Wednesday
July 23, 2014

Homework Help: Chemistry

Posted by Amphee on Sunday, April 25, 2010 at 1:04pm.

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.

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