[nhatvy]

I did 3 Experiments and got 3 rate as the following:
Exp1: (S2O8)2-:0.05M; I-:0.05M rate:1.7x10^-5 M/s

Exp2: (S2O8)2-:0.10M; I-:0.05M rate:3.4x10^-5 M/s

Exp3:(S2O8)2-:0.05M; I-:0.10M rate:3.6x10^-5 M/s

what effect does doubling [(S208)2-] have on the rate? By what factor does the rate increase?

what effect does doubling [I-] have on the rate? By what factor does the rate increase?

How do I calculate x, y for the rate law: Rate=K[(S208)2-]^x times [I-]^y? Do I substitute the value of[(S208)2-]and [I-] from each Exp?

How do I calculate the value of the rate constant K for each Exp and the average value?

[Unknown008]

To find what happens to the rate, compare the two values you've got with the concentration of being kept constant, that is compare the two first experiments.

When you double the concentration of S2O8^2-, what happens to the rate? Does it stay the same? Does it double? Triple?

Post what you get! :)

[nhatvy]

double the concentration of S2O8^2- -->the rate doubles.

"By what factor does the rate increase?" Is the answer increasing the concentration of S2O8^2- by a factor of 0.05 increased the rate from 1.7x10^-5 M/s to 3.4x10^-5 M/s? Does "By what factor" mean numbers or factors such as temperature, pressure, catalyst.

According to my lab manual, for Exp 1: rate=K [S2O8^2-]^x [I^-]^y, Exp 2:rate=2^x times Rate 1; Exp 3:rate=2^y times Rate 1, so I find the rate law is Rate=K [S2O8^2-]^0.5 [I^-]^0.5, but the rate law's general form is Rate=K [S2O8^2-]^x times [I^-]^y, so I find the rate law is Rate=K [S2O8^2-] [I^-]. Which one is correct to find x and y??

I find k is 6.8x10^-3 for exp1, 6.8x10^-3, 7.2x10^-3 for exp 2 and 3. Do I add up them, then divided by 3 to find the average value?

[nhatvy]

double the concentration of S2O8^2- -->the rate doubles.

"By what factor does the rate increase?" Is the answer increasing the concentration of S2O8^2- by a factor of 0.05 increased the rate from 1.7x10^-5 M/s to 3.4x10^-5 M/s?

[nhatvy]

According to my lab manual, for Exp 1: rate=K [S2O8^2-]^x [I^-]^y, Exp 2:rate=2^x times Rate 1; Exp 3:rate=2^y times Rate 1, so I find the rate law is Rate=K [S2O8^2-]^0.5 [I^-]^0.5,

[nhatvy]

I find k is 6.8x10^-3 for exp1, 6.8x10^-3, 7.2x10^-3 for exp 2 and 3. Do I add up them, then divided by 3 to find the average value?

[Unknown008]

Hm... don't rush up too much. One thing at a time.

What does double mean?
Double means 2, hence in the rate equation, you get:



x here is equal to 2.

The order of the reaction with respect to S2O8^2- is 2, not 0.05 or 0.5.

Now, post what you get for the second part, that is how the doubling of the concentration of iodide ions affect the rate of the reaction.
~~~~~~~~~~~~~~~~~~~~

Factor here refers to the number of times. Not a catalyst, or a condition for the reaction to occur.

I'm surprised you are doing this question without knowing this... :(

[Fr_Chuck]

You seemed to mis use the comment feature, I have deleted all of the comments, please "answer" to give follow up