Rate/ order question (a2)

The question asked you to find the order with respect to H+;

The concentration of H+ (mol dm-3) and rates are;

Experiment 1- 0.005. / rate- 5.75x10^-6
Experiment 2- 0.005. / rate- 1.15x10^-5
Experiment 3- 0.010. / rate- 2.30x10^-5

I said that H+ was 2nd order because if you double the concentration from 0.005 to 0.010, the rate quadruples from 5.75x10^-6 to 2.30x10^-5?

However the mark scheme says that H+ is zero order so I'm just really confused atm.

The paper is Ocr Jan 2010/ the old spec!

Please can someone help!

You also need to take into account the data from the other components of the reaction. You have not included these in the question.

Ok,
When I- concentration quadruples from 0.010 to 0.040, the rate also quadruples. So would the answer be that as I- is first order, it has caused the rate to quadruple rather than the H+ being involved?

But I still can't see how you would deduce that H+ is zero order!

Ok,
When I- concentration quadruples from 0.010 to 0.040, the rate also quadruples. So would the answer be that as I- is first order, it has caused the rate to quadruple rather than the H+ being involved?

But I still can't see how you would deduce that H+ is zero order!

In the second and third experiment numbers that you listed above, I am guessing that the concentration of iodine also changed (and I'm guessing the order with respect to iodine was first order), so that means that the rate changed due to the change in iodine concentration rather than the change in hydrogen ion concentration. Therefore, you can conclude that the hydrogen ion concentration has no effect on initla rate of reaction --> Order with respect to H+ = 0.

That makes sense, but how would you work out which one caused the rate to change when both concentrations increased?