Calculate δ h for the reaction: no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3(g) ↔ no2(g) + o2(g) δh=- 200 kj

To calculate the δ h, we must balance first the reaction: 

NO + 0.5O2 > NO2

Then we write all the reactions,

2O3 > 3O2    δ h = -426 kj        eq. (1)

O2 > 2O    δ h = 490 kj             eq. (2)

NO + O3 > NO2 + O2    δ h = -200 kj          eq. (3)

We divide eq. (1) by 2, we get

O3 > 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 >  NO2 + O2   δ h = -200 kj
-       (O3 > 1.5 O2         δ h = -213  kj)
NO > NO2 - 0.5O2        δ h = 13  kj               eq. (5)

eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O > 0.5O2  δ h = -245  kj         eq. (6)

Add eq. (6) to eq. (5), we get

NO > NO2 - 0.5O2        δ h = 13  kj 
+  O > 0.5O2                 δ h = -245  kj
NO + O > NO2               δ h = -232 kj

ANSWER: NO + O > NO2               δ h = -232 kj

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