Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 4 NO(g) + 2 O2(g) → 4 NO2(g) ΔH°rxn = ? Given: N2(g) + O2(g) → 2 NO(g) ΔH°rxn = +183 (in text format)

Answer 1

The standard enthalpy change (ΔH°rxn) for the reaction 4 NO(g) + 2 O₂(g) → 4 NO₂(g) is calculated using Hess's Law and the given reaction enthalpies, resulting in -594.2 kJ.

Step-by-step explanation:

To calculate the standard enthalpy change (ΔH°rxn) for the reaction 4 NO(g) + 2 O₂(g) → 4 NO₂(g), we can use the provided enthalpy changes from given reactions and apply Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how it is carried out, provided the initial and final conditions are the same.

Given:

1. 2NO(g) + O₂(g) → 2NO₂(g), ΔH° = -114.1 kJ (reaction is multiplied by 2 to match the required 4 NO molecules)
2. N2(g) + O₂(g) → 2NO(g), ΔH° = +183 kJ (reaction is multiplied by 2 to match the required 4 NO molecules)

First, we double the first reaction to match the number of NO molecules in the desired equation:

4NO(g) + 2O₂(g) → 4NO₂(g), ΔH° = 2(-114.1 kJ) = -228.2 kJ

Then, since the second reaction forms NO(g) from N₂(g) and O₂(g), and we need it in the opposite direction, we reverse the second reaction and change the sign of ΔH°:

4NO(g) → 2N2(g) + 2O₂(g), ΔH° = 2(-183 kJ) = -366 kJ

Next, we add the reversed reaction of forming NO(g) from N2(g) and O₂(g) to the reaction of forming NO₂(g) from NO(g) and O₂(g):

4NO(g) - 2N2(g) - 2O₂(g) + 4NO(g) + 2O₂(g) → 4NO2(g), ΔH°rxn = -366 kJ + (-228.2 kJ) = -594.2 kJ

Therefore, the standard enthalpy change for the reaction 4 NO(g) + 2 O₂(g) → 4 NO₂(g) is -594.2 kJ.

Answer 2

The standard reaction enthalpy for the given reaction is -456.4 kJ/mol.

Step-by-step explanation:

The standard reaction enthalpy for the reaction 4 NO(g) + 2 O2(g) → 4 NO2(g) can be determined using the given information:

Standard enthalpy change for the reaction 2 NO(g) + O₂(g) → 2 NO₂(g) is -114.1 kJ/mol

To find the standard reaction enthalpy, we need to multiply the enthalpy change of the given reaction by the stoichiometric coefficients:

4 NO(g) + 2 O2(g) → 4 NO₂(g)

1. ΔH°rxn = 4 * (-114.1 kJ/mol) = -456.4 kJ/mol

Therefore, the standard reaction enthalpy (ΔH°rxn) for the given reaction is -456.4 kJ/mol.