Posted by **Anonymous** on Tuesday, April 10, 2012 at 8:19pm.

When 0.5 mole of CO2 and 0.5 mole of H2 were forced into a litre reaction container, and equilibrium was established:

CO2(g) + H2(g) <--> H2O (g) + CO (g)

Under the conditions of the experiment, K=2.00

a) Find the equilibrium concentration of each reactant and product.

K = [H20][CO]/[CO2][H2]

2.00 = X2/(0.5-X)^2

2(0.5-X) = X

1-2X = X

1-3X = 0

-3X = -1

X = 1/3

^ is this right?

b) how would the equilibrium concentrations differ is 0.50 mole of H2O and 0.50 moles of CO had been introduced into the reaction vessel instead of the CO2 and H2?

please help :)

- CHEM -
**DrBob222**, Tuesday, April 10, 2012 at 9:16pm
Is this K = 2.00 Kp or Kc? I went with Kc.

a is not right. You have made two errors.

1. You took the square root of the left side but you didn't take the square root of 2. I get x = 0.293 mols.

2. You calculated mols but not M

3. You solved for x but not 0.5-x (but since this is in 1L mols = M.

For b, equilibrium is the same whether approached from the left or right as long as moles on the left = moles on the right.

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