i posted this question a few minutes ago i tried to solve it on my own can someone check if my work is okay?
The data in the table below were obtained for the reaction:
A + B → P
3 Experiments
1 (A) (M): 0.273 (B) (M): 0.763 Initial Rate
(M/s): 2.83
2 (A) (M): 0.273 (B) (M): 1.526 Initial Rate
(M/s): 2.83
3 (A) (M): 0.819 (B) (M): 0.763 Initial Rate
(M/s): 25.47
1) The order of the reaction in A is __________.
25.47 k[0.819][0.763]
----- = ----------------
2.83 k [0.273][0.763]
9 = 3x
log of each:
0.95 = x
-----
0.48
2=x
2) The order of the reaction in B is __________.
2.83 k[0.273][1.526]
---- = ------------------
2.83 k[0.272][0.763]
1=2y
log
0
---=y
0.3
0=y
6 answers
also, there is one more question that is asking me what the magnitude of the rate constant is how do i figure that out?
Jack, your work is confusing because = 1=2y you actually mean 1 = 2^y. Written that way, y = 0 and you are correct. However, for the x, I think you have an error.
What you have written is
9=3x. I'm sure you mean
9 = 3^x in which case x = 3 since 3 squared is 9.
By the way I worked the y part and told you how to do the x part in your earlier post.
To calculate k, take any one of the experiments, write it as
rate = k*(A)^3 and substitute the concn and rate from that particular experiment. Solve for k. Why didn't I include (B)^0 in the equation above? Because B^0 is 1 and it just clutters the equation. :-).
What you have written is
9=3x. I'm sure you mean
9 = 3^x in which case x = 3 since 3 squared is 9.
By the way I worked the y part and told you how to do the x part in your earlier post.
To calculate k, take any one of the experiments, write it as
rate = k*(A)^3 and substitute the concn and rate from that particular experiment. Solve for k. Why didn't I include (B)^0 in the equation above? Because B^0 is 1 and it just clutters the equation. :-).
oh, im sorry our teacher told us to use log in problems like these he didn't mention anything about exponents.
but my book says x=2 ...?
First, Jack, let me say you're right but your reasoning is wrong. My mistake, and you probably should have jumped all over it, is I said 9 = 3^x; therefore, x had to be 3. Nuts! 3^3 = 27 so it CAN'T be 3. If 3^x = 9, then x must be 2 because 3 squared = 9. So you'r right, x = 2. To do the k thing, set rate = k*(A)^2, substitute concn A and rate from any of those experiments, and solve for k.
Now, when your prof told you to use logs, this is what s/he meant. The rate equation is to be set up like this.
(24.7/2.83) = (0.819/0.273)^x (Note your equation has no x in it. You added it later out of the thin blue sky.)
9.0 = 3^x
If we take the log of both sides we get this.
log 9.0 = x*log 3
0.954 = x*0.477
x = 0.954/0.477 = 2
However, when it is simple, like 9.0 = 3^x, most people (except me :-)) can see that x must be 2 and they don't need to resort to logs.
I hope this has been helpful.
Now, when your prof told you to use logs, this is what s/he meant. The rate equation is to be set up like this.
(24.7/2.83) = (0.819/0.273)^x (Note your equation has no x in it. You added it later out of the thin blue sky.)
9.0 = 3^x
If we take the log of both sides we get this.
log 9.0 = x*log 3
0.954 = x*0.477
x = 0.954/0.477 = 2
However, when it is simple, like 9.0 = 3^x, most people (except me :-)) can see that x must be 2 and they don't need to resort to logs.
I hope this has been helpful.
thanks for clearing it up, i understand it now.
0 answers