# Chemistry

posted by Amphee

Consider the following equilibrium process at 686 C
C02(g)+H2(g)=CO(g)+H20(g)
The equilibrium concentration of the reacting species are [CO]=0.050M, [H2]=0.045M, [CO2]=0.086M, and [H20]=0.040M. (a)Calculate the Kc for the reaction at 686 C. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the concentration of all the gases be when equilibrium is reestablished.

I got (a) Kc= 0.52 but I'm having hard time getting part (b) because the answers are way different then what I'm getting. The answers should be:
[CO]=0.075M, [H2]=0.020M, [CO2]=0.48M, [H20]=0.065M.

At certain temperature the following reactions have the constant shown:

S(s) + O2(g) = SO2(g)
K'c=4.2*10^52

2S(s) + 3O2(g) = 2SO3(g)
K"c=9.8*10^128

Calculate the equilibrium constant Kc for the following reaction at that temperature:

2S02(s) + O2(g) = 2SO3(g)

This one was specially hard because time to time my calculator couldn't do it because of overflow.

Thanks.

1. DrBob222

I worked part b and obtained the answers you gave as correct.
(CO)(H2O)/(CO2)(H2) =
CO2 = 0.5-x
(H2O) = 0.045-x
(CO( = 0.05+x
(H2O) = 040+x
Solve for x, THEN add or subtract from the starting point to arrive at the equilibrium amounts. Post your work if you can't find the error and I'll give it a go.

2. DrBob222

I have just looked at the second problem. Is that SO2(s) a typo? Must be, huh?

3. DrBob222

ASSUMING that SO2(s) is a typo, then multiply equation 1 by 2 and reverse it. That new k1 = (1/k^2) and multiply that by k2.
That will be, if I haven't goofed in the arithmetic, is
k2/k1^2 = 9.8 x 10^128/(4.2 x 10^52)^2 = ??
You need not have a problem with overflow. Just do 9.8/(4.2)^2 on your calculator to obtain 0.555 or so and you do the exponent on paper. It is 128-52-52 = 24 (or 5.55 x 10^23). Check my work carefully.

4. Amphee

Oh yes it is a typo my bad. But you got the right answer though. Thank you and thank you for the short cut on the calculator too, I was doing it just like you did it but my the calculator was having overflow problem

5. Rand

Thank you so much for your help!!!

6. kim

can someone help me with part A of this question?

Thank you so much

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