Question

A proposed mechanism for the preparation of COBr₂ from CO(g) and Br₂ is:

Br₂ ⇄ 2 Br (g) fast

Br(g) + CO (g) ⇄ COBr fast

COBr + Br₂ → COBr₂ + Br slow

The correct rate law for this mechanism is:

• A.

  k [CO][Br₂]

• B.

  k [CO] / [Br₂]^1/2

• C.

  k[Br₂]

• D.

  k [CO]²[Br₂]

• E.

  k [CO][Br₂]^3/2

Answer 1

Solution:-

As, we know that the slow step is always rate-determining. Thus, from step 3,

Rate = k₁ [COBr] [Br₂] .....................(i)

From the fast equilibrium step 1,

K' = [Br]² / [Br₂]

or, [Br] = (K')^1/2 [Br₂]^1/2 ........................(ii)

And, from fast equilibrium step 2,

K" = [COBr] / [Br​​​​​​​] [CO]

[COBr] = K" [Br​​​​​​​] [CO] ...........................(iii)

Substituting for [COBr] in rate expression [equation (i)], we have

Rate = k₁ K" [Br] [CO] [Br​​​​​​​₂] ............................(iv)

Substituting for [Br] in the equation (iv), we have

Rate = k₁ K" (K')^1/2 [Br₂]^1/2 [CO] [Br₂]

or, Rate = k [CO] [Br₂]^3/2

(Where, k = k₁ K" K​​​​​​​^1/2)

hence, Option E) : k [CO] [Br​​​​​​​₂]^3/2 is correct answer.

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