Question

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

Answer 1

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