Sciencemadness Discussion Board

Equilibrium constant question

SunriseSunset - 11-1-2016 at 02:11

So my book has an equilibrium constant of = 1.5

The equilibrium favors the products.

Then it says, calculate the percentage of each at equilibrium. In the answer key it has: reactants = 40% and products = 60%.

I'm wondering how you figure out these percentages; based off only having an equilibrium constant to use? Because they don't show me the equation. I'm feeling pretty stupid.

The reaction is this: Keq = [products]/[reactants]. Unimulecular reaction if I'm not mistaken. It's simply using Axial and Equatorial isomers as the example to work with.

blogfast25 - 11-1-2016 at 02:43

Quote: Originally posted by SunriseSunset  
So my book has an equilibrium constant of = 1.5

The equilibrium favors the products.

Then it says, calculate the percentage of each at equilibrium. In the answer key it has: reactants = 40% and products = 60%.

I'm wondering how you figure out these percentages; based off only having an equilibrium constant to use? Because they don't show me the equation. I'm feeling pretty stupid.

The reaction is this: Keq = [products]/[reactants]. Unimulecular reaction if I'm not mistaken. It's simply using Axial and Equatorial isomers as the example to work with.


It seems like a pretty poorly formulated question.

Assume unimolecular: A < === > B

Say you start from 100 mol A, then at equilibrium x mol have reacted away, and B = x mol.

Then:

$$K=\frac{[B]}{[A]}$$

$$\frac{x}{100-x}=1.5$$

$$x=150-1.5x$$

$$x=\frac{150}{2.5}$$

$$x=60$$

That is 60 % conversion.

[Edited on 11-1-2016 by blogfast25]

SunriseSunset - 11-1-2016 at 10:40

That would mean that the starting material is in 1 part and the product is 1.5 parts. (2.5 total)

If the Keq was any other number, is the starting material always 1 part?

Keq = 1.5

1.5+1 = 2.5

1/2.5 = 0.4 (starting material)
1.5/2.5 = 0.6 (product)

Etaoin Shrdlu - 11-1-2016 at 10:47

They are in a ratio which is what matters; there is no set "one part." Could be 0.5 parts starting material and 0.75 parts product. Could be 2 parts starting material and 3 parts product.

For any Keq you could choose to call the quantity of starting material "one part."

EDIT: If you're asking can you call the leftover starting material 1 part and then the amount of product will be equal to Keq, then yes, in the case of unimolecular reactions specifically.

[Edited on 1-11-2016 by Etaoin Shrdlu]

SunriseSunset - 11-1-2016 at 11:05

Does anyone know a youtube video that shows the steps of solving the equation blogfast25 posted? Because I want to make sure I can do it and practice it without getting ahead of myself by assuming I can always rely on:
Quote:

If the Keq was any other number, is the starting material always 1 part? Keq = 1.5 1.5+1 = 2.5 1/2.5 = 0.4 (starting material) 1.5/2.5 = 0.6 (product)

Etaoin Shrdlu - 11-1-2016 at 14:23

Just search for "equilibrium constant" and you will find all manner of them.

blogfast25 - 11-1-2016 at 14:45

Quote: Originally posted by SunriseSunset  
Does anyone know a youtube video that shows the steps of solving the equation blogfast25 posted? Because I want to make sure I can do it and practice it without getting ahead of myself by assuming I can always rely on:
Quote:

If the Keq was any other number, is the starting material always 1 part? Keq = 1.5 1.5+1 = 2.5 1/2.5 = 0.4 (starting material) 1.5/2.5 = 0.6 (product)



Say you start from A0 moles of A and B0 moles of B.

Assume now that x of a was converted to B during reaction, then:

$$K=\frac{B_0+x}{A_0-x}=1.5$$

Insert values A0,B0 and solve for x.

[Edited on 11-1-2016 by blogfast25]