Physical Chemistry , 1st ed.

Here, we can see that k 1 is much larger than k 2 (that is, 10^6 is two orders of magnitude higher than 10^8 ), so we might ex ...

constant,k, of the reaction. Using the symbol EAfor activation energy, equa- tion 20.48 becomes   ( ( l 1 n /T k ) )E R A ...

Solution First, we need to determine the values that we will be plotting: they aren’t the values right from the table above! We ...

   ( ( l 1 n /T k ) )  E R A —that is, that this derivative has some constant value given by the expression EA/R—it i ...

atoms are not arranged properly as they collide. In the other case, orientation factors are more favorable, and the reaction mig ...

reaction. However, we can collect experimental evidence to supporta proposed mechanism, or to show that some proposed mechanism ...

Example 20.10 What is the expected rate law of the equation below, assuming that it is an el- ementary process? What is the mole ...

The individual steps in the mechanism are proposed to be Cl 2 →2 Cl ClCH 4 →HCl CH 3  CH 3 Cl 2 →CH 3 Cl Cl ClCH 4 ...

At this point, we need to keep track of what process we are speaking of when we are referring to a rate. For elementary processe ...

A→B (fast) B→C (slow) Here, B represents an intermediate product whose concentration is difficult to determine experimentally. ...

which is a rate law in terms of A, a species whose concentration is knowable (as supposed a few paragraphs above). Now: does thi ...

Notice that this is the same overall rate law that we got when we used the equi- librium constant of the first elementary proces ...

By defining Vand Kas Vk 2 [E 0 ] and K k 1 k  1 k 2  (20.67) we can write the reciprocalof the rate, 1/rate, as  ra 1 te ...

reactions.The reactions d represent a loss of the intermediates that propagate the chain reaction. They are called termination s ...

theory that there is a range of energies for molecules at any given temperature). More formally, an initiation reaction should i ...

will occur. Figure 20.20 shows a diagram of some of those conditions for H 2 /O 2 mixtures. Another example of interesting kinet ...

concentration is low, the intermediate-producing pathway is favored; and when the intermediate’s concentration is high, the inte ...

Both pathways react bromate ion to products, but path I has Brions as a re- actant whereas path II has Brions as a product. Th ...

into account the molecular orientations that are thought necessary to promote a reaction (and so take into account what collisio ...

Thus, if we can determine Kc and k, we can calculate the rate of the elemen- tary process. The rate constant k for the second st ...