Macrophysical Chemistry Answer each questions clearly and correctly Safari ..ooo
ID: 500872 • Letter: M
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Macrophysical Chemistry Answer each questions clearly and correctly Safari ..ooo WE 9:24 PM DN. 6 Yet MORE Steady State Practice [Compatibility Mode] Home Insert Draw Layout Review View Cambria Yet MORE Steady State Practice consider the mechanism for the conversion of A to P A A B A+ A B A a.) Use the steady State Approximation to derive the rate law. Show all six steps. b) The conversion of A to P was observed to be second order in A. Under what conditions, if any, does the result you got for a match the experiment? c.) Describe an experiment (or set of experiments) which would let you test the mechanism. 31% aExplanation / Answer
a) The steady state approximation says that the rate of formation of the intermediate B is equal to the rate of consumption of B.
Rate of formation of B = d[B]/dt = k1[A]2
Rate of consumption of B = -d[B]/dt = k-1[B] + k2[A][B]
At the steady state, we must have for B,
d[B]/dt = -d[B]/dt
===> k1[A]2 = k-1[B] + k2[A][B]
===> k1[A]2 = [B](k-1 + k2[A])
===> [B] = k1[A]2/(k-1 + k2[A]) …..(1)
The rate of production of the product P is given by d[P]/dt = k2[A][B]
Plug in the value of [B] from (1) to obtain the rate of the reaction as
d[P]/dt = k2[A]*k1[A]2/(k-1 + k2[A]) = k1k2[A]3/(k-1 + k2[A]) ……(2) (ans)
b) The conversion of A to P was found to second order in A so that the rate law for the conversion is of the form
d[P]/dt = kobs[A]2 ….(3)
The rate law obtained in part (a) above is given in expression (2). Expression (2) will be of the form of expression (3) when k-1 << k2 such that k-1 + k2[A] k2[A].
With this approximation, expression (2) reduces to
d[P]/dt = k1k2[A]3/(k2[A]) = k1[A]2
Therefore the condition is k-1 << k2 and also k1 such that there is no equilibrium involving reactant A and the intermediate B (ans).
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