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April 25, 2014

Homework Help: CHEM

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 :)

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