2.07a Finding Equilibrium Concentrations, part 1

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From the course by University of Kentucky

Advanced Chemistry

316 ratings

University of Kentucky

Advanced Chemistry

316 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

The equilibrium constant for this reaction is 60.2 at 250 kelvin.

If the initial concentrations of hydrogen and iodine are both 0.250,

what is the equilibrium concentration of HI?

And we're also given the balanced equation H2 plus I2 yields 2HI.

Since this is an equilibrium problem, the first thing I want to do is set up my ice

table that gives me a way to summarize the information that's given and

figure out how I'm going to get to the equilibrium concentration to our product.

So I put in I for initial, C for change and E for equilibrium.

When I look back at the problem I see that my initial concentration

of H2 0.0250 my initial concentration of I2 is also 0.250 and

there's no mention of HI being present initially.

And so I assume that concentration initially is 0.

Now I need to look at the change row and to do that I have to look at two things.

One, the stoichiometry of the balance chemical equation and

I have to determine which side is going to be losing substance and

which side is going to be gaining substance.

Because I start with no HI, the only thing I can do on that side is gain.

So that's going to have to be the gain side.

And what I'm going to see is on the reactive side,

I'm going to lose H2 and I'm going to lose I2.

Then I put this in terms of the balance chemical equation,

and I say well if I lose one M of H2 and I represent that as X,

I will also lose one M of I2 and I will gain 2M of HI.

Now I can take my initial row and my change row, and

add them together to get the equilibrium row,

so I end up with 0.250- x for H2,

0.250- x for I2 and 2x for HI.

Now, I want to set up the Kc expression and so I say that Kc

equals the concentration of HI, and

I square that because I have a 2 in my balance chemical equation over

the concentration of H2 times the concentration of I2.

Now I can plug in the values that I know here.

I know my value of KC is 60.2.

I know my equilibrium concentration of HI is

represented with 2x and I have to square that.

Then I'm going to put that over 0.250- x and

this term is also squared because what I see is that the equilibrium

concentration of H2 and I2 are exactly the same.

Because the value of K is so large I can't make any simplifying assumptions.

However, I can make a mathematical simplification

by taking the square root of both sides.

And this works because I have a square on both the numerator and the denominator.

And when I do that, I find that I get 7.76

equals 2x over 0.250 minus x.

And I can use my algebra skills to solve for x, and so I end up with

1.94 minus 7.76x equals 2x.

Then 1.94 equals 9.76x.

And finally, the value of x if equal to 1.99.

Now I'm not quite finished with the problem because I see that the value

of x is 1.99 and that gives me the x that in my ice table, but I'm asked for

the equilibrium concentration of the HI and I noticed that when I look at

my ice table, the equilibrium concentration of HI is equal 2 times x.

So the concentration of HI = to 2x and

therefore the concentration of HI

will be = to 2 times 0.199 and

that will make it equal to 0.398 M.

And these types of moles you can also check them work we can go back if we find

a concentration of HI, and we can substitute the value on xn for

what the HI, and we can solve and we should get a value fairly close to 60.2.

There may be a slight variation due to rounding along the way.

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