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
To determine the order of the reaction in A and B, we can use the method of initial rates. Let's go through the calculations you provided and check if they are correct.
1) The order of the reaction in A:
The equation you used is correct:
(25.47 / 2.83) = (k * [0.819] * [0.763]) / (k * [0.273] * [0.763])
9 = 3x
However, your next step seems to have an error. It should be:
9 / 3 = x
x = 3
So, the order of the reaction in A is 3.
2) The order of the reaction in B:
The equation you used is correct:
(2.83 / 2.83) = (k * [0.273] * [1.526]) / (k * [0.273] * [0.763])
1 = 2y
Again, your next step seems to have an error. It should be:
1 / 2 = y
y = 0.5
So, the order of the reaction in B is 0.5.
To summarize:
- The order of the reaction in A is 3.
- The order of the reaction in B is 0.5.
Let me know if you have any further questions or if you need additional assistance!
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. :-).
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.