Question

A scientist measures the standard enthalpy change for the following reaction to be -888.4 kJ : 4NH3(9) +5 02(9 *4NO(g) + 6 H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of NO(g) is kJ/mol. Submit Answer

Answer 1

Given:
ΔHo rxn = -888.4 KJ/mol
Hof(NH3(g)) = -46.11 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol
Hof(H2O(g)) = -241.818 KJ/mol

Balanced chemical equation is:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)

ΔHo rxn = 4*Hof(NO(g)) + 6*Hof(H2O(g)) - 4*Hof( NH3(g)) - 5*Hof(O2(g))
-888.4 = 4*Hof(NO(g)) + 6*(-241.818) - 4*(-46.11) - 5*(0.0)
Hof(NO(g)) = 94.517 KJ/mol
Answer: 94.52 KJ/mol