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### Shortcut method (finding concentration change)

Posted: Sun Jan 20, 2019 9:27 pm
Can someone explain the reasoning behind the shortcut method which can be used to bypass the quadratic equation when solving for the change in molarity(x)?

### Re: Shortcut method (finding concentration change)

Posted: Sun Jan 20, 2019 9:34 pm
I think you might mean ignoring the -x subtraction in the change. So if Kc=[x][x]/0.5-x, and let's say our kc= 25, to find x you'd set up the equation as
[x][x]/0.5-x= 25. In order to avoid the quadratic equation calculations, you can ignore the -x in the denominator, which won't affect the entire value as long as it satisfies the 5% rule.

### Re: Shortcut method (finding concentration change)

Posted: Sun Jan 20, 2019 9:45 pm
Keep in mind that you can only use the approximation method/shortcut if K < 10^-3.

### Re: Shortcut method (finding concentration change)

Posted: Sun Jan 20, 2019 10:40 pm
An analogy that would be easy to understand that highlights this concept is as follows: If a millionaire gave someone who had no money \$100, that loss of the \$100 wouldn't affect the millionaire.

In the same way, removing the -x from the denominator when doing the shortcut method doesn't change the answer much... To highlight this numerically:
(using the example from Week 2 Wednesday's lecture):
if x = 1.3x10^(-3) and (x^2)/(.10-x),
the difference of (.10 - 1.3x10^(-3)) is very small...so small that it doesn't change the answer much if -x is removed.

(1.3x10^(-3))^2/(.10 - 1.3x10^(-3)) = 1.71x10^(-5)
(1.3x10^(-3))^2/(.10) = 1.69x10^(-5)

The difference is minuscule, so this shortcut allows you to get an almost similar answer without going through the quadratic formula.

### Re: Shortcut method (finding concentration change)

Posted: Sun Jan 20, 2019 11:44 pm
It's also important to remember that we are not saying that X=0, but that it is a small enough number so that when subtracted from a larger number, that larger number remains relatively the same amount.

### Re: Shortcut method (finding concentration change)

Posted: Mon Jan 21, 2019 11:09 am
If K is less than 10^-3, then the reaction heavily favors product. Thus, x, or the change in molar concentration, is so low that it can be considered negligible. Essentially, the equilibrium concentration of reactant is the same as its initial concentration.