[Violet89]

Consider the reaction with the rate law, Rate = k[BrO3-][Br-][H+]2

By what factor does the rate change if the concentration of H+ is decreased by a factor of 3? Just put in the number as a whole number or fraction.

Would it be 2/3?

[UG]

Have a think about this one, the rate law tells us that the rate of reaction is proportional to the concentration of H⁺ squared.
That is, rate ∝ [H⁺]². If the concentration of H+ was initially 1 unit, the rate is proportional to 1 squared. Changing the concentration to 3 units would increase the rate by 3 squared and so on. It is exactly the same case for a decrease in concentration.

[toeknee2120]

Have a think about this one, the rate law tells us that the rate of reaction is proportional to the concentration of H⁺ squared.
That is, rate ∝ [H⁺]². If the concentration of H+ was initially 1 unit, the rate is proportional to 1 squared. Changing the concentration to 3 units would increase the rate by 3 squared and so on. It is exactly the same case for a decrease in concentration.

So it this [1/3H⁺]²?

[Borek]

[tex]R_1=k\times(...)\times[H^+]^2[/tex]

[tex]R_2=k\times(...)\times(\frac{[H^+]}{3})^2[/tex]

[tex]\frac {R_1} {R_2} = ...[/tex]