Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: Equilibrium constant question
SunriseSunset
Hazard to Self
**




Posts: 82
Registered: 9-6-2015
Location: USA
Member Is Offline

Mood: No Mood

[*] posted on 11-1-2016 at 02:11
Equilibrium constant question


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.




Why do chemists call helium, curium and barium the medical elements?
because if you cant helium or curium, you barium! - Heimerdinger
View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 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]




View user's profile View All Posts By User
SunriseSunset
Hazard to Self
**




Posts: 82
Registered: 9-6-2015
Location: USA
Member Is Offline

Mood: No Mood

[*] posted on 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)




Why do chemists call helium, curium and barium the medical elements?
because if you cant helium or curium, you barium! - Heimerdinger
View user's profile View All Posts By User
Etaoin Shrdlu
National Hazard
****




Posts: 724
Registered: 25-12-2013
Location: Wisconsin
Member Is Offline

Mood: Insufferable

[*] posted on 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]
View user's profile View All Posts By User
SunriseSunset
Hazard to Self
**




Posts: 82
Registered: 9-6-2015
Location: USA
Member Is Offline

Mood: No Mood

[*] posted on 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)





Why do chemists call helium, curium and barium the medical elements?
because if you cant helium or curium, you barium! - Heimerdinger
View user's profile View All Posts By User
Etaoin Shrdlu
National Hazard
****




Posts: 724
Registered: 25-12-2013
Location: Wisconsin
Member Is Offline

Mood: Insufferable

[*] posted on 11-1-2016 at 14:23


Just search for "equilibrium constant" and you will find all manner of them.
View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 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]




View user's profile View All Posts By User

  Go To Top