Question

initially 1 mol each of gaseous carbon dioxide, co2(g) and hydrogen h2(g) is inhected into a 10.0L reaction chamber at 986 degrees. What is the predicted concentration of each entity at equilibrium?
the equation is as follows:
co2(g)+ h2(g)= co(g)+h20 (g)
k=1.60 at 986 degrees

Answer 1

You need to locate the caps key on yur keyboard and use it.
The concn is 1mol/10 L = 0.1M
........CO2 + H2 --> CO + H2O
I.......0.1...0.1....0.....0
C.......-x....-x.....x.....x
E......0.1-x..0.1-x..x.....x

Substitute the E line into the K expression and solve for x, then evaluate 0.1-x for the others.

Answer 2

thanks for replying back
Keq=(x)(x)/(0.1-x)(0.1-x)
Keq=x^2/x^2-0.2x+0.01
do i take the square root of each side and then...

Answer 3

Yes you would; however, you have expanded the (0.1-x)(0.1-x) unnecessarily. I would have done it this way. It's less work.
Keq = (x)(x)/(0.1-x)(0.1-x)
Keq = x^2/(0.1-x)^2
sqrt (Keq) = x/0.1-x and go from there.