Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
The ΔG°rxn for the following reaction 3N2O(g) + 3NO2(g) → 9 NO(g) = 69 kJ
From the information given:
[tex]\mathbf{N_2O_{(g)} + NO_{2(g)} \to 3 NO_{(g)} \ \ \ \Delta G^0_{rxn}=-23.0 \ kJ}[/tex]
To determine the [tex]\mathbf{ \Delta G^0_{rxn}}[/tex] for the reaction [tex]\mathbf{3N_2O_{(g)} +3NO_{2(g)} \to 9 NO_{(g)}}[/tex], we need to reverse the [tex]\mathbf{ \Delta G^0_{rxn}}[/tex] of the following reaction given
So;
[tex]\mathbf{ 9 NO_{(g)} \to 3N_2O_{(g)} +3NO_{2(g)} \ \ \ \ \ \Delta G = ??? }[/tex]
Since we can get three times the moles of the given reaction in the following reaction, then:
The [tex]\mathbf{ \Delta G^0_{rxn}}[/tex] for the following reaction = (23 × 3) kJ
The [tex]\mathbf{ \Delta G^0_{rxn}}[/tex] for the following reaction is = 69 kJ
Learn more about the change in the free energy ΔG°rxn here:
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