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2.07c Finding Equilibrium Concentrations, part 3

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Química Avançada

Universidade de Kentucky

4.7 (422 classificações) | 51K alunos inscritos

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 (422 classificações)

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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!

Na lição

Chemical Equilibrium

This unit introduces the concept of chemical equilibrium and how it applies to many chemical reactions. The quantitative aspects of equilibrium are explored thoroughly through discussions of the law of mass action as well as the relationship between equilibrium constants with respect to concentrations and pressures of substances. Much of the discussion explores how to solve problems to find either the value of the equilibrium constant or the concentrations of substances at equilibrium. ICE (initial-change-equilibrium) tables are introduced as a problem-solving tool and multiple examples of their use are included. From a qualitative standpoint, Le Châtelier’s principle is used to explain how various factors affect the equilibrium constant of a reaction along with the concentrations of all species.

2.01 Dynamic Equilibrium6:28

2.02 Law of Mass Action7:34

2.03 Law of Mass Action for Combined Reactions4:36

2.04 Relationship Between Kc and Kp6:51

2.04a Relationship Between Kc and Kp2:34

2.05 Calculating the Equilibrium Constant13:09

2.05a Finding Kc5:04

2.06 Reaction Quotient5:55

2.06a Predicting Reaction Progress with the Reaction Quotient1:55

2.07 Calculating Equilibrium Concentrations12:54

2.07a Finding Equilibrium Concentrations, part 14:59

2.07b Finding Equilibrium Concentrations, part 23:20

2.07c Finding Equilibrium Concentrations, part 34:22

2.08 Le Chatelier's Principle Part A9:55

2.08a Le Chatelier's Principle Part B19:50

2.08b How Changes Shift the Equilibrium5:18

Ministrado por

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

Transcrição

Let's look at another equilibrium example.

The equilibrium constant for the given reaction is 0.561 at 800 K.

If the initial concentration of HI is 0.500 M,

what are the equilibrium concentrations of each species?

Now this is an equilibrium problem.

The first thing we need to do is set up an ice table.

So we have I for initial, C for change, E for equilibrium.

We're told we have an initial concentration of HI = 0.500 M.

And since it doesn't mention anything about initial concentrations of I2 and H2,

we're going to assume they're 0.

Now this problem is a little bit different because here,

we're starting with the product,

or something on the right side of the equation, rather than the left side.

It really doesn't matter what we start with because it's an equilibrium reaction,

the process is actually going in both directions.

For our change row, we still have to do it in terms of x.

We'll still be gaining H2, so +x, and gaining I2 so +x there as well.

On the HI, we'll be losing HI, but we have to put -2x.

Because if we look at the coefficients in our balanced chemical equation,

we see that we have a coefficient of 2 in front of the HI.

Our equilibrium row is simply the sum of the initial and change row.

So we have x, x, and 0.500-2x.

Now, we can set up our log mass

action, so K = [HI].

Remember we have to square that because we have

a coefficient of 2, over [H2] and [I2].

Now because all of these are in pressures, we could have done this as a KP or KC.

However, the values were given in terms of molarity, and

our K is given in terms of concentration.

So we want to make sure we do this in terms of molarity.

Now I can substitute in what I know,

I have 0.561 = my HI concentration at equilibrium,

which is 0.500-2x, and then I need to square that.

And on the bottom, I have x and x, so I'm going to write that as x squared.

Now this might the point at which you're saying,

where can I make that simplifying assumption?

And I can't make this simplifying assumption because of the value of K.

However, there is a solution that makes this problem a little bit easier.

Notice that on the fraction, we have both of our terms squared.

That means I can take the square root of both sides.

This will greatly simplify the calculation I need to do.

When I take the square root of 0.561,

I get 0.749.

And what that leaves me with on the right

side is (0.500-2x) / x.

Because I took the square root of everything.

Now I can rearrange, and simplify my

expression a little bit, combine some terms.

And then I can solve for x and found that x = 0.182.

Now, I can use that information to figure out what

my equilibrium concentrations are.

For H2 and I2, my equilibrium concentrations will be just that, 0.182 M.

For HI, I'll have to do a little bit of a calculation,

I have to do 0.500-2x.

And I end up with 0.464 M for

my HI concentration.

So on those problems where you can't make the simplifying assumption,

always look to see if there's a mathematical solution,

which will make the problem a little bit easier.

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