The chemical reaction, 2O3 → 3O2 proceeds as follows: Step 1: O3 leftharpoons O2 + O .......(fast) Step 2: O + O3 → 2O2 .......(slow) The rate law expression should be

Question

The chemical reaction, 2O3 → 3O2 proceeds as follows: Step 1: O3 leftharpoons O2 + O .......(fast) Step 2: O + O3 → 2O2 .......(slow) The rate law expression should be (A) r = k'[O3][O2] (B) r = k'[O3]2[O2]-1 (C) r = k'[O3]2 (D) unpredictable

Tardigrade Admin · Accepted Answer

As the slowest step is the rate determining step, hence from step 2,

r = k [O_{3}] [O] ... (i)

From step 1,

K_{e q} = \frac{[ O _{2} [ O ] ]}{[ O _{3} ]}

or

[O] = \frac{K _{e q} [ O _{3} ]}{[ O _{2} ]} ... (ii)

From eq. (i) and (ii),
r = k

K_{e q} \frac{[ O _{3} ] ^{2}}{[ O _{2} ]} = k^{'} [O_{3}]^{2} [O_{2}]^{- 1} [∵ k^{'} = k K_{e q}]

Q. The chemical reaction, 2O₃ $\to$ 3O₂ proceeds as follows :
Step 1 : O₃ $⇌$ O₂ + O .......(fast)
Step 2 : O + O₃ $\to$ 2O₂ .......(slow)
The rate law expression should be

$r = k^{'} [O_{3}] [O_{2}]$

$r = k^{'} [O_{3}]^{2} [O_{2}]^{- 1}$

$r = k^{'} [O_{3}]^{2}$

unpredictable

Q. The chemical reaction, 2O3 → 3O2 proceeds as follows : Step 1 : O3 ⇌ O2 + O .......(fast) Step 2 : O + O3 → 2O2 .......(slow) The rate law expression should be

r=k′[O3​][O2​]

r=k′[O3​]2[O2​]−1

r=k′[O3​]2

unpredictable

As the slowest step is the rate determining step, hence from step 2, r=k[O3​][O]...(i) From step 1, Keq​=[O3​][O2​[O]]​ or [O]=[O2​]Keq​[O3​]​...(ii) From eq. (i) and (ii), r = k Keq​[O2​][O3​]2​=k′[O3​]2[O2​]−1[∵k′=kKeq​]

Q. The chemical reaction, 2O₃ $\to$ 3O₂ proceeds as follows :
Step 1 : O₃ $⇌$ O₂ + O .......(fast)
Step 2 : O + O₃ $\to$ 2O₂ .......(slow)
The rate law expression should be

$r = k^{'} [O_{3}] [O_{2}]$

$r = k^{'} [O_{3}]^{2} [O_{2}]^{- 1}$

$r = k^{'} [O_{3}]^{2}$