Question

Part 1: Warm Up (12 minutes) Almost every chemical reaction that occurs in a living thing is catalyzed by an enzyme. The purpose of this exercise is for you to: 1) get a better understanding of the principles of catalysis; 2) gain experience and practice drawing and interpreting reaction coordinate diagrams, and 3) developing a better understanding of how enzymes behave as catalysts. Try to answer questions 1-4 without consulting notes, books, etc. Use each other's knowledge 1. With respect to catalysis of chemical reactions, what are some ways that enzyme catalysts differ from small molecule catalysts? 2. In terms of the thermodynamics of chemical reactions, how are enzymes and small molecule catalysts alike? 3. In terms of reaction rates and equilibrium constants, how do catalysts affect chemical reactions? 4. The table below contains some experimental data for a reaction between reagents A and B. The reaction rates were measured in seconds. Rate [A] (MM) [B] (MM) 10 10 100 10 100 15 10 15 | 150 a. What order reaction is this? zero order first order second order Otake-out order b. What would be appropriate units for the rate constant for this reaction? c. What would be appropriate units for the rate of this reaction?

Answer 1

1) Difference;

A) The difference between enzyme and catalyst is that enzyme are organic compounds and are natural bio-catalyst where are molecular catalysts are generally inorganic compounds.

B) Enzymes have high molecular weight whereas molecular catalyst has lower molecular weight.

2) Alike ;

A) Both enzyme and molecular catalysts increase the rate of reaction.

B) Enzymes and chemical catalysts bind their substrates, not permanently, but transiently—for a brief time. There is no action at a distance involved.

3) Catalyst increases the rate of reaction and catalysts don't impact equilibrium constant.

4)

Let rate law is;

r = k * [A]^x [B]^y

Using trial data;

100 = k *(10)^x (10)^y ...(1)

100 = k *(15)^x (10)^y ....(2)

150 = k *(10)^x (15)^y ....(3)

Dividing equation (2) by (1);

100/100 = (15/10)^x

1 = (1.5)^x

(1.5)^0 = (1.5)^x

x = 0

Now, divoding equation (3) by (1),

150/100 = (15/10)^y

1.5 = (1.5)^y

y = 1

A) Thus, order of reaction = x + y = 0 + 1 = 1 ...Answer

B) Unit of rate constant K is second^-1

C) unit of rate = mM/second