A proposed mechanism for the oxidation of nitric oxide to
nitrog...

Question

A proposed mechanism for the oxidation of nitric oxide to 
nitrogen dioxide was described in Problem 14.29. Another 
possible mechanism for this reaction is 
2 NO1g2  k1 
k-1 
N2O21g2 Faster, reversible 
N2O21g2 + O21g2 ¡ k2 2NO21g2 Slower, rate-determining 
(c) Relate the rate constant k to the rate constants for the 
elementary reactions.

Accepted Answer

Hello. Everyone in this video are trying to establish a relationship between the rate constants came an elementary reaction rate constants. So when the reaction mechanism is given, the rate law can be determined using the slow step in the reaction mechanism. So for a rate law this only involves our rackets. So that's going to be our starting materials. And the coefficients of the reactant in this slow step are the order of the reaction with respect to that reactant. So that's our rate determining step. Alright, so for our rate law equation that's equal to K. Multiplied by the concentration of a race to power its coefficient to note that as X multiplied by the concentration of B raised to power of its coefficient. Alright, so for our slow step which is already labeled for us here, our slow step we're gonna have to rewrite, the equation is equal to i negative and it's a great state plus H. Oh cl also in his state and this yields H. O. I. And cl minus. So the rate for this Is that it's equal to K2, multiplied by the concentration of I minus, multiplied by the concentration of H. O. C. L. So H. L. Cl its concentration is from step one. So for equilibrium the brakes forward is equal to the rate reversed. So we go ahead and write this again that K. Of one multiplied by O. C. L minus, multiplied by H 20. Is equal to K. Of negative one multiply the concentration of H. O. C. L. And the concentration of O. H minus. So here trying to isolate our H. O. C. L. So this is right over here on the right side of the equation. So let's go ahead scroll down. So again we're isolating that. So let's do this in the blue color. Let's isolate concentration of H. L. C. L. If we do. So we get got the concentration of H. O. C. L. Is equal to. So basically just dividing both sides by K. F. Negative one and concentration about H minus. So that gives us K. Of one, multiplied by the concentration of O. C. L minus, multiplied by H 20 divided by K. Of negative one, multiplied by the concentration of O. H minus. So the rate is equal to K. Two multiplied by one. Multiplied by this whole thing here. So that is cave one multiplied by O. C. L minus, multiplied by R. H. 20. Divided by K. Of negative one. Multiplied by the concentration of O. H minus. Now just putting in more things here we have K. F one times K. Of two divided by K. Of negative one. It's going to be multiplied by the concentration of I negative multiplied by the concentration of O. C. L minus, multiplied by the concentration of H 20 divided by the concentration of O. H minus. So the concentration of H 20. Is liquid and that's the concentration is constant and it can be included with the rate constant value. So then finally our rate is equal to The K. F one multiplied by KF two modified by the concentration of H 20. And then this is going to be all divided by the concentration of K -1. And then this is multiplied by the concentration of I minus, multiplied by the concentration of O C l minus, divided by the concentration of O H minus, Sir. Final answer here is that K is equal to K F one times K. Of two, multiplied by the concentration of H 20 over K of negative one. So this right here is going to be my final answer for this problem.

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