5% rule


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Briana Lopez 4K
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Joined: Wed Nov 22, 2017 3:01 am

5% rule

Postby Briana Lopez 4K » Thu Jan 17, 2019 4:08 pm

How do you use the 5% rule if there is more than one reactant? In the example in class it was
CH3COOH-> H30+ + CH3COO-
and the percentage is calculated by dividing (1.3 x 10^-3)/(0.10) * 100% which equals 1.3%

but how would you do it if it were this for instance??

2N2 + O2 —> 2N2O
and our x= 7.15 x 10^-14
and the ice box is (3.21-2x) for 2N2 , and for the other reactant it is O2 (6.21-x)

Chem_Mod
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Re: 5% rule

Postby Chem_Mod » Thu Jan 17, 2019 5:01 pm

We normally apply the 5% rule to acid/base types of reactions. Essentially though, the idea is to divide the change in dissociation (usually "x") by the original concentration the dissociation came from (whatever quantity you subtract x from).

Emily Kennedy 4L
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Re: 5% rule

Postby Emily Kennedy 4L » Sun Jan 20, 2019 5:15 pm

if the approximation is above 5% then is the complete answer void?

lizettelopez1F
Posts: 20
Joined: Thu Sep 27, 2018 11:19 pm

Re: 5% rule

Postby lizettelopez1F » Sun Jan 20, 2019 5:51 pm

Emily Kennedy 4L wrote:if the approximation is above 5% then is the complete answer void?

If it is greater than 5%, the quadratic equation should be used instead to find x.

almaochoa2D
Posts: 54
Joined: Thu Sep 27, 2018 11:23 pm

Re: 5% rule

Postby almaochoa2D » Mon Jan 21, 2019 9:28 am

you can still do the 5% rule he sort of did an example like this in lecture on Wednesday I believe.

405211415
Posts: 37
Joined: Mon Oct 08, 2018 11:16 pm

Re: 5% rule

Postby 405211415 » Mon Jan 21, 2019 3:16 pm

You use the 5% rule to determine if X is negligible, if it is over 5% then you just need to include (#-x) in your calculations when trying to find the value of X.

Jason Ye 2I
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Re: 5% rule

Postby Jason Ye 2I » Mon Jan 21, 2019 4:26 pm

I think each component of X should comply to the 5% rule.

Luc Lorain 1L
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Joined: Thu Sep 27, 2018 11:18 pm

Re: 5% rule

Postby Luc Lorain 1L » Mon Jan 21, 2019 4:42 pm

As a rule, Dr. Lavelle states that it is generally safe to use the 5% rule when the equilibrium constant <= 10-3. For most reaction scenarios, this number is so small that it will lead to a final value of x that does not affect the concentration or partial pressure of an equilibrium species in a noticeable way (only visible after many sig figs). The 5% rule is checked after we complete an ICE table to find the equilibrium expressions for each species in the reaction, and applied only to expressions that add or subtract x; if the proportion of the (initial condition-x)/ (initial condition) is <= 0.05, then it is safe to apply the rule.

Xingzheng Sun 2K
Posts: 49
Joined: Thu Sep 27, 2018 11:29 pm

Re: 5% rule

Postby Xingzheng Sun 2K » Wed Feb 20, 2019 3:17 am

I think it simply means that if x (the change) is less than 5% than it is ok. It is important that the formula x=(-b+/-(b2-4ac)^1/2)/2a always work.


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