1.08a Activation Energy

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En provenance du cours de University of Kentucky

Advanced Chemistry

414 notes

University of Kentucky

Advanced Chemistry

414 notes

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

À partir de la leçon

Kinetics

The study of chemical kinetics is the study of change over time. It answers questions like: How fast are reactants consumed? How fast are products formed? This unit is dedicated to the exploration of how these questions are answered. We will look at the experimental evidence of how concentration affects these rates. We will also examine what occurs on the molecular level, especially with respect to the motion of molecules, that affects rates of reactions.

1.01 The Rate of Chemical Reactions10:12

1.02 Comparing Rate of Change for Reactants and Products13:52

1.02a Obtaining a Rate Law from Experimental Data Equations4:23

1.03 The Rate Law9:01

1.04 Obtaining a Rate Law from Experimental Data16:38

1.04a Rate Law Calculations8:20

1.05 First-Order Kinetics and the Integrated Rate Law0:00

1.05a Graphic 1st Order7:10

1.06 First-Order Kinetics and the Half-Life9:32

1.07 Second-Order Reactions8:18

1.07a Graphics 2nd Order6:02

1.08 Collision Theory13:13

1.08a Activation Energy3:48

1.09 The Arrhenius Equation15:32

1.09a Graphic Activation Energy2:57

1.10 Reaction Mechanisms12:45

1.10a RDS Fast Equilibrium4:38

1.10b Rate Determining Step2:41

1.11 Catalysis4:23

Rencontrer les enseignants

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this problem, we are given information about temperature, and

information about the rate of reactions.

We are told that the reaction is 4 times faster at 25 degrees Celsius,

than it is at 0 degrees Celsius.

And from this information,

we're going to be able to determine the activation energy.

The Arrhenius equation has the form,

natural log of k2/k1 = negative E

sub a/R (1/T2- 1/T1).

We're not given rate constants but we do know that since it's a first-order

reaction, that the rate at the lower temperature,

I'll call it 1 is equal to the rate constant 1 times the concentration of A.

And if we were to elevate the temperature, the rate would change and

we would have a different rate constant.

Now, assuming that the reactions are taking place with the same concentrations,

the change in rate has gotta be due solely to the change in the K.

Now, if rate 2 as the higher temperature,

is 4 times faster then rate 1,

It is due to, and we can replace the rate

2 with the K2[A] and the rate 1 with K1[A].

What this tells me is that K2, And

K1, all I've done is divided both sides by this portion, those would cancel.

And K2/K1 = 4.

All of this is to say is if the rate is 4 times faster,

then the K's are 4 times larger.

And we can replace K2/K1 with 4.

E sub a is what we're trying to determine.

R, we're going to use the joule one, joule per mole Kelvin.

And we will be sure that we use the temperatures in Kelvin.

So, the K2 is the higher temperature, that would be 298 Kelvin.

The K1 is the lower temperature, it's 273 Kelvin.

So the natural log of 4 (8.314),

Divided by this portion,

which is equal to 3.07

x 10 to the minus 4.

And actually this portion is negative, and when we change the sign it become

positive, so it's a positive 3.07 x 10 to the minus 4.

. The units will be 1 over Kelvin.

That will equal the activation energy.

This gives me a value of 3.75 x 10 to the 4th.

The units will be joules per mole because the Kelvin has cancelled.

That's the activation energy in joules per mole, and more commonly,

activation energy is reported in kilojoules per mole.

We divide by 1,000, we get 37.5 kilojoules per mole.

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