Question

The rate law for the reaction between catalase and hydrogen peroxide can be written as Rate = k [H2O2]x [catalase]y. A catalase solution of unknown, but constant concentration is prepared. Can the partial order with respect to the peroxide (x) be found using this catalase solution? If so, how would you do it? Describe in detail the steps you would take or show a calculation.

Answer 1

You could use a "pseudo" first order rate equation concept.

Here's the idea.. let's say we have this reaction
.. 1 A + 1 B ---> 1 AB

for every 1 A consumed, we consume exactly 1 B.. right?
so.. our rate equation will look like this
.. rate = k * [A]^x * [B]^y

and as we consume A and B (let's say we consume Z A's) it will look like this
.. rate = k * [A - Z]^x * [B - Z]^y

now
.. IF.. [B] >>>> [A]...
.. THEN [B - Z] ≈ [B]
(because very little of the total amount of B was actually consumed)

let's think of an example.. let's say
.. [A] = 10 mol / L
.. [B] = 1000 mol / L
if we consume 9 mol / L of A, we'll consume 9 mol / L of B as well (see the balanced equation)
so that the final concentration of A and B will be
.. [A] = 10 - 9 = 1 mol / L
.. [B] = 1000 - 9 = 991 mol / L

and we can see that the concentration of A dropped significantly... by an order of magnitude. 10x
BUT.. the concentration of B barely changed. 1000M to 991M that's a 0.9% change

and that's the idea. if [B] >>>> [A], we assume [B] is essentially a constant!

*********
this problem..
.. rate = k * [H2O2]^x * [catalase]^y

is approximately .. (assuming [catalase] >>> [H2O2])
.. rate = k * [H2O2]^x * constant

which can be rewritten like this
.. rate = k' * [H2O2]^x

**********
so we run an experiment like this
.. run #.. .. [H2O2].... [Catalase].. . rate (measured)
.. .. 1.. .. . ..0.1M.. .. . .. 10M.. .. . ... rate1
.. .. 2.. .. . .0.05M.. .. . . 10M.. .. . .. .rate2

now we know that
.. rate1 = k' * [H2O2 #1]^x
.. rate2 = k' * [H2O2 #2]^x

dividing
.. (rate1 / rate2) = (k' / k') * ([H2O2 #1] / [H2O2 #2])^x

canceling k/k.
.. (rate1 / rate2) = ([H2O2 #1] / [H2O2 #2])^x

and we can then sub in the data
.. (rate1 measured / rate2 measurement) = (0.1M / 0.05M)^x
.. (rate1 measured / rate2 measurement) = (2)^x

and we can readily then determine "x"

**********
now we know "x" in this rate equation
.. rate = k * [H2O2]^x * [catalase]^y

and all we need is "k" and "y". So to determine "y", we do 2 more runs of our reaction with a diluted [catalase].
Let's say we dilute it say by a factor of 100 and 200 to finish out our experiment

.. run #.. .. [H2O2].... [Catalase].. . rate
.. .. 1.. .. . ..0.1M.. .. . .. 10M.. .. . ... rate1
.. .. 2.. .. . .0.05M.. .. . . 10M.. .. . .. .rate2
.. .. 3.. .. . .0.05M.. .. . . 0.1M.. .. . .. .rate2
.. .. 4.. .. . .0.05M.. .. . . 0.05M.. .. . .. .rate2

and then
.. rate3 = k * [H2O2 #3]^x * [catalase #3]^y
.. rate4 = k * [H2O2 #4]^x * [catalase #4]^y

dividing
.. (rate3 / rate4) = (k/k) * ([H2O2 #3] / [H2O2 #4])^x * ([catalase #3] / [catalase #4])^y

this time we hold [H2O2] constant.. 0.05M.. so that term drops out (along with k/k) leaving
.. (rate3 / rate4) = ([catalase #3] / [catalase #4])^y

and subbing in the data yields "y"
.. (rate3 measurement / rate4 measurement) = (0.1M / 0.05M)^y
.. (rate3 measurement / rate4 measurement) = 2^y

**********
now we know x and y in this equation
.. rate = k * [H2O2]^x * [catalase]^y

pick the data from run 3 or 4 and plug it in to get "k"