Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

Thank you so much for your help!!!

Answer 6

can someone help me with part A of this question?

Thank you so much