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22 April, 00:38

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?

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  1. 22 April, 02:31
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    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.
  2. 22 April, 02:59
    0
    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)
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