- [Instructor] The reaction quotient
is symbolized by the capital letter Q.
And it tells us whether a reaction is at equilibrium or not.
If the reaction is not at equilibrium,
it also allows us to predict
which direction the net reaction will go
to reach equilibrium.
For example, let's look at the hypothetical
reaction where A gas turns into gas B.
Gas A will be represented by red circles or red spheres,
and gas B will be represented by blue spheres.
The equilibrium constant for this hypothetical reaction
is equal to 3 at 25 degrees Celsius.
Let's start by writing out the expression
for the reaction quotient.
So we would write out here QC is equal to,
and this has the same form
as the equilibrium constant expression.
So we would put concentration of B to the first power,
divided by the concentration of A, also to the first power.
Let's look at our first particulate diagram here,
and let's think about each particle
representing 0.1 moles of a substance.
And the volume of the container is one liter.
So let's first find the concentration of B
so we can plug it into our expression for Q.
B are the blue spheres.
So we count up one, two, three blue spheres.
Each sphere or each particle represents 0.1 moles.
So three times 0.1 is 0.3.
And then we divide that by the volume
of one liter to get a concentration of 0.3 moles.
So we can go ahead and plug in 0.3 mole
for the concentration of B.
Next, we divide that by the concentration of A,
and since there are five red particles,
and each particle represents 0.1 moles,
five times 0.1 is 0.5 moles of A
divided by a volume of one liter,
it gives a concentration of 0.5 mole for A.
Notice that we could have just counted
the number of particles, three blues and five reds,
and just done three divided by five
and get the same value for the reaction quotient Q.
So QC is equal to three fists or 0.6.
And KC remember was equal to three.
So Q is not equal to K.
In this case, QC is less than KC.
Since Q is not equal to K, at this moment and time,
the reaction is not at equilibrium.
In order for this reaction to reach equilibrium,
Q needs to be equal to K.
And since Q is a lot smaller than K,
if you think about it,
we need to increase the numerator
and decrease the denominator.
So that means, we have too many reactants
and not enough products.
And so the net reaction, if go back up here
to the equation here,
the net reaction is going to move to the right
to make more products.
So the net reaction moves to the right
to make more blue particles
and therefore the number of red particles would decrease.
We can see that comparing these
first two particulate diagrams here.
So let's compare the first particular diagram
to the second particular diagram.
And the first particulate diagram,
we had three blues and five reds.
And the second particulate diagram,
we have five blues and only three reds.
So that shows the reaction has moved to the right
to increase the amount of products
and to decrease the amount of reactants.
Let's calculate QC at this moment and time
for our second particular diagram
to see if we've reached equilibrium yet.
Well, we have five blue particles and only three reds.
So QC would be equal to five over three or five thirds.
Remember that the equilibrium constant KC,
is equal to three.
Therefore, QC is still not equal to KC.
And therefore the reaction is not at equilibrium.
And Q is actually still less than K.
Therefore the net reaction
is going to move to the right again
to increase the amount of products
and to decrease the amount of reactants.
Let's compare our second particulate diagram
to our third particulate diagram.
And the second particulate diagram,
we had five blues and three reds.
And then the third one here,
we have one, two, three, four, five, six blues,
and two reds.
So we've increased in the amount of blue
and we've decreased in the amount of red.
Let's calculate QC for the moment of time
represented by our third particulate diagram.
Well, there are six blues and only two reds.
So QC is equal to six divided by two,
which is equal to three.
So QC is equal to three
and remember K is also equal to three.
So QC is equal to KC and therefore this reaction
is now at equilibrium.
And at equilibrium,
the reactions turn into the products
at the same rate the products turn back into the reactants.
And therefore the concentration of both reactants
and products remained constant at equilibrium.
So when Q is less than K,
the reaction is not at equilibrium.
There are too many reactants and not enough products.
Therefore the net reaction goes to the right
to increase the amount of products.
The net reaction continues to go to the right
until Q is equal to K and the reaction
has reached equilibrium.
At that point, the concentrations of reactants and products
stop changing and they remain constant.
It's also possible for Q to be greater than K.
And if that's true, the reaction is not at equilibrium,
but in this case you have too many products
and not enough reactants.
Therefore the net reaction goes to the left
to increase the amount of reactants
and to decrease the amount of products.
The net reaction will continue to go to the left
until Q is equal to K and the reaction reaches equilibrium.
Let's look at another reaction,
which is the decomposition of phosgene
to form carbon monoxide and chlorine gas.
KC for this reaction is equal to 2.2 times
10 to the negative 10th at 100 degrees Celsius.
And let's say we're given concentrations
of phosgene carbon monoxide and chlorine
at a moment in time,
and asked if the reaction is at equilibrium or not.
And if the reaction's not at equilibrium,
we need to predict which direction the net reaction will go
to reach equilibrium.
Our approach is gonna be to calculate QC
at that moment in time, and then compare QC to KC.
So first we need to write our QC expression,
and this is equal to the concentration of carbon monoxide
raised to the first power
times the concentration of chlorine
raised to the first power
and that's divided by the concentration of phosgene.
So the concentration of COCL2.
So at this moment of time,
the concentration of carbon monoxide is 3.4 times
10 to the negative six mole.
The concentration of chlorine is 6.0 times,
times 10 to the negative six mole,
and the concentration of phosgene
is equal to 2.0 times 10 to the negative third mole.
When we plug all those into our Q expression and solve,
we get that QC is equal to 1.0
times 10 to the negative eighth.
So in this case, QC is greater than KC
because QC is equal to 1.0 times 10 to the negative eight
and KC is equal to 2.2 times 10 to the negative 10th.
And when QC is greater than KC,
we have too many products and not enough reactants.
Therefore the net reaction is going to go to the left
and there's going to be an increase
in the amount of phosgene.
The reaction will continue to go to the left
until Q is equal to K, and the reaction reaches equilibrium.