1.

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 steady state aproximateion is applied to a reaction intermediate but never to a reactant or product because

Answer»

the concentration of the intermediate is very small
the concentration of the reactants and PRODUCTS are variable quantities
reactants are only consumed and products are only PRODUCED whereas intermediate are both produced and consumed
both ( A) and ( C ) are correct

Answer :D


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