ΔH°rxn = -967.2 kJ/mol + 9.16 kJ/mol

ΔH°rxn = -958.04 kJ/mol

Therefore, the enthalpy change for the reaction is -958.04 kJ/mol.