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Posted: Sun Dec 03, 2017 9:41 pm
by Juanalv326
For the reaction PCl5(g) Δ PCl3(g) Cl2(g), Kc 1.1 x10^-2 at 400. K. (a) Given that 1.0 g of PCl5 is placed in a 250.-mL
reaction vessel, determine the molar concentrations in the mixture at equilibrium.

How would you solve, I did convert, however answer doesn't appear to be correct.

Re: 11.47

Posted: Sun Dec 03, 2017 10:15 pm
by Julian Krzysiak 2K
Because we are given grams of the reactant, we have to convert to concentration by converting the grams to moles, and then dividing that by the vessel volume.
1.0g/208.22 moles --> .00480 mole/.250 Liters ---> .0192 mol./L.

Then we would insert this concentration (.0192) into the ICE box, in the "Initial" row for PCL5, while the other initial concentrations are 0.

The "Change" row would contain -x, +x, +x, in that order.

And the "Equilibrium" row would contain, (.0192-x), +x,+x, in that order.

Then you would solve for the Equilibrium equation, plugging in the values you got for the equilibrium constant , and the equilibrium values for the reactant and products.

Then you would have to solve by using the quadratic formula.

You then find x, plug back into the ICE box, and then you would get the correct molar concentration of the reactant and product.