Question

to determine the order of a reaction do we multiply the intial reaction of first value with the second one in a given data?

Answer 1

How to determine the order of reaction in respect to each reactant species:

1 0.10 M 0.10 M 1.23 x 10-3 M/s
2 0.10 M 0.20 M 2.46 x 10-3 M/s
3 0.20 M 0.10 M 4.92 x 10-3 M/s
4 ? 0.30 M 8.00 x 10-3 M/s

Answer 2

thanx

Answer 3

Let's first write the general rate equation:\[Rate=k[A]^m[B]^n\]Let's ignore k for now. Since it's constant, it won't affect rates based on concentrations of reactants.\[Rate=[A]^m[B]^n\]In the case of the second row of the table, [A] is constant with respect to the first row, so we can ignore that too. We are doubling [B] and by doing that, the rate is doubled, so:\[2(Rate)=(2[B])^n \rightarrow 2(Rate)=2^n[B]^n\]Rate and [B] are constant for this particular reaction, so we can drop them off:\[2=2^n\]We can see that n=1 here.

Now let's look at the 1st and 3rd rows and go through a similar process. k and [B] are constant with respect to the two rows, so:\[Rate=[A]^m\]When the concentration is doubled, the rate is quadrupled:\[\frac{4.92 \times 10^{-3}}{1.23 \times 10^{-3}}=4\]So we can write the rate equation as:\[4(Rate)=(2[A])^m \rightarrow 4(Rate)=2^m[A]\]Since Rate and [A] are constant for this particular case, we can drop them off and we are left with:\[4=2^m\]Which leaves "m" to be what?

Answer 4

2

Answer 5

so to find the order of a reaction in a given data we have to choose any two experiments?

Answer 6

Yes, and then the final rate reaction equation is:\[Rate=k[A]^2[B]\]

Answer 7

after taking any two rows we use this formula

Answer 8

Yes, you need to choose two experiments where one of the concentrations is constant.

Answer 9

or equation
rate =k [a][b]

Answer 10

For example, you can't determine the powers if you have, for instance, these two sets of data:

1 0.10 M 0.10 M 1.23 x 10-3 M/s
2 0.20 M 0.20 M 9.84 x 10-3 M/s

In this case, it is impossible to determine how the concentrations of each reactant affect the rate because there are no constants to compare to.

Answer 11

You'll always compare two experiments where at least one variable is constant between the two, whether it's [A], [B], or rate.

Answer 12

oh so in ur given example which is constant?

Answer 13

No variables are constant. The orders of reaction can't be determined by just that data.

Answer 14

ahan

Answer 15

my teach used the log method to find the order

Answer 16

If you use the previous method described:\[Rate=k[A]^m[B]^n\]k is constant, so we drop it.\[Rate=[A]^m[B]^n\]If we double both concentrations, we see that the rate is multiplied by 8:\[8(Rate)=(2[A])^m(2[B])^n \rightarrow 8(Rate)=2^m2^n[A]^m[B]^n\]Drop off the variables to get:\[8=2^m2^n\]To which there are an infinite number of solutions.

Answer 17

thats easier than the log method

Answer 18

so this is the order of the reaction.

Answer 19

Yeah, it can be easier. The log method is basically a shortcut of doing the calculation once you understand the concept behind it.

Answer 20

now how can we find the overall order?

Answer 21

becuase we found the order of the reaction not each reactant

Answer 22

Well, we found this to be true, right?\[Rate=k[A]^2[B]\]

Answer 23

ohh so we found the order of A

Answer 24

And the order of B. It was just implied as 1:\[Rate=k[A]^2[B]^1\]

Answer 25

2+1 = 3 overall order is 3

Answer 26

Yes.

Answer 27

so every time we use log or any method to find the order we are actually finding the order of A only.

Answer 28

how B is 1? same method?

Answer 29

Yes. Same method. We look at the original table:

1 0.10 M 0.10 M 1.23 x 10-3 M/s
2 0.10 M 0.20 M 2.46 x 10-3 M/s
3 0.20 M 0.10 M 4.92 x 10-3 M/s
4 ? 0.30 M 8.00 x 10-3 M/s

And we choose the two rows in which B changes, so we can see how the rate varies with B. These are rows 1 and 2. The general rate equation:\[Rate=k[A]^m[B]^n\]Drop off k and [A]^m because they are constant between rows 1 and 2:\[Rate=[B]^n\]When we double [B], we see that the rate doubles as well, so:\[2(Rate)=(2[B])^n \rightarrow 2(Rate)=2^n[B]^n\]Drop Rate and [B]^n:\[2=2^n \rightarrow n=1\]

Answer 30

oh so in this reaction we did not find the value of A because it was constant

Answer 31

we only find the order of B becuase it was not constant?

Answer 32

I mean the order of A.

Answer 33

so the order which we find out , is it the order of B ?

Answer 34

Yes. It is impossible to determine the order of A only using rows 1 and 2. We were able to find the order of [B] by knowing how rate changes with [B].

Answer 35

becuase A was constant then we only check the order of B. But in case if we take the 3 and 1 in which the rate is changing in A then what ?

Answer 36

It's what I outlined before. The third post in the question. My second post.

Answer 37

so even if I take row 1 or 2 or row 3 and 1. I will come up with the same ordeR?

Answer 38

Yes. As long as you take into account which values are constant when you're working with your equations.

Answer 39

ok. U mean between A or B atleast one value has to be constant

Answer 40

Yes. Otherwise you end up with an equation like this (which I posted before), which cannot be solved because there are two unknown variables which are not functions of eachother:\[8=2^m2^n\]

Answer 41

ook . u can see that there is a question mark in the table. which means the last initial concentration of A is missing ? How will you find that? what equation applies here?

Answer 42

Since we've found the orders of A and B, we know that:\[Rate=k[A]^2[B]\]We can choose a specific row on the table where we know rate, [A], and [B] and that leaves only k to be solved for. So go ahead and solve for k by choosing any row -- it doesn't matter which one you choose. They will all give the same answer.

Answer 43

I choosed 1st row and I came up with 25.92

Answer 44

into 10^-2

Answer 45

I applied the equation k = rate/[a][b]

Answer 46

I get 1.23(M^-2)(s^-1)

Answer 47

ohh yes yes . sorry .

Answer 48

\[Rate=k[A]^2[B] \rightarrow k=\frac{Rate}{[A]^2[B]} \rightarrow k=\frac{1.23 \times 10^{-3}\frac{M}{s}}{(0.10M)^2(0.10M)}=1.23\frac{1}{M^2s}=1.23M^{-2}s^{-1}\]

Answer 49

yes yes.

Answer 50

now how will you find the intial rate of A which is missing in the table

Answer 51

Back to the rate equation:\[Rate=k[A]^2[B]\]From the 4th row of the table, we know everything except [A], so we can calculate it, yes?

Answer 52

yes

Answer 53

And we'll get?

Answer 54

so we can calculate the 8.00/3.0?

Answer 55

\[Rate=k[A]^2[B] \rightarrow [A]=\sqrt{\frac{Rate}{k[B]}}\]We know rate and [B] for row 4 and we just calculated k. Plug the numbers in an go.

Answer 56

so where the sq root came from?

Answer 57

Algebraic manipulation of the rate equation:\[Rate=k[A]^2[B] \rightarrow \frac{Rate}{k[B]}=[A]^2 \rightarrow \sqrt{\frac{Rate}{k[B]}}=[A]\]Make sense?

Answer 58

ohh . so I just add the values and that will be my missing rate

Answer 59

I didnot get it ?

Answer 60

In the above equation do I have to add any values?

Answer 61

You aren't missing a rate, you are missing a concentration. You don't add any values, you just substitute the values that you know from row 4 of the table, which are rate, [B], and we've calculated k above. Plug those numbers in and solve for [A].

Answer 62

rate of the same row in which the A is missing?

Answer 63

Yes.

Answer 64

yes I found it .it is 0.215

Answer 65

0.021

Answer 66

am I right?

Answer 67

I'm getting 0.147M.

Answer 68

do I ALSO have to take under root of 0.021

Answer 69

Yes. 0.021 is [A]^2

Answer 70

ok .. Thanx alot u taught me a great concept.

Answer 71

how will you find the rate if its missing?

Answer 72

You would need to be given the concentrations of both reactants, and you could just use the rate reaction equation in its standard form. Plug in k and both concentrations to get rate:\[Rate=k[A]^2[B]\]

Answer 73

ok.

Answer 74

when we calculate the log why it came up with 0?

Answer 75

The log of what?