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- [Instructor] An equilibrium constant can be calculated

from experimentally measured concentrations

or partial pressures of reactants and products

at equilibrium.

As an example, let's look at the reaction

where N2O4 in the gaseous state turns into 2NO2

also in the gaseous state.

And let's say we do an experiment

and we allow this reaction to come to equilibrium

and the temperature is 100 degrees Celsius.

And at equilibrium,

the concentration of NO2 0.017 molar

and the concentration of N2O4 is 0.00140 molar.

To calculate the equilibrium constant for this reaction

at 100 degrees Celsius, we first need to write

the equilibrium constant expression.

We can write the equilibrium constant expression

by using the balanced equation.

We start by writing the equilibrium constant,

which is symbolized by K.

And since we're dealing with concentrations,

we're calculating Kc.

And Kc is equal to, we do products over reactants.

So this would be the concentration of NO2.

And since there is a coefficient of two in front of NO2,

this is the concentration of NO2 raised to the second power

divided by the concentration of our reactant, N2O4.

And since there's an implied one in front of N2O4,

this is the concentration of N2O4 raised to the first power.

Next, we plug in our equilibrium concentrations.

So the equilibrium concentration of NO2 is 0.0172.

So let's plug that in.

So this is equal to 0.0172 squared

divided by the equilibrium concentration of N2O4,

which was 0.00140.

So we plug that in as well.

So 0.00140.

When we solve this, we get that Kc is equal to 0.211,

and this is at 100 degrees Celsius.

It's important to always give the temperature

when you're giving a value for an equilibrium constant,

because an equilibrium constant is only constant

for a particular reaction at a particular temperature.

And it's also important to note

that the equilibrium constant doesn't have any units.

So we would just say that Kc is equal to 0.211

at 100 degrees Celsius for this particular reaction.

Let's calculate the equilibrium constant

for another reaction.

In this reaction, carbon dioxide reacts with hydrogen gas

to produce carbon monoxide and H2O.

And since everything is in the gaseous state,

experimentally, it's easier to work with partial pressures

than it is to work with concentrations.

So instead of calculating Kc,

we're gonna calculate Kp or the p stands for pressure.

So we're trying to find Kp at 500 Kelvin for this reaction.

To help us find Kp, we're going to use an ICE table

where I stands for the initial partial pressure

in atmospheres, C stands for the

change in the partial pressure, also in atmospheres

and E is the equilibrium partial pressure.

Let's say that a mixture of carbon dioxide,

hydrogen gas and H2O are placed

in a previously evacuated flask

and allowed to come to equilibrium at 500 Kelvin.

And let's say the initial measured partial pressures

are 4.10 atmospheres for carbon dioxide,

1.80 atmospheres for hydrogen gas

and 3.20 atmospheres for H2O.

And since we didn't add any carbon monoxide

in the beginning, the initial partial pressure

of that would be zero.

And after the reaction comes to equilibrium,

we measure the partial pressure of H2O

to be 3.40 atmospheres.

So that's why we have 3.40 in the equilibrium parts

on the ICE table under H2O.

So the initial partial pressure of H2O is 3.20 atmospheres

and the equilibrium partial pressure is 3.40.

So H2O has increased in partial pressure.

We can go ahead in here and write plus X

for an increase in the partial pressure of H2O

and 3.20 plus X must be equal to 3.40.

So X is equal to 0.20.

So the partial pressure of water increased by 0.20.

And we could either write plus X in here on our ICE table,

or we could just write plus 0.20.

Now that we know that change in the partial pressure

for H2O, we can use this information

to fill out the rest of our ICE table.

For example, the mole ratio of carbon monoxide

to H2O is 1:1.

So if we gained plus 0.20 for H2O,

we're also gonna gain plus 0.20 for carbon monoxide.

And if we're gaining for our two products here,

the net reaction is moving to the right

to increase the amount of products,

which means we're losing reactants.

And we can figure out by how much by looking

at the mole ratios again.

So for both of our reactants,

we have ones as coefficients in the balanced equation.

So if it's plus X for both of our products,

it must be minus X for both of our reactants.

And since X is 0.20, it'd be minus 0.20

for the change in the partial pressure

for both of our reactants.

Therefore the equilibrium partial pressure of carbon dioxide

would be 4.10 minus 0.20, which is 3.90

and for H2, it'd be 1.80 minus 0.20, which is 1.60.

And for carbon monoxide, we started off with zero

and we gained positive 0.20.

Therefore the equilibrium partial pressure is 0.20.

So as the net reaction moved to the right,

we lost some of our reactants

and we gained some of our products

until the reaction reached equilibrium

and we got our equilibrium partial pressures.

In our equilibrium, the rate of the forward reaction

is equal to the rate of the reverse reaction

and therefore these equilibrium partial pressures

remain constant.

Now that we know our equilibrium partial pressures,

we're ready to calculate the equilibrium constant Kp.

So we need to write an equilibrium constant expression.

So Kp is equal to, we think about products over reactants.

And for our products,

we would have the partial pressure of carbon monoxide.

And since the coefficient is a one

in front of carbon monoxide in the balanced equation,

it would be the partial pressure of carbon monoxide

raised to the first power times the partial pressure

of our other product, which is H2O.

And once again, the coefficient is a one.

So that's the partial pressure raised to the first power.

All of this is divided by,

we think about our reactants next,

and they both have coefficients of one

in the balanced equation.

So it would be the partial pressure of carbon dioxide

times the partial pressure of hydrogen gas.

The partial pressures in our equilibrium constant expression

are the equilibrium partial pressures,

which we can get from the ICE table.

So the equilibrium partial pressure of carbon monoxide

is 0.20, the equilibrium partial pressure of H2O is 3.40.

We can plug in the equilibrium partial pressures

for carbon dioxide and the equilibrium partial pressure

for hydrogen gas as well.

And here we have the equilibrium partial pressures

plugged into our equilibrium constant expression.

And when we solve this,

we get that Kp is equal to 0.11 at 500 Kelvin.