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
