A
(g)(g)
2B
(g) +C
(g)When 1.00 mol of A is placed in a 4.00 L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t?
I've been able to solve questions like this when I've been given K
e or the concentrations. Now I'm stuck because I'm given 1 mol and the amount of the container. How do I find the initial concentration of A and B. I've been using the I.C.E method so once from there I can calculate this problem its just getting to that point I'm confused.
I've been looking online also and I've found this info.
C = 0.050 mol/L x 4.00L = 0.2L /4.00L = 0.05 (+x) and this I understand its from there finding B and A?
Next I would guess to solve B. B being two parts of A does this mean I would just multiple 0.05 by 2. which then I would get 0.10(+2x)
So really I guess I just need help solving for A.
Thank you