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 4? Just put in the number as a whole number or fraction?

Respuesta :

Answer:

The answer to this is the rate of the reaction will be reduced by a factor of 16

or  [tex]\frac{1}{16} Rate[/tex]

Explanation:

Reaction rate is influence by the concentration of the reacting reagents. Te rate law is a mathematical relationship that relates the rate of a chemical to the concentration of the the reacting reagents this can be expressed in the form

Rate of a chemical reaction = [tex]k[A]^{m} [B]^{n} ^{2} [C]^{p}[/tex]

In the above rate reaction [A], [B] and [C] are the concentration of the reactants in moles therefore considering the given reaction rate law, we have

Rate = [tex]k[BrO^{-} _{3} ][Br^{-} ] [H^{+} ]^{2}\\[/tex]

If the concentration of the hydrogen ion in decreased by a factor of 4 then we have New Rate = [tex]k[BrO^{-} _{3} ][Br^{-} ] [\frac{H}{4} ^{+} ]^{2}\\[/tex] = [tex]\frac{1}{16}k[BrO^{-} _{3} ][Br^{-} ] [H^{+} ]^{2}\\[/tex] = [tex]\frac{1}{16} (Initial Rate)[/tex]

Hence the rate of the reaction will be reduced by a factor of 16