2.05a Finding Kc

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Do curso por University of Kentucky

Química Avançada

316 ratings

University of Kentucky

Química Avançada

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.

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

Conheça os instrutores

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

In this problem, we're asked to find the Kc or the equilibrium constant for

this particular reaction.

We're given some information about the initial concentrations of our reactants,

H2 and I2, and

information about the equilibrium concentration of HI, our product.

So we know this is an equilibrium process because we see we have our equilibrium

arrow here.

We know that because it's an equilibrium reaction,

we are not going to completely convert from our reactants to our products.

So what we need to find is, what are the equilibrium concentrations of the H2 and

I2, so that we can solve for the value of Kc.

For equilibrium problems, the way that we kind of organize and

sort our data is through the use of an ICE table.

And I stands for initial, change, and equilibrium.

We can also set up a column here that says H2, I2, and HI.

So now we can take the information given in our problem and

sort it into this table.

So the first thing I notice is that my initial concentrations of H2 and

I2 are both 0.100, so I can put this in.

And in this case, they happen to be the same, they don't have to be.

Sometimes, you'll see different values.

We also see that there's no mention of our initial concentration of H!, so

we can assume that that is 0.

The next thing we need to look at is our change row, and

we do this in terms of x, so I know that if I lose some x amount of H2,

I'm going to lose the same amount of I2.

And I'm going to gain twice as much of my HI because as we lose reactants,

we're going to products.

And if I look back at the stoichiometry of my equation,

what I see is that it's a 1 to 1 to 2 ratio in my balanced chemical equation.

Now I can look at my equilibrium row, and

all I need to do is sum the initial and change rows.

So my equilibrium concentration of H2 is 0.100-x.

It's the same for I2, and for HI, it's equal to 2x.

So now I can go back and say, well, wait a minute, in the initial problem,

it gave me what the actual equilibrium concentration of HI was.

So I can use that, along with my value of 2x, to determine my value of x.

So I can say HI is, at equilibrium,

is equal to 0.134, and it's also equal to 2x.

And so I can solve for x, and find that it equals to 0.0670.

Now I can use that value of x to figure out what my

equilibrium concentrations are for H2 and I2.

So let's set that up to solve for those, and so, I look back at my ICE table.

I look at the equilibrium row, and I see that the concentration

of H2 at equilibrium was equal to 0.100-x.

I substitute in the value I have for x,

which is 0.0670, and what I find is that

my equilibrium concentration of H2 is 0.033 M.

I can do the same thing for I2, and I'm actually going to get the same value

because I have the same initial concentration.

And I'll also have, they're both losing x, so

that value is going to be the same as well.

And I get my equilibrium concentration of I2 is equal to 0.033 M.

Now what I can do is I have all of my equilibrium concentrations

of my substances.

So now I can actually solve for the value of Kc,

and I do that by setting up my law of mass action.

So Kc = [HI] squared,

because I have a coefficient of 2 here.

So I have to put that in to my law of mass action.

And I'm going to be dividing by the concentration of H2 times

the concentration of I2.

And because the coefficients of both of these are 1,

I'm not going to have a power on either of those numbers.

So now I can substitute in my concentration.

So Kc equals my equilibrium concentration of HI,

which is 0.134, I need to square that.

And on the bottom, I’m going to have my concentration,

(0.033)(0.033).

And those are for the concentrations of H2 and I2.

And when I solve this, I end up with a value of 16 for

the equilibrium concentration.

And this tells me that I'm going to have more products favor because my K

value is greater than 1.

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