2.07b Finding Equilibrium Concentrations, part 2

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From the course by University of Kentucky

Advanced Chemistry

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University of Kentucky

Advanced Chemistry

388 ratings

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.

From the lesson

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

Meet the Instructors

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

Let's look at an example dealing with pressures, instead of concentrations.

If I2 has an initial pressure of 1.00 atm,

what are the equilibrium pressures of I2 and I?

We're given our balanced chemical equation and the value of Kp.

Just like we have for other equilibrium problems, we're going to set up

an ice table, I for initial, C for change, and E for equilibrium.

I know my initial pressure of iodine is 1.00.

There's no mention of atomic iodine.

So I'm going to assume that the pressure of that is 0.

Now for the change row,

I have to use my coefficients from my balanced chemical equation, and

in doing so, I'll see I'm going to lose some of the I2 and I'm going to gain I.

But I'm going to gain 2 times the amount of I because of the 2 in my balanced

chemical equation.

For my equilibrium, I simply add up the initial and

change rows, so I have 1.00- x and 2x.

So now I have my equilibrium values with respect to x,

my equilibrium pressures with respect to x.

Now I can set up my KP expression,

so I know that KP = PI,

elemental iodine, squared / PI2.

Now I can substitute in the values that I know.

I have 2.91 x 10 to the -4th for my KP value.

That's going to be equal to the pressure of iodine, which note that is 2x.

I still have to square that because the pressure is 2x.

The law of mass action tells me that whatever that term is,

it must also be squared.

Note that the squared does go outside the parenthesis, so

that we square both the 2 and the x.

That's a common mistake that students make.

And on the bottom, we have the pressure by 2, which is 1.00- x.

And we're going to, again, make our simplifying assumption that x is much,

much less than 1.00.

And when we do, we can write out 2.91

x 10 to the -4th = 4x squared.

We could show the 1 in the denominator, but

it doesn't change the value in our problem, so we're going to leave it out.

Now I can divide both sides by 4 and take the square root.

And when I do that, I find that x

= 8.53 x 10 to the -3rd.

Now I can use this value of x to find my equilibrium pressures

by substituting this value m back into what I have in my equilibrium row.

And what I find is the equilibrium pressure of I2

is equal to 0.991, and the equilibrium

pressure of elemental iodine is 0.0171.

And these are both in units of atmospheres.

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