The elementary reactions involved in a mechanism are given...

The elementary reactions involved in a mechanism are given below A + A -> A* + A A* + A -> A + A A* + bA -> products Where b = 0 for unimolecular reaction steps b = 1 for bimolecular reaction steps b = 2 for termolecular reaction steps Carry out a steady state analysis to show that: (i) unimolecular reactions are first order at high pressures, moving to second order at low pressures. State the rate determining steps in each region. (ii) In contrast, show that bimolecular reactions are second order at all pressures

Transcript

00:01 Hello students, we have been given a question on collision theory.

00:05 So collision theory of gases tells us that this a mole of gas is reacting with another molecule of gas in a reversible step in which we are getting an activated molecule along with an deactivated molecule.

00:29 Now this activation step has a rate constant k1 and deactivation step has a rate constant k -1.

00:38 In deactivation step the activated molecule is colliding with unactivated molecule and both molecules get deactivated.

00:48 Now the second possibility is that this activated molecule of gas gives our final product by k2.

01:01 Now for this reaction rate of reaction is given in terms of rate of disappearance of a that is minus da by dt.

01:11 Now that is given k2 k1 a square divided by k minus 1 a plus k2.

01:26 We have to tell at what conditions rate of overall reaction will be of second order.

01:37 So let's consider the first condition.

01:42 In the first option we have given a condition of high pressure.

01:49 Now if pressure of gas is very high then dissociation of a star into product will be very high.

01:58 So we can say that k2 will be very high.

02:04 Now at high pressure condition the deactivation process will be slow because at high pressure dissociation is favored.

02:16 So a star will be converted to product.

02:19 Hence at high pressure k2 will be much more higher than k -1.

02:25 So we can ignore the term of k -1 in this equation.

02:31 Hence our total equation will become minus da by dt will be equal to k2 k1 a square divided by k2.

02:45 Hence this k2 will be cancelled out and our rate of reaction will be k1 a square.

02:54 So from here we can see that the power of a concentration is 2.

02:59 So this reaction is of second order.

03:05 That's why we can say that at high pressure the reaction will be of second order.

03:11 Now let's consider other options that are given as well.

03:15 In second condition we have given low pressure.

03:19 So if pressure is very low then activated molecules of gas won't be able to dissociate into products.

03:27 So this will be very slow and most of the molecules will undergo deactivation.

03:34 So deactivation will be very fast.

03:37 So we can say that at low pressure k -1 will be very higher than k2...

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