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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.

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joseaaronlara

Answer:

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

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