Be sure to answer all parts. Consider the following mechanism: (1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast] (2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow] (3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

a) The Overall Equation is

ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)

b) The intermediates include

HClO(aq)

OH⁻(aq)

HIO(aq)

H₂O(l)

c) Step 1

rate = k [ClO⁻] [H₂O]

Moleculiarity = 2, Bimolecular

Step 2

rate = k [I⁻] [HClO]

Moleculiarity = 2, Hence, it is Bimolecular

Step 3

rate = k [OH⁻][HIO]

Moleculiarity = 2, Hence, it is Bimolecular too.

d) No, this mechanism isn't consistent with the actual rate law.

Explanation:

(1) ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq) [fast]

(2) I⁻(aq) + HClO(aq) → HIO(aq) + Cl⁻(aq) [slow]

(3) OH⁻(aq) + HIO(aq) → H₂O(l) + IO⁻(aq) [fast]

a) For the overall reaction, summing the elementary reactions up.

ClO⁻(aq) + H₂O(l) + I⁻(aq) + HClO(aq) + OH⁻(aq) + HIO(aq) → HClO(aq) + OH⁻(aq) + HIO(aq) + Cl⁻(aq) + H₂O(l) + IO⁻(aq)

We then eliminate those that appear on both sides of the eqn. Those ones are the intermediates.

HClO(aq)

OH⁻(aq)

HIO(aq)

H₂O(l)

Then we have

ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)

Hence, this is the overall equation.

b) Like we pointed out in (a), all those that appeared on both sides of the overall equation, are intermediates.

HClO(aq)

OH⁻(aq)

HIO(aq)

H₂O(l)

c) Moleculiarity and Rate Law for each step

Moleculiarity is expressed as the number of molecules that react in the elementary reaction. It is simply the sum of their coefficients in a balanced elementary equation.

Step 1

rate = k [ClO⁻] [H₂O]

Moleculiarity = 2, Bimolecular.

Step 2

rate = k [I⁻] [HClO]

Moleculiarity = 2, Hence, it is Bimolecular

Step 3

rate = k [OH⁻][HIO]

Moleculiarity = 2, Hence, it is Bimolecular too.

d) We first try to write the overall rate law from the elementary steps given

(1) ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq) [fast]

(2) I⁻(aq) + HClO(aq) → HIO(aq) + Cl⁻(aq) [slow]

(3) OH⁻(aq) + HIO(aq) → H₂O(l) + IO⁻(aq) [fast]

The slow step is usually the rate determining step

So, from step 2,

Rate = k [I⁻] [HClO]

But from step 1,

rate = k [ClO⁻] [H₂O]

But k₁ [ClO⁻] [H₂O] = k₂ [HClO] [OH⁻]

[HClO] = (k₁/k₂) [ClO⁻] [H₂O]/[OH⁻]

Rate = K [I⁻] [HClO] =K [I⁻] [(k₁/k₂) [ClO⁻] [H₂O]/[OH⁻]] = (kk₁/k₂) [I⁻][ClO⁻] [H₂O]/[OH⁻]

Then from step 3

rate = k[OH-][HIO]

k₃ [OH-][HIO] = k₄ [H₂O] [IO⁻]

[H₂O]/[OH-] = (k₃/k₄) [HIO]/[[IO⁻]

Rate = (Kk₁/k₂) [I⁻][ClO⁻] [H₂O]/[OH⁻] =(kk₁/k₂) [I⁻][ClO⁻] [(k₃/k₄) [HIO]/[IO⁻]

Rate = (kk₁k₃/k₂k₄) [I⁻][ClO⁻][HIO]/[IO⁻]

(kk₁k₃/k₂k₄) = K

Rate = K [I⁻][ClO⁻][HIO]/[IO⁻]

Actual rate law: rate = k[ClO-][I -]

Rate Law obtained from this mechanism

Rate = K [I⁻][ClO⁻][HIO]/[IO⁻]

Hence, this mechanism isn't consistent with the actual rate law.

Hope this Helps!!!

d. temperate deciduous forest

can you list the answers so i can answer