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