## 5% rule

Shanzey
Posts: 120
Joined: Wed Sep 18, 2019 12:20 am

### 5% rule

When approximating, how do we test to make sure that our approximation violates the 5% rule? Do we have to do (equilibrium concentration)/(initial concentration)?

Ariana Iranmahboub1G
Posts: 114
Joined: Fri Aug 09, 2019 12:17 am

### Re: 5% rule

Yes, you calculate (equilibrium concentration)/(initial concentration) times 100 and if it is less than 5%, the approximation is valid.

Bryce Ramirez 1J
Posts: 120
Joined: Sat Aug 24, 2019 12:16 am

### Re: 5% rule

If the value you get is more than 5%, you would know because the value should be significantly higher so that you are sure that it's not a valid approximation. If it's close around the 5% range, then you should redo the problem and make sure not to round any of the numbers until the final answer.

Brian J Cheng 1I
Posts: 115
Joined: Thu Jul 11, 2019 12:15 am

### Re: 5% rule

If the equilibrium concentration divided by the initial reactant concentration is less than 0.05, then the equilibrium constant is valid. Basically this means: the change in concentration of product can be ignored while solving the equilibrium constant equation as long as it is less than 5% of the initial starting concentration.

Long Luong 2H
Posts: 51
Joined: Thu Sep 19, 2019 12:16 am

### Re: 5% rule

Do you only check with the 5% rule if you make the assumption about ignoring the change in concentration of product? In other words, if you solve chemical equilibrium problems with the quadratic formula, do you check with the 5% rule?

Owen-Koetters-4I
Posts: 50
Joined: Fri Sep 28, 2018 12:16 am

### Re: 5% rule

You calculate initial concentration/equilibrium concentration by plugging in your approximated x-value. If this value is greater than 0.05 you must solve the cubic equation.

KDang_1D
Posts: 127
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 1 time

### Re: 5% rule

Yes, that expression refers to "percent protonation." Take note that these are the concentrations of the "primary" compounds involved in the reaction. For example, regarding NH3 + H2O --> NH4+ + OH-, you would only use [NH4+] formed /[NH3] initial, not including [H2O] and [OH-]

Also note that if the reaction equation has stoichiometric coefficients, the concentration of the product will be the x-value (from an ICE table) multiplied by that number. For example: HA + H2O --> 2A- + H3O; [A-] = 2x

Jacob Villar 2C
Posts: 105
Joined: Sat Aug 17, 2019 12:18 am

### Re: 5% rule

It’s equilibrium concentration divided by initial concentration.

205154661_Dis2J
Posts: 109
Joined: Wed Sep 18, 2019 12:21 am

### Re: 5% rule

We test it out by dividing equilibrium concentration/initial concentration x 100. if the value is less than 5%, the approximation is valid.