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.  

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)

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