Calculate ΔH for the reaction
NO + O → NO2
given the following information:
NO + O3 → NO2 + O2 ΔH = -198.9 kJ
O3 → 3/2 O2 ΔH = -142.3 kJ
O2 → 2O ΔH = 495 kJ
You need NO on the left, so write equation (1)
(1) NO + O3 ===> NO2 + O2 /\H -198.9 kJ
You need O on the left, so divide equation (3) by 2, write it backwards and reverse the sign of 1/2 /\H
((4) O ===> 1/2 O2 -247.5 kJ
You need to cancel O3 on the left when you add all these equations up, so write equation (2) backwards and reverse the sign odf /\H
(5) 3/2 O2 ===> O3 /\H = +142.3 kJ
Add equations (1), (4), and (5) and add the /\H's
-198.9 -247.5 + 142.3 = -304.1 kJ
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Verified answer
You need NO on the left, so write equation (1)
(1) NO + O3 ===> NO2 + O2 /\H -198.9 kJ
You need O on the left, so divide equation (3) by 2, write it backwards and reverse the sign of 1/2 /\H
((4) O ===> 1/2 O2 -247.5 kJ
You need to cancel O3 on the left when you add all these equations up, so write equation (2) backwards and reverse the sign odf /\H
(5) 3/2 O2 ===> O3 /\H = +142.3 kJ
Add equations (1), (4), and (5) and add the /\H's
-198.9 -247.5 + 142.3 = -304.1 kJ