## 5% rule

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Briana Lopez 4K
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### 5% rule

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

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

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

lizettelopez1F
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### Re: 5% rule

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
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### Re: 5% rule

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

Carlos De La Torre 2L
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### Re: 5% rule

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

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

Luc Lorain 1L
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### Re: 5% rule

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
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### Re: 5% rule

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.

jlinwashington1B
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### Re: 5% rule

What is the 5% rule, and what does it refer to?

David Sarkissian 1K
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### Re: 5% rule

After reading this entire thread I've somehow only become even more confused. How can you divide the (initial condition-x)/(initial condition), if we are looking for x in the first place. There is no change in value, its still the same initial amount being divided by the initial amount, except now there is just a variable x involved? I just don't see how you would get anything but 100% every time, which I know is clearly wrong. I just want to see an example of the actual 5% rule being applied with numbers, not having the concept of it be repeated over and over again without any of the math being shown.

Pritish Patil 1K
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### Re: 5% rule

If the change based on the approximation is less than 5% then an approximation can be made. Normally, this works for k values that are less than 10^-3.

jocelyntzeng
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### Re: 5% rule

you use the 5% rule to check to make sure that your approximation is correct and can be used