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## AP®︎/College Chemistry

### Unit 7: Lesson 4

Magnitude and properties of the equilibrium constant# Magnitude of the equilibrium constant

AP.Chem:

TRA‑7 (EU)

, TRA‑7.C (LO)

, TRA‑7.C.1 (EK)

, TRA‑7.F (LO)

, TRA‑7.F.1 (EK)

The magnitude of the equilibrium constant provides information about the relative amounts of reactants and products at equilibrium. A large

*K*value (greater than 1) indicates that there are more products than reactants at equilibrium, while a small*K*value (less than 1) indicates that there are more reactants than products at equilibrium. Created by Jay.## Video transcript

- [Instructor] The magnitude
of the equilibrium constant tells us the relative amounts of products and reactants at equilibrium. For example, let's look
at a hypothetical reaction where gas A turns into gas B. And for the first example, let's say that gas A is
represented by a red sphere and gas B is represented by a blue sphere. And down here, we have
a particulate diagram showing an equilibrium mixture
of our hypothetical reaction. Let's write the equilibrium
constant expression for this hypothetical reaction. So we're gonna write Kc is equal to, and we think products over reactants. So our product is B. So this can be the concentration of B. And since the coefficient is a
one in the balanced equation, it's the concentration of
B raised to the first power divided by the concentration
of our reactant, which is A. And A in the balanced equation also has a coefficient of one. So this is the concentration
of A raised to the first power. If we assume that each particle
in our particulate diagram represents 0.1 moles of a substance, and the volume is one liter, we can calculate the
concentration of both A and B. For example, for B, there
are five blue spheres. So that'll be five times
0.1 moles or 0.5 moles. So for the concentration of B, we have 0.5 moles divided
by a volume of one liter. So 0.5 divided by one is 0.5 molar. So we can go ahead and plug that in for our concentration of B. It's 0.5 molar. Next, we can do the same thing for A. There are also five red spheres. And so therefore the concentration
of A is also 0.5 molar. So we can plug that into our equilibrium constant expression. 0.5 divided by 0.5 is equal to one. So therefore, Kc, the equilibrium constant is equal to one at whatever
temperature we have for our hypothetical reaction. So our equilibrium constant
Kc is equal to one. And we saw in our particulate
diagram at equilibrium, we have equal amounts of
reactants and products. Therefore just by knowing the value for the equilibrium constant, we know about the relative
amounts of reactants and products at equilibrium. Let's look at another
hypothetical reaction, which also has gas A turning into gas B. However, this time gas A
is green and gas B is red. And let's calculate the
equilibrium constant Kc for this reaction. And once again, our particulate diagram shows an equilibrium mixture. So Kc is equal to the concentration of B over the concentration of A. And it's a lot faster to
simply count our particles. So for B, B is red, we
have one red particle here, so we can go ahead and put in one. And then for gas A, we
have one, two, three, four, five, six, seven, eight,
nine, 10 particles. So one divided by 10 is equal to 0.1. So Kc is equal to 0.1 for
this hypothetical reaction at a certain temperature. So the magnitude of the
equilibrium constant tells us about the reaction mixture at equilibrium. For this reaction, Kc is equal to 0.1. So K is less than one. And if we think about what that means, K is equal to products over reactants. So if K is less than one, that means we have a smaller
number in the numerator and a larger number in the denominator, which means there are more
reactants than products at equilibrium. Let's look at another
hypothetical reaction where gas A turns into gas B. This time gas A is
yellow and gas B is blue. If we look at our particulate diagram, showing our reaction
mixture at equilibrium, there are 10 blue particles
and only one yellow particle. So when plug into our
equilibrium constant expression, this time it's going to be 10 over one. Therefore the equilibrium
constant Kc is equal to 10 for this particular reaction
at a certain temperature. Once again, the magnitude
of the equilibrium constant tells us something about
the reaction mixture at equilibrium. For this hypothetical
reaction, Kc is equal to 10. So K is greater than one. And when K is greater than one, once again, we have
products over reactants. So the numerator must be
larger than the denominator, which means we have a lot more products than reactants at equilibrium. Let's look at the reaction
of carbon monoxide and chlorine gas to form phosgene. At 100 degrees Celsius,
the equilibrium constant for this reaction is 4.56
times 10 to the ninth. Since the equilibrium constant
K is greater than one, we know there are more
products than reactants at equilibrium. And with the extremely large value for K, like 10 to the ninth, we could even assume this reaction essentially goes to completion. For the reaction of
hydrogen gas and iodine gas to form hydrogen iodine,
the equilibrium constant Kc is equal to 51 at 448 degrees Celsius. Since the equilibrium constant
is relatively close to one. This means at equilibrium, we have appreciable amounts
of both our reactants and our products. Let's look at the reaction of
nitrogen gas plus oxygen gas plus bromine gas to form NOBr. At 298 Kelvin, the equilibrium
constant for this reaction is 9.5 times 10 to the negative 31st. Since the equilibrium
constant K is less than one, we know at equilibrium, there are more reactants
than there are products. And with an extremely small K value, like 9.5 times 10 to the negative 31st, this reaction barely proceeds at all. So at equilibrium, you're gonna have almost all nitrogen, oxygen and bromine and very little NOBr.