why do you square equilibrium constant when you multiply whole rxn by 2?

[sixpence]

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im in tbr. p.175 of chemistry. when you multiply a rxn by 2 in order to combine two reactions, why do you square the Keq value. Conversely when you multiply by 1/2 you take the square root of the Keq?

 

[DendWrite]

im in tbr. p.175 of chemistry. when you multiply a rxn by 2 in order to combine two reactions, why do you square the Keq value. Conversely when you multiply by 1/2 you take the square root of the Keq?

Basically say you have this reaction:

aA + bB ----> cC + dD Keq = 5

a, b, c, and d are the mole coefficients and A, B, C, and D are species.

The Keq expression here is 5 = ([C]^c)([D]^d) / ([A]^a)(^b)

If you double everything, you double a, b, c, and d. Say that the coefficients are initially all 1. When you double them you get 2, so you effectively square the right side of the equation. Since you are squaring the right side you also square the left side, so you get:

5^2 = ([C]^2)([D]^2) / ([A]^2)(^2)

It's the same thing for multiplying by 1/2, except in this case it's Keq^(1/2) or the square root of Keq.

 

[sixpence]

thanks. i'm pretty bad with equations

 

[DendWrite]

thanks. i'm pretty bad with equations

No problem. A little trick that I use to remember it is that it's basically the same as the laws of logarithms (if you don't know these then don't worry about it, just memorize the chemistry stuff).

Basically:

Add two equations: multiply the Keq's (just like when you add two logarithms).

Subtract equations: you flip one around and add it to the other. The negative of a logarithm is the same thing as the logarithm of the inverse, so whenever you flip a chemical equation around, you take the inverse of the Keq. Then to add, you just do as above -- multiply them both.

Multiply equations: Whatever number you multiply an equation by, the Keq is raised to that power (because nlog(2) = log(2^n)).

Divide equations: Same thing, except the power n above is a fraction.

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This all works because if you take the log of both sides of the equilibrium expression:

Keq = ([C]^c)([D]^d) / ([A]^a)(^b)

You get:

log(Keq) = c*log[C] + d*log[D] - a*log[A] - b*log

You can then manipulate it from there to solve for what you need. I think this approach works better for me because it's easier for me to think about the math than the chemistry. Fastest is just to memorize the chemistry, but if you end up forgetting, just remember you can work it out like this ^^.