Question

Calculate ΔG∘rxn for the following reaction:
4CO(g)+2NO2(g)→4CO2(g)+N2(g).
Use the following reactions and given ΔG∘rxn values:
2NO(g)+O2(g)→2NO2(g), ΔG∘rxn= - 72.6 kJ
2CO(g)+O2(g)→2CO2(g), ΔG∘rxn= - 514.4 kJ
12O2(g)+12N2(g)→NO(g), ΔG∘rxn= 87.6 kJ

Answer 1

Calculate ΔG∘rxn for the following reaction:
4CO(g)+2NO2(g)→4CO2(g)+N2(g).

Use the following reactions and given ΔG∘rxn values:
2NO(g)+O2(g)→2NO2(g), ΔG∘rxn= - 72.6 kJ
2CO(g)+O2(g)→2CO2(g), ΔG∘rxn= - 514.4 kJ
12O2(g)+12N2(g)→NO(g), ΔG∘rxn= 87.6 kJ

Solution :-

We need to rearrange the equations so that we can get the desired equation

We need CO2 on the product side so need to multiply eq 2 by factor 2

To get the NO2 on the reactant side we need to reverse equation 1

Reverse equation 3 and multiply it by 2

New equations are as follows

2NO2(g) ----- > 2NO(g)+O2(g),               ΔG∘rxn= 72.6 kJ
4CO(g)+2O2(g) ----- > 4CO2(g),                 ΔG∘rxn= - 514.4 kJ*2
2NO(g) ---- > O2(g)+ N2(g) ,              ΔG∘rxn= -87.6 kJ*2

Now add all equations to get the final equation then we get

Therefore the enthalpy change for the reaction is -1131.4 kJ