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### 11.67 Why do we use a new initial value?

Posted: Mon Nov 16, 2015 7:40 pm
can someone please explain why there is an original and a new initial?

### Re: hw problem 11.67

Posted: Mon Nov 16, 2015 8:04 pm
There are three initial pressures. The 1.0 mol of H2 refers to H2 when it is at equilibrium. If you are confused about how to proceed, you would find the pressure of H2 at equilibrium and use PV=nRT to figure out the volume of H2 at equilibrium.

### Re: hw problem 11.67

Posted: Mon Nov 16, 2015 10:36 pm
i think they are referring to the solutions manual for this question. In the ICE box there is an initial and a new initial. Um im not sure why that is I asked my TA and she doesnt even know so I think you can solve the problem without doing it that way

### Re: hw problem 11.67

Posted: Tue Nov 17, 2015 9:40 pm
Oh, I understand now. I feel like the reason the manual did that was that it could use an approximation of zero to estimate x, since K is so small. If you just did the regular way, each equilibrium concentration would have a number "+x" or "-x" within it. When you push the reaction as far to the left as possible, the equilibrium concentration of H2 would be x, making it possible to now approximate that change in some kind of "x" amount to zero. Otherwise, if you were to do the regular way, you would deal with numbers so small, since K is so small, that the calculator would just act as if those numbers are zero. This way, you can yield the really small value of that "x."

### Re: 11.67 Why do we use a new initial value?

Posted: Wed Nov 18, 2015 3:10 am
We use a new initial value because it makes our calculation easier to perform. We can do this because a reaction doesn't "care" from what point it starts from as its equilibrium constant at the given conditions will be constant. If we write a new initial value, our calculation would become $K=\frac{(2.0+2x)^{2}}{(1.0-x)(3.0-x)}$ and we would not be able to make the approximations necessary to isolate x. In this case, solving for x would be a nightmare without a good calculator.