Question

A scientist measures the standard enthalpy change for the following reaction to be -133.1 kJ:

NH4NO3 (aq) ---> N2O (g) + 2 H2O (l)

based on this value and the standard enthalpies if formation for the other substances, the standard enthalpy of formation of NH4NO3 (aq) is what? (kJ/mol)

Answer 1

Given:

ΔHo rxn = -133.1 KJ/mol

Hof(N2O(g)) = 82.05 KJ/mol

Hof(H2O(l)) = -285.83 KJ/mol

Balanced chemical equation is:

NH4NO3(aq) ---> N2O(g) + 2 H2O(l)

ΔHo rxn = 1*Hof(N2O(g)) + 2*Hof(H2O(l)) - 1*Hof( NH4NO3(aq))

-133.1 = 1*(82.05) + 2*(-285.83) - 1*Hof(NH4NO3(aq))

Hof(NH4NO3(aq)) = -356.51 KJ/mol

Answer: -356.51 KJ/mol