Answer to Question #292629 in General Chemistry for Ash

Question

Q1. Catalase is an important enzyme that breaks down its substrate, H2O2, into water and oxygen in nearly all living cells. The enzyme has k2 = 1.1*107 sec-1 and Km = 2.0 mM. An experiment was set up under the conditions when the reaction rate is independent of [H2O2]. In this experiment, at what concentration of catalase (use SI units and scientific notation) will the reaction have a rate of 850 mM/min?

Q2. The degradation of ethyl acetate (EA) is a first order process with t1/2 = 110 min.

a) Fill in the blank: For this reaction, a graph of _______________ vs. time will give a straight line (linear plot).

b) What is the rate constant for the degradation of EA ?

c) What percentage of EA will remain after 440 minutes?

Answer 1

Question 1)

[A]/[n]n=kt.

Where [A] is the concentration of the reactants at time t.

[n] n is the concentration of the reactants at t=0.

K is the rate constant.

Therefore:

1.1×107sec-¹×2.0=850/[n]n.

=850/1.1×107×0.2

Question 2)

a)the graph of degradation of ethyl acetate vs time will give a straight line.

b)t½=110min

K=0.693/110min=6.3×10-³

=0.693/110×60=0.693/6600sec=1.05×10-⁴sec-¹

c)After 4 half lives, percentage of EA will remain

=100%/2⁴

=100%/16

=6.25%