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Use the data provided along with Hess's law to determine the [tex]\Delta G^\circ_{\text{rxn}}[/tex] for the reaction:

\[ 2 \text{N}_2\text{O}(g) + 3 \text{O}_2(g) \rightarrow 4 \text{NO}_2(g) \]

Given reactions:

1. \[ \text{N}_2(g) + 2 \text{O}_2(g) \rightarrow 2 \text{NO}_2(g) \quad \Delta G^\circ_{\text{rxn}} = 66.4 \, \text{kJ} \]

2. \[ 2 \text{N}_2\text{O}(g) \rightarrow 2 \text{N}_2(g) + \text{O}_2(g) \quad \Delta G^\circ_{\text{rxn}} = -164.2 \, \text{kJ} \]

Choose the correct value for [tex]\Delta G^\circ_{\text{rxn}}[/tex]:

a) -97.8 kJ
b) -31.4 kJ
c) -230.6 kJ
d) 31.4 kJ
e) 97.8 kJ

Answer :

Final answer:

Using Hess's law and the given reactions, we manipulate and add them to get the target reaction's Gibbs free energy change, leading to a result of -31.4 kJ.

Explanation:

To determine the standard Gibbs free energy change (ΔG°rxn) for the reaction 2 N2O(g) + 3 O2(g) → 4 NO2(g) using Hess's law, we must manipulate and combine the given reactions to match the target equation:

  • N2(g) + 2 O2(g) → 2 NO2(g) ΔG°rxn = 66.4 kJ
  • 2 N2O(g) → 2 N2(g) + O2(g) ΔG°rxn = -164.2 kJ

We'll use these two reactions to find the Gibbs free energy change for the target reaction by adding them in such a way that extra substances cancel out:

First reaction times 2:

2 [N2(g) + 2 O2(g) → 2 NO2(g)]

→ 2 N2(g) + 4 O2(g) → 4 NO2(g) ΔG°rxn = 2 * 66.4 kJ = 132.8 kJ

Second reaction as it is:

2 N2O(g) → 2 N2(g) + O2(g) ΔG°rxn = -164.2 kJ

Now add the two equations:

[2 N2(g) + 4 O2(g) → 4 NO2(g)] + [2 N2O(g) → 2 N2(g) + O2(g)]

→ 2 N2O(g) + 3 O2(g) → 4 NO2(g) ΔG°rxn = 132.8 kJ - 164.2 kJ

The final ΔG°rxn for our target reaction is:

ΔG°rxn = 132.8 kJ - 164.2 kJ = -31.4 kJ

The correct answer is (b) -31.4 kJ.

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Rewritten by : Jeany

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