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The chemical reaction $2O_3 \to 3O_2$ proceeds as follows : $ O_3 \rightleftharpoons O_2+O ......(fast) \qquad O+O_3 \to 2O_2 ....(slow)$. The rate law expression should be :

$(A)\;r=K [O_3]^2 \\ (B)\;H=K [O_3]^2[O_2]^{-1} \\(C)\;r= K [O_3][O_2] \\(D)\;unpredictable $
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For slow step, rate $=K [O][O_3]$.....(i)
Also for equilibrium $K_c=\large\frac{[O][O_3]}{[O_3]}$........(ii)
$\therefore [O]=\large\frac{Kc[O_3]}{[O_2]}$ .......(iii)
By (iii) and (i), the intermediate [O] is eliminated as rate,
$\large\frac{K.K_c[O_3][O_3]}{[O_2]}=\frac{K'[O_3]^2}{[O_2]}$
Hence b is the correct answer.
answered Apr 3, 2014 by meena.p
 

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