Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 2 NO(g) + O2(g) → 2 NO2(g) ΔH°rxn = ? Given: N2(g) + O2(g) → 2 NO(g) ΔH°rxn = +183 kJ 1/2 N2(g) + O2(g) → NO2(g) ΔH°rxn = +33 kJ Which molecules or compounds in this question have ΔH°f,298 =0? Explain.

Answer 1

To determine the standard reaction enthalpy for the given reaction, 2 NO(g) + O2(g) → 2 NO2(g), we can use Hess's law. By manipulating the given reaction enthalpies and using stoichiometric coefficients, we can find the desired reaction enthalpy to be +432 kJ.

Step-by-step explanation:

In order to determine ΔH°rxn for the reaction 2 NO(g) + O2(g) → 2 NO2(g), we can use Hess's law and the given standard reaction enthalpies:

1) N2(g) + O2(g) → 2 NO(g) ΔH°rxn = +183 kJ

2) 1/2 N2(g) + O2(g) → NO2(g) ΔH°rxn = +33 kJ

To find the desired reaction, we can multiply equation 2 by 2 and equation 1 by 2, and then sum the two reactions:

2 NO(g) + O2(g) → 2 NO2(g) ΔH°rxn = 2 * (+33 kJ) + 2 * (+183 kJ) = +432 kJ

Therefore, ΔH°rxn for the reaction 2 NO(g) + O2(g) → 2 NO2(g) is +432 kJ.