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1.05a Graphic 1st Order

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Chimie avancée

Université du Kentucky

4.7 (419 notes) | 50K étudiants inscrits

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.

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4.7 (419 notes)

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KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

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

Enseigné par

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

Transcription

This problem seems fairly lengthy and it has lots of parts, but

we're going to learn a lot about chemical kinetics by working through this problem.

First, let's look at the reaction that's given.

It says it's a decomposition reaction so

I'm going to write the reaction up here in this space.

It's SO2Cl2 decomposing, that means it's breaking apart.

And it gives me the products since breaking apart in two.

Both gases, and it's balanced as written.

We're wanting to figure out the order of this reaction, and many times students

will look at a overall reaction and think that they can get the order from that, but

you cannot determine an order by looking at a reaction.

You have to experimentally determine the reaction, and

that is what is taking place here.

They monitored this reaction at a certain temperature.

Now we're not told what the temperature is.

But it's some temperature they're monitoring SO2CL2 as a function of time,

meaning as time goes by, they're taking readings of how much SO2CL2 is present and

they found that when they plotted

the natural log of that concentration versus time, they got a straight line.

So along the y-axis is a natural log of SO2 Cl2, and

along the x-axis is time in seconds.

And they tell me it is a linear plot so it's a straight line.

This is a slope intercept form of a line so it's y equals mx plus b.

We are seeing right here the slope.

It's a negative slope so I want to draw my line in this direction, and

the slope of this line is a -3.10.

Times 10 to the minus 4 and I'm going to give it units.

It's changing Y over change in X to get a slope.

Y has no units but X is second so the units of that slope is 1/seconds.

So all the information in the first half of this problem is obtained and

given in that little graph.

Now we can start answering the questions.

The first question is what is the order of the reaction?

Any time the natural log of concentration versus time gives a linear plot,

this tells me it is first order.

If we know it's first order, we can write the rate law.

Ray is equal to K.

Reactant S02CL2 to the 1st power.

Question number two says what is the rate constant.

Well that is obtained from the slope.

The slope of the line is equal to -K.

So K will equal 3.10 times 10 to the minus 4.

Rate constants are always positive, and that is the value of that rate constant.

So that's question two.

Now let's move on to question number three.

It starts off by saying at this temperature.

The reason they say that is because k changes with temperature.

So if we want to know something about this reaction and we want to use this k

that we have here we better make sure we're running it the same temperature.

We're given time information and concentration information and

that tells me that I would like to use the integrated law for first order reaction.

There would be the natural law that the concentration of SO2 Cl2 at same time

divided by concentration of SO2 Cl2 Initially is equal to minus k times time.

Now let's plug in what we know.

We know, we don't know this that's what we're trying to determine.

But we know we're starting with 0.2 molar.

The value of k was 3.10 times 10 to the minus 4 1 over seconds.

And the time is five minutes, but I'm going to need to convert that five minutes

to seconds because the rate constant was in seconds.

So sixty seconds is the same as one minute.

The right hand side of this equation is equal to a negative zero point 0 930.

And that will have no units.

And that's equal to the natural log of X over 0.200 molar.

Trying to solve for X, so I'm going to take e to both sides.

That will cancel that part and

have X over 0.200 m is equal to 0.911.

If I multiply both sides by 0.200,

I will have X equal to 0.180 molar so

this is a concentration of SO2 CO2 after five minutes has gone by.

Part four, part four we want to know the concentration

of the products at five minutes.

Now a lot of times students will look at the balanced equation and

say well they all have the same coefficients so

their concentrations all have to be the same at the end.

Well that is not true we know that

however much of SO2 reacts that's how much of SO2 and

CO2 are produced and 0.180 is not how much reacted it's how much is still there.

So let's write the reaction with some room to work under it to explain how

you will figure out the amount of the s02 and the sl2 that are produced.

Let's think about this reaction at the start.

At the start of the reaction we were told that we had 0.200 molar of this.

And of course there's not going to be any product.

After 5 minutes has gone by,

we learned from the previous problem that we have 0.180 molar still there.

So right here let's think about how much reactant or

formed, depending on whether you're talking about reactants or products.

Now, if I start with 0.2 and I finish with 0.18,

did I not have to use up .020 molar?

I'm going to use up that much, think about that.

If I started with 0.2 and finish with 0.18, 0.02 had to of reacted.

Now this is how much actually reacted; so we can go up to this balanced equation

now and say every time one of these reacts I produced one of these.

But one of these didn't react, 0.02 did.

So I have to produce 0.02 of this.

Same thing for the Cl2, the amount of Cl2 matches the amount of SO2 so

it is also 0.020 and those are all in molar.

So, if I start with none and I added 0.02 then,

after five minutes these are the amounts of the SO2 and

the Cl2 that get produced at the end of the reaction.

So now we know how much of this remains or is produced once the reaction takes place.

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