Consider the exothermic reaction 4nh3(g)+5o2(g)→4no(g)+6h2o(g) calculate the standard heat of reaction, or δh∘rxn, for this reaction using the given data. also consider that the standard enthalpy of the formation of elements in their pure form is considered to be zero.

The answer to your question is:  ΔHrxn = - 903.6 kJ/mol

Explanation:

Reaction

                4NH3 (g) + 5O2 (g)   →    4NO(g)   +   6H2O(g)

H NH3 = -46.2 kJ/mol

H O2 = 0 kJ7mol

H NO = 90.3 kJ/mol

H H2O = -241.6 kJ/mol

Equation

               ΔHrxn = ∑ nH products   -  ∑ nH reactants

               ΔHrxn = 4(90.3) + 6(-241.6) - 4(-46.2) - 5(0)

              ΔHrxn = 361.2 - 1449.6 + 184.8

              ΔHrxn = - 903.6 kJ/mol