CH147-2: Chemical Reaction Engineering 2
Exam 1.1 PSSH
Problem 1.
Solution:
Overall reaction:
CO + Cl2 → COCl2
The hints indicate that the active intermediates are Cl* and COCl*.
Postulated mechanism:
Cl2 in the numerator dissociates into the active intermediate Cl*:
Cl2 → 2Cl∗
𝑘1
(1)
The reverse of reaction (1):
2Cl∗ → Cl2
𝑘2
(2)
𝑘3
(3)
CO in the numerator collides with active intermediate Cl* to form
another active intermediate COCl*:
CO + Cl∗ → COCl∗
The reverse of reaction (3):
COCl∗ → CO + Cl∗
(4)
𝑘5
(5)
𝑘4
Cl2 in the numerator collides with active intermediate COCl* to
form COCl2 and the other active intermediate Cl*:
Rate laws:
𝑟1 = 𝑘1 𝐶Cl2
2
∗
𝑟2 = 𝑘2 𝐶Cl
𝑟3 = 𝑘3 𝐶CO 𝐶Cl∗
𝑟4 = 𝑘4 𝐶COCl∗
𝑟5 = 𝑘5 𝐶COCl∗ 𝐶Cl2
COCl∗ + Cl2 → COCl2 + Cl∗
Rate of product formation:
𝑟COCl2 = 𝑟5 = 𝑘5 𝐶COCl∗ 𝐶Cl2
Net rate of formation of active intermediates:
𝑟COCl∗ = 𝑟3 − 𝑟4 − 𝑟5
𝑟Cl∗ =
𝑟1
− 𝑟2 − 𝑟3 + 𝑟4 + 𝑟5
2
Pseudo-steady state hypothesis: Net rate of formation of active intermediates ≈ 0
𝑟COCl∗ = 0 = 𝑘3 𝐶CO 𝐶Cl∗ − 𝑘4 𝐶COCl∗ − 𝑘5 𝐶COCl∗ 𝐶Cl2
𝑘4 𝐶COCl∗ + 𝑘5 𝐶COCl∗ 𝐶Cl2 = 𝑘3 𝐶CO 𝐶Cl∗
𝐶COCl∗ (𝑘4 + 𝑘5 𝐶Cl2 ) = 𝑘3 𝐶CO 𝐶Cl∗
𝐶COCl∗ =
𝑘3 𝐶CO 𝐶Cl∗
𝑘4 + 𝑘5 𝐶Cl2
𝑟Cl∗ = 0 ⇒ 𝑟COCl∗ + 𝑟Cl∗ = 0
𝑟COCl∗ + 𝑟Cl∗ = 𝑟3 − 𝑟4 − 𝑟5 +
𝑟1
− 𝑟2 − 𝑟3 + 𝑟4 + 𝑟5 = 0
2
𝑟1
𝑘1
2
∗ = 0
− 𝑟2 = 0 ⇒ 𝐶Cl2 − 𝑘2 𝐶Cl
2
2
𝑘1
2
∗
𝐶 = 𝑘2 𝐶Cl
2 Cl2
𝑘1
𝐶Cl∗ = √
𝐶
2𝑘2 Cl2
The rate of product formation becomes:
𝑟COCl2 = 𝑘5 𝐶COCl∗ 𝐶Cl2 = 𝑘5 [
1
𝑘3 𝐶CO
𝑘1
∙√
(𝐶Cl2 )2 ] 𝐶Cl2
𝑘4 + 𝑘5 𝐶Cl2
2𝑘2
Let:
𝑘1
𝑘 ′ = 𝑘3 𝑘5 √
2𝑘2
The rate of product formation simplifies into:
3
𝑘 ′ 𝐶CO 𝐶Cl2 2
𝑟COCl2 =
𝑘4 + 𝑘5 𝐶Cl2
Assuming that 𝑘4 ≫ 𝑘5 𝐶Cl2 :
𝟑
𝒓𝐂𝐎𝐂𝐥𝟐 = 𝒌𝑪𝐂𝐎 𝑪𝐂𝐥𝟐 𝟐
where:
𝑘′
𝑘=
𝑘4