Question

find the order of a reaction with respect to H^+ ion?

Answer 1

You have to do it experimentally or by graphs to find the order of a reaction

Answer 2

Do you have the rate equation?

Answer 3

Example: \[rate=k[H^+]^x\]

Answer 4

I have a data of experiment with three values of H . which one I have to put?

Answer 5

So you have 3 experimental values, 3 rate values, so,
\[\frac{Rate~of~Experiment_1}{Rate~of~Experiment_2}=\frac{k[H^+]_1^x}{k[H^+]_2^x}\]
[
[H+] at the top is for concentration of H+ for experiment 1
[H+] at the bottom is for concentration of H+ for experiment 2
k will cancel out, then find x from there

Answer 6

can I take any two experiments from the data?

Answer 7

so I have to put the initial concentration of H ion?

Answer 8

You have to, the equation I wrote above requires 2 experiment results

Answer 9

so every time I have to find the order of a reaction this is how I have to do it?

Answer 10

if there are 4 values in an experimental data which 2 values I should choose?

Answer 11

Yes, but if you were to find two orders of reactions at the same time, then
\[\frac{Rate~of~Experiment_1}{Rate~of~Experiment_2}=\frac{k[X]^x[Y]^y}{k[X]^x[Y]^y}\]
then, find one of the concentration that will cancel out either X or Y concentration, then you will be able to solve for Y.

Example,
[X1]=0.1 [Y1]=0.5
[X2]=0.1 [Y2]=0.2
From here you can see that X will cancel out, leaving concentrations of Y, from here, use the concentrations of Y and solve for the order of reactions.

Answer 12

You should "not" choose both same values,
Example,
[X1]=0.1 [Y1]=0.2
[X2]=0.1 [Y2]=0.2
Then all will cancel out and nothing to solve for

Answer 13

I have to find the order of H ion only

Answer 14

after that I have to find the order of I ion and then overall order

Answer 15

if there's two ions, use
\[\frac{Rate~of~Experiment_1}{Rate~of~Experiment_2}=\frac{k[H^+]^x[I^-]^y}{k[H^+]^x[I^-]^y}\]
then, find one of the concentration that will cancel out either H+ or I- concentration, then you will be able to solve for either x or y.

Answer 16

If you have to find H ion then find any experiments that has same values of I- so that I- will cancel each other leaving you with
\[\frac{Rate~of~Experiment_1}{Rate~of~Experiment_2}=\frac{[H^+]^x}{[H^+]^x}\]

Answer 17

ok the value of H is 0.40 and 0. 20 and I has 0.0010 and 0.0010 .. correct

Answer 18

?

Answer 19

Hmm can you construct the table here? using the draw tool or attach a file, because it will be confusing to do it like this.