Welcome to Chemical Equilibrium Made Easy brought to you by Ketzbook.
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In this video we are going to learn what a chemical equilibrium is,
learn how to calculate an equilibrium constant, and learn the RICE method for determining
the concentrations of chemicals at equilibrium.
Let’s start from the beginning. What is equilibrium?
Equilibrium occurs when opposing forces or actions are balanced.
It’s kind of like a tug of war when the two sides pull equally. Nobody goes anywhere
even though they are pulling with a lot of force. But tug of war is a static equilibrium.
A chemical equilibrium is a dynamic equilibrium. That is, a chemical equilibrium involves constant motion,
kind of like a water fountain. The water is constantly in motion, but it stays
in the same place more or less.
A chemical equilibrium occurs when the forward and reverse reaction rates are equal.
That is, reactants constantly react to form product, and products constantly react to form reactant,
and the two things happen with the same speed so that the amount of each chemical stays the same.
Consider the following reaction. Nitrogen reacts with hydrogen to form ammonia.
This reaction is known as the Haber-Bosch process, and like nearly all reactions, it can also
go in reverse. That is, ammonia will also decompose to reform nitrogen and hydrogen.
What does that look like? Suppose that we have a 1 liter container, and we fill it with
2.8 moles of nitrogen, 4.4 moles of hydrogen, and an appropriate catalyst. We heat the mixture
to 450 degrees Celsius, and we observe that the concentration of nitrogen and hydrogen
decrease as they turn into ammonia. After a while, however, the concentrations of nitrogen,
hydrogen, and ammonia no longer change. There are 2 moles of nitrogen, 2 moles of hydrogen,
and 1.6 moles of ammonia. Those amounts do not change no matter how long you wait.
You might think that the reaction has stopped, but it hasn’t. Rather, the reaction has
reached equilibrium. That is, the chemicals are still reacting and never stop reacting.
The amounts do not change because new ammonia is formed at the same rate old ammonia decomposes.
Keep in mind that equilibrium is not a special type of reaction. In fact, nearly all chemical
reactions are equilibrium reactions. They just might not appear that way because they
strongly favor the products, they are very slow, or product is removed from the reaction mixture.
Now, we can use our knowledge of kinetics to describe an equilibrium in a very useful way.
For the Haber process, the rate of the forward reaction is a rate constant times
the molarity of nitrogen times the molarity of hydrogen raised to the third power.
Remember that a rate law is typically just a constant times the reactants raised to the power of
their coefficients. Likewise, the rate of the reverse reaction is another rate constant
times the molarity of ammonia squared.
Remember that at equilibrium, the forward and reverse reactions occur at the same rate,
so these two expressions are equal. I am going to erase the words, so that we can focus on
the two equal rate laws. Now what I want to do is to put all the variables on one side
and the constants on the other, so let’s see what happens if I divide both sides of
the equation by k2[N2][H2]3. On the left side, N2 over N2 cancels out as does H2 over H2.
On the right side, k2 over k2 cancels out. That leaves me with k1 over k2 equals [NH3]2
over [N2][H2]3. Because k1 and k2 are both constants, their ratio is also a constant,
and that is a special constant that we call the equilibrium constant.
We can call this constant Keq for “equilibrium constant,” or Kc for the “concentration”
equilibrium constant, or simply just “K,” capital K that is. Lowercase k is for rate
constants, and uppercase K is for equilibrium constants. The equilibrium constant is a physical
constant for a specific reaction at a specific temperature, and it tells you about the reaction.
In general, the equilibrium expression of any reaction is that the equilibrium constant
is equal to the molarity of the products raised to the power of their coefficients divided
by the molarity of the reactants raised to the power of their coefficients.
Let’s calculate the equilibrium constant of the Haber process at 450 degrees Celsius.
To do that, we are going to use the equilibrium expression of the Haber process and plug in
the molarities at equilibrium. Kc equals 1.6 squared divided by 2 times 2 cubed,
which equals 0.16. Interestingly enough, equilibrium constants are unitless because the concentration
values we use in the calculation are all relative to the standard concentration—that is, relative
to 1 mole per liter.
But what does the equilibrium constant even mean?
A large equilibrium constant, that is if K > 1, means that the reaction is product favored.
It will tend to make more products than reactants. A small equilibrium constant, that is if K < 1,
means that the reaction is reactant favored, and it will tend to make less products than reactants.
Let’s consider another reaction. Carbon tetrachloride decomposes at 700 Kelvin to
make elemental carbon and chlorine gas. The concentration equilibrium constant of that
reaction should be equal to the molarities of the products raised to the power of their
coefficients divided by the molarity of the reactant raised to the power of its coefficient.
However, carbon is a solid, so what is its molarity? The molarity (or more correctly,
activity) of any solid or non-homogeneous reactant or product has a value of one.
That means that we can exclude all solids or non-homogeneous chemicals from the equilibrium expression,
so the molarity of carbon should be erased. In addition, solvents typically have an activity
of one and are not written in the equilibrium expression.
Let’s try another problem. Suppose that 1 mole of carbon tetrachloride is added to
a 5 L flask and heated to 700 K. After the reaction reaches equilibrium, there are 0.88
moles of carbon tetrachloride remaining. Calculate the equilibrium constant of the reaction.
When faced with a problem like this, the best way to solve it is to set up a rice chart.
That’s right, R I C E. Whenever you solve an equilibrium problem, just picture a big
bowl of rice. The “R” stands for “reaction,” and we already have our balanced reaction written.
The “I” stands for initial molarity, the “C” stands for change in molarity,
and the “E” stands for equilibrium molarity.
Be sure to actually write out R I C E so that you don’t skip anything or make a mistake.
The question tells us that we are starting out with an initial amount of 1 mole of carbon
tetrachloride in a 5 L flask, so we can calculate its initial molarity using the equation M = n/V
which is 1 mole divided by 5 liters, or 0.2 molar. We put the 0.2 directly under the CCl4.
We can ignore carbon because it is a solid, and there is initially no chlorine gas,
so we put a zero underneath Cl2.
As for the change in molarity, we know that the reactants are decreasing by a certain amount,
so we put –x for CCl4, and the coefficients tell us that 2 Cl2 are made for every 1 CCl4
that is lost, so we put plus 2x for Cl2.
The equilibrium molarity is simply the initial molarity plus the change in molarity, so that
is 0.2 – x for carbon tetrachloride, and 2x for chlorine.
Now, how do we figure out x? Looking back at the problem, it says that there are 0.88
moles of CCl4 at equilibrium, so we can calculate the equilibrium concentration of carbon tetrachloride,
which is 0.88 moles divided by 5 L, or 0.176 molar. The equilibrium molarity of CCl4 is
also equal to 0.2 – x. If we add x to both sides and subtract 0.176 from both sides,
we can calculate x to be 0.2 minus 0.176, which equals 0.024.
Now the only thing left is to plug in the numbers and calculate the equilibrium constant.
Kc equals the equilibrium molarity of chlorine squared divided by the equilibrium molarity
of carbon tetrachloride. The equilibrium molarity of chlorine is 2x, and we can substitute 0.024
in for x. We already calculated the equilibrium molarity of CCl4 to be 0.176. Plugging this
into our calculator gives us an equilibrium constant of 0.013. Does this number make sense?
Yes! The concentration of reactants is larger than the concentration of products at equilibrium,
so an equilibrium constant less than 1 makes sense.
Okay, let’s try one more equilibrium problem. Nitrogen dioxide dimerizes at 100 degrees
Celsius to make dinitrogen tetroxide. Under these conditions, Kc equals 4.7. If 0.95 moles
of dinitrogen tetroxide is added to a sealed one liter container and kept at 100°C, calculate
the equilibrium concentrations of nitrogen dioxide and dinitrogen tetroxide.
Like most equilibrium problems, it is best to start by makings a RICE table.
We already have the balanced reaction.
This problem states that we are starting with N2O4, which is our product,
so the initial molarity of NO2 is zero, and the initial molarity of N2O4 is 0.95.
As this reaction progresses, the amount of nitrogen dioxide should increase, so the
change in NO2 should be plus 2x. The two comes from the coefficient in the balanced reaction.
Because the reaction will progress to the left, the N2O4 molarity will decrease,
so its change should be minus x. To get the equilibrium molarities, simply add the initial and the change.
The result is 2x for NO2 and 0.95 minus x for N2O4.
Next, let’s write out the equilibrium expression. Kc equals the molarity of N2O4 (on the top
because it is the product, and there is no exponent, because N2O4 has no coefficient
in the balanced reaction) divided by the molarity of NO2 squared (because NO2 has a coefficient
of two in front of it in the balanced reaction). We already know that the equilibrium constant
equals 4.7. Now, we can plug in the equilibrium values and solve for x.
Substitution gives us 0.95 minus x divided by 2x squared equals 4.7. We can eliminate
the fraction by multiplying both sides by 2x squared, which yields, 0.95 minus x equals
4.7 times 2x squared. By now, we can see that this is a quadratic equation, so let’s put
it in a form that enables us to plug it into the quadratic formula. Subtract 0.95 and add
x to both sides of the equation. This gets us zero equals 18.8 x squared plus x minus 0.95.
The 18.8 comes from 4.7 times 2 squared. Now, we can plug those coefficients into the
quadratic formula. Hopefully, you have a calculator that can solve this, or you can program an
equation into your calculator to solve it. If not, just use the quadratic formula to
calculate it yourself. In this case, a is 18.8, b is one, and c is negative 0.95.
The result from the quadratic formula is that x equals 0.2 or negative 0.25.
Now pause here and think about those two values. Which is the correct value for x? The negative
value does not physically make sense because that would mean a negative concentration of
NO2 at equilibrium, which of course is impossible. So, x equals 0.2. Last of all, we plug this
value of x back into our equilibrium values on the RICE table.
The nitrogen dioxide molarity is two times 0.2, which is 0.4 molar.
The dinitrogen tetroxide molarity is 0.95 minus 0.2, or 0.75 molar.
That’s it. Remember that all reactions move towards equilibrium, which is the point at
which the forward and reverse reactions occur at the same rate, and remember to use a RICE
table to solve equilibrium problems.
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