Calculate Delta H for the reaction: NO (g) + O (g) ——> NO2 (g)


Given the following data: 
2O3 (g) ——> 3O2 (g) Delta H= -427 kJ
O2 (g) ——–> 2O (g) Delta H= +495 kJ
NO (g) + O3 (g) —–> NO2 (g) + O2 (g) Delta H= -199 kJ

Calculate Delta H for the reaction: NO (g) + O (g) ——> NO2 (g)

 

Use the third reaction by not changing it because the NO (g) is already on the left side of the reaction and the NO2 (g) is already on the right side of the reaction.

NO (g) + O3 (g) —–> NO2 (g) + O2 (g) Delta H= -199 kJ

Because O (g) is on the left side of the wanted reaction, use the second reaction and write the reaction in the reverse direction while divided by two.

O (g) ——–> 1/2 O2 (g) Delta H= -247.5 kJ

Add these two reactions.

NO (g) + O3 (g) —–> NO2 (g) + O2 (g) Delta H= -199 kJ
O (g) ——–> 1/2 O2 (g) Delta H= -247.5 kJ
—————————————…
NO (g) + O3 (g) + O (g) —–> NO2 (g) + 3/2 O2 (g) Delta H= -446.5 kJ

This isn’t the wanted reaction, but it should be after using the first reaction by writing it in the reverse direction and dividing it by two.

3/2 O2 (g) ——> O3 (g) Delta H= 213.5 kJ

Add this reaction to the reaction above.

NO (g) + O3 (g) + O (g) —–> NO2 (g) + 3/2 O2 (g) Delta H= -446.5 kJ
3/2 O2 (g) ——> O3 (g) Delta H= 213.5 kJ
—————————————…
NO (g) + O (g) ——> NO2 (g) Delta H= -233 kJ 


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