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Considering the following general mechanism , which applies to many thermal decompositions and isomerisations. A + M rarr underset( K_(-1))overset(K_(1))(rarr) A^(*) + M A^(*) overset( K_(2))(rarr) B + C Here A^(*) represent the intermediate species formed during the process. if the first step is slow and rate determining, the rate law is ( d [B])/(dt) = K_(1) [ A][M] On the other hand, if K_(-1) is large, and thus A^(*) is in rapid equilibrium with A, we have, ([A^(*)])/([A]) = (K_(1))/( K_(-1)) ( d[B])/( dt) = K_(2) [A^(*)] = ( K_(1) K_(2))/(K_(-1)) [A] Using steady state approximation, the general rate law for the system can be found out as follows : Rate of production of A^(*) = rate of destruction of A^(*) K_(1) [A][M] = K_(-1) [A^(*)][M]+K_(2) [ A^(*)] or [A^(*)] = ( K_(1) [A][M])/(K_(-1)[M]+K_(2)) Since the rate of reaction is ( d[B])/( dt) = K_(2) [A^(*)] :. ( d [B])/( dt) = ( K_(1) K_(2) [A][M])/(K_(-1) [ M ] + K_(2)) A practical example of the application of the steady state approximation is the decomposition of N_(2)O_(5) as follows : 2N_(2) O_(5) rarr 4NO_(2) + O_(2) It follows the following mechanism : N_(2)O_(5) underset( K_(-1))overset( K_(1))(rarr)NO_(2) + NO_(3) NO_(3) + NO_(2) overset( K_(3))(rarr) 2NO_(2) The overall order of the reaction , 2N_(2)O_(5) rarr 4NO_(2) O_(2) is :

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2
1
`3//2`
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Answer :B


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