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AP®︎/College Chemistry
Course: AP®︎/College Chemistry > Unit 7
Lesson 3: Calculating the equilibrium constantWorked examples: Calculating equilibrium constants
AP.Chem:
TRA‑7 (EU)
, TRA‑7.B (LO)
, TRA‑7.B.1 (EK)
In this video, we'll calculate equilibrium constants using measurements of concentration and partial pressures at equilibrium. First, we'll find Kc for an equilibrium system using equilibrium concentrations. Then, we'll find Kp for a different system using equilibrium partial pressures. Created by Jay.
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Q.
At 400°C a 50L container has 2 mole N2 and 6 mole H2.
If at equilibrium, 2 mole NH3 is produced, what's the value of Kc(3 votes)
For the last question when finding the Kp I got .28 instead of .11 when I plugged (.2)(3.4)/(3.9)(1.6). I don't know what I did wrong(2 votes)
The other replier is correct. Computers, like calculators, are stupid so they’ll only know to perform the calculations in the order you input them into the calculator. So if you tell it to do the operation you stated, the calculator will read it as 0.2 x 3.4 ÷ 3.9 x 1.6, and do it in that order (from left to right like PEMDAS). Which is why you get 0.28 instead of the actual answer of 0.11.
You have to specify to the calculator which order you’d like the operations to be done in with parentheses. If you input it as (0.2 x 3.4) ÷ (3.9 x 1.6), you’re telling the calculator you want it to calculate the products in the parentheses first, then divide those two products to get a quotient, which is actually what we want to do. When doing math on a computer you have to be very exact about what you want or it’ll misinterpret it.
Hope that helps.(2 votes)
- didn't yall say if we have gas we use pressure to get like kp so how come we have gas and we get the concentration and you solve to get kc(1 vote)
- For gases we can express their concentration in molarity as well as pressure units like pascals or bars. So they have the opportunity of having both a Kc (using molarity) and a Kp (using pressure units). They’ll have different numerical values, but they still express the same reaction’s equilibrium. Depending on the information given we would calculate one equilibrium constant as opposed to the other. The first reaction has the concentrations in molarity so Kc is more convenient to calculate, but for the second reaction at2:43the chemicals are in atms so Kp is more convenient.
We can convert between the two types of equilibrium constants using the formula: Kp = Kc(RT)^(Δn), where R is the gas constant, T is temperature, and Δn is the change in gas molecules per reaction (Δn = moles of gaseous product moles – moles of gaseous reactant moles). Notice that if Δn is 0 (same moles of gas on both sides of reaction), then RT is 1 and Kp = Kc. This is true for the second reaction.
Hope that helps.(2 votes)
- At4:58, what would happen if the mole ration was not 1-1?(1 vote)
- The change corresponds to their coefficients in the chemical equation. If CO has a 2 coefficient, and water still had a 1, the ratio would be 2:1. This means water would increase by x amount, but CO would increase by 2x amount since it forms at twice the rate that water does.
Hope that helps.(2 votes)
2:43Video transcript
- [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.