Consider the following three-step mechanism for a reaction: Cl2 (g) ⇌2Cl (g) Fast Cl (g) + CHCl3 (g) →HCl (g) + CCl3 (g) Slow Cl (g) + CCl3 (g) →CCl4 (g) Fast What is the overall reaction?

Following steps are involved in the reaction

Step 1: Cl2 (g) ⇌2Cl (g)

Step 2: Cl (g) + CHCl3 (g) →HCl (g) + CCl3 (g)

Step 3: Cl (g) + CCl3 (g) →CCl4 (g)

...

Overall reaction : Cl2 (g) + CHCl3 (g) → HCl (g) + CCl4 (g)

Note: Overall reaction is obtained by adding step 1-3 and canceling common terms that are present on reactant and product sides.

Answer: Cl₂ (g) + CHCl₃ (g) ⇌ HCl (g) + CCl₄ (g)

Explanation:

To find the overall reaction you need to add the three equations and find the net reactants and products

1) First add the reactants (left side) and the products (right side):

Cl₂ (g) + Cl (g) + CHCl₃ + Cl (g) + CCl₃ (g) ⇌ 2Cl (g) + HCl (g) + CCl₃ (g) + CCl₄ (g)

2) Next, note what you can cancell out:

- two Cl (g) of the left side with two Cl (g) of the right side

- CCl₃ (g) of the left side with CCl₃ (g) of the right side

3) That leads to the overall equation: Cl₂ (g) + CHCl₃ (g) ⇌ HCl (g) + CCl₄ (g)