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 
(b) Show that this mechanism is consistent with the experimental 
rate law, Rate = k3NO423O24.

Accepted Answer

Hello. In this problem, we are told the contact process and method for manufacturing sulfuric acid has the following mechanism, we are told the overall reaction for the process has the following experimental rate law. And we are asked is the mechanism given above consistent with this experimental rate law. So let's begin by writing the overall reaction from these elementary steps. So when we look at the first step, we see that we produce two molds of sulfur trioxide, and in the second that we are consuming sulfur trioxide. Let's make the moles same for both reactions. So multiply everything to buy through by A two. And we see then in the second step of the reaction mechanism, we are producing two moles now of di sulfuric acid and it's being consumed in the third step. So let's again get our moles to be the same. Will multiply everything through by two For our 3rd step in this reaction mechanism, if we then simplify things, we see that we have two moles of sulfur trioxide again being produced and then consumed in the next step, and two moles of sulfuric acid being consumed and produced. So it's being produced in the second step and consumed in the third step. So these two are referred to as intermediate. So they are produced in one step and then consumed in a subsequent step. We also see that we have sulfuric acid that can be simplified. So we have two moles of sulfuric acid on the reactive side. So we can eliminate two on the product side. What's left over then provides our overall reaction. So we have two moles of sulfur dioxide gas reacts with oxygen, gas and two moles of water To produce two moles of sulfuric acid. And so we can write the general form of the rate law based on this overall reaction, the rate is equal to the reaction rate constant times the concentration of sulfur dioxide to some order X. And concentration of oxygen to some order Y. The water is not included because it is a pure liquid. Next let's write the rate law expression in terms of great determinate step from our reaction mechanism. So our right then be equal to our reaction rate constant times the concentration of sulfur track side squared. The sulfuric acid would not be included because it again is a pure liquid. And so we see that the sulfur trioxide is shown in this great law, as determined by the rate determining step, and the sulfur trioxide is a intermediate, generally intermediate. Um do not appear in the rate law for the overall reaction. In our next step, make use of the first reaction. So the one that is fast and reversible to determine then the relationship between our sulfur trioxide and our reactant, so that we can eliminate the sulfur dioxide from our reaction rate law. So based on the first step in the reaction mechanism, then, since it's fast and reversible, it will come to equilibrium equilibrium, says. Then that the rate of the forward reaction will equal the rate of the reverse. And so for our forward reaction, we would have our reaction rate constant times concentration of sulfur dioxide squared, times the concentration of oxygen. And that would be pulled in the reaction rate law for the reverse. So that would be our reaction rate constant for the reverse reaction times the concentration of sulfur trioxide squared. And so we can then isolate sulfur trioxide. So sulfur dioxide squared, it's equal to them K forward over K reverse times the concentration of sulfur dioxide squared times the concentration of oxygen. And so we can substitute this into then our rate law expression based on the rate determining step That we found in step two. So our right then is equal to our reaction rate constant times the concentration of sulfur trioxide, which is everything that's shown above. So we actually constant for the forward reaction divided by that for the reverse times the concentration of sulfur dioxide squared times the concentration of oxygen, combining all our reaction rate constants into one single constant. We'll just call that K. Prime times the concentration of sulfur dioxide squared times the concentration of oxygen. And so if we return to problem statement it says, then the experimental rate law is equal to the reaction rate constant, times the concentration of sulfur dioxide squared times the concentration of oxygen. And so that this is what we determined the rate law expression to be. We also see that this is consistent with the format that we would have arrived at from our overall reaction. And so we conclude then that, yes, the mechanism is consistent with the experimental rate law, and this corresponds then to answer a thanks for watching.

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