Consider this reaction: CO (g) + H2O (g0 <---> CO2(g)+H2(g)

kc=102 at 500 K

A reaction mixture initially contains 0.145 M of CO and 0.145 M of H20. What would be the equilibrium concentration of [CO]?

I so far have
kc= [H2][CO2]/[CO][H2O]= 102

I know i have to set up an ice table so i have:
CO H2O CO2 H2
I 0.145 M 0.145M 0 0

C -x -x +x +x

E

what do i do now? i'm kind of stuck. thank you!

Add the columns so that initial column + change column = equilibrium column.

Equilibrium:
CO = 0.145-x
H2O = 0.145-x
CO2 = x
H2 = x
Now substitute those values into your Kc expression and solve for x.

To solve this problem, you are on the right track by setting up an ICE Table. Here's how you can proceed:

1. Set up your ICE Table and write down the initial concentrations:
CO H2O CO2 H2
I 0.145 M 0.145 M 0 M 0 M

2. Write the change in molar concentration for each species (using -x for reactants and +x for products):
CO H2O CO2 H2
C -x -x +x +x

3. Express the equilibrium concentrations in terms of the initial concentrations and the values of x (representing the change in concentration):
CO H2O CO2 H2
E 0.145 - x 0.145 - x 0 + x 0 + x

4. Plug these expressions for equilibrium concentrations into the equilibrium constant expression:
Kc = [CO2][H2]/[CO][H2O] = (0 + x)(0 + x) / (0.145 - x)(0.145 - x) = 102

5. Solve for x by rearranging the equation and then solving a quadratic equation:
(0 + x)(0 + x) / (0.145 - x)(0.145 - x) = 102

(x^2) / (0.145 - x)^2 = 102

x^2 = (102)(0.145 - x)^2

Take the square root of both sides to eliminate the square:
x = sqrt((102)(0.145 - x)^2)

This equation is quadratic, and to solve for x, you'll need to expand and rearrange the equation to isolate x.

6. Once you find the value of x, substitute it back into the equilibrium concentration expressions and calculate the equilibrium concentration of CO ([CO] = 0.145 - x).

Note: Remember to consider the appropriate significant figures and units in your final answer.

I hope this helps you proceed with solving the problem. Let me know if you have any further questions!