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Byju's Answer

Standard XII

Chemistry

Elementary and Complex Reactions

The reaction ...

Question

The reaction between chloroform, $C H {C l}_{3} (g)$ and chlorine ${C l}_{2} (g)$ to form $C {C l}_{4} (g)$ and $H C l (g)$ is believed to occur by this series of steps:
Step 1: ${C l}_{2} (g) ⟶ C l (g) + C l (g)$
Step 2: $C H {C l}_{3} (g) + C l (g) ⟶ C {C l}_{3} (g) + H C l (g)$
Step 3: $C {C l}_{3} (g) + C l (g) ⟶ C {C l}_{4} (g)$
If this reaction is first order in $C H {C l}_{3}$ and half order in ${C l}_{2}$ , which statement about the relative rates of steps 1, 2 and 3 is correct?

Step 1 is the slowest

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Steps 1 and 2 must both be slow

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Step 2 must be slower than step 1

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Step 3 must be the slowest

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Solution

The correct option is C Step 2 must be slower than step 1
Given rate= $k [C H C l_{3}] [C l_{2}]^{1 / 2}$
This rate can be obtained if step 2 is slower than step 1 and that step 1 is an equilibrium reaction.

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Similar questions

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O_{3}

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O_{3} (g) + C l (g) ⟶ O_{2} (g) + C l O (g);

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C l O (g) + O (g) ⟶ C l (g) + O_{2} (g);

k_{2} = 2.6 \times 10^{10} L {m o l}^{- 1} {s e c}^{- 1}

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Δ H^{o} = - 104 k J

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, the experimentally determined rate equation is,

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The correct option is C Step 2 must be slower than step 1Given rate=k[CHCl3][Cl2]1/2This rate can be obtained if step 2 is slower than step 1 and that step 1 is an equilibrium reaction.

The correct option is C Step 2 must be slower than step 1
Given rate= $k [C H C l_{3}] [C l_{2}]^{1 / 2}$
This rate can be obtained if step 2 is slower than step 1 and that step 1 is an equilibrium reaction.