If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 7

Lesson 3: Calculating the equilibrium constant

# Worked examples: Calculating equilibrium constants

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.

## Want to join the conversation?

• 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 • 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 • 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.
• 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 at the 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. 