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Le Chatlier's principle from Chads

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neppywise

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Taken from Chad's vids:
Which of the following changes to the following reaction at equilibrium would result in a shift to the right?

COCl2(g) + HCl(g)
3_b6357340a4139ce0922f4b306ea70acd.png
CHCl3(l) + ½O2(g)

none of these
a decrease in pressure
removal of COCl2
an increase in the volume of the reaction vessel

It says the answer is none of these. And in the explanation it says that in general, removing a reactant and adding a product shifts the rxn left.

But I was looking at the number of moles of gas on each side, (since liquids usually don't affect le chatlier's principle). If you remove CoCl2 wouldn't it still move to the right because you would still have less moles on that side (1/2 mole o2) ? Please help me to understand this one!
 

clathrin413

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The number of moles of gas can only be used to determine the new equilibrium position when the pressure or volume is changed; you are not necessarily trying to balance the number of moles of gas (notice here, there are 2 moles of gas vs. 1/2 moles of gas at equilibrium). The decrease in pressure has the same effect as an increase in volume, which will shift equilibrium to the side of the reaction with more moles of gas (the left).

Removing CoCl2 just causes the 'usual' LCP behavior, in which removing a reactant causes the reaction to shift to the left. Here, we look at the affected position of Q vs. K, where Q is the effective equilibrium expression. Since CoCl2 is removed and in the denominator of the equilibrium expression, K < Q, so to adjust for this, equilibrium will shift to the left.
 

neppywise

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Perfect explanation, thank you! I never knew that we only considered moles of gas for pressure and volume changes.
 

nomdeplume1234

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I think the important thing to note is that the question stem states that the reaction is at equilibrium. I think that means that there is some Keq = [Chcl3][O2] / [Cocl2]. If you remove Cocl2 you make the denominator smaller so then there would be too many products vs reactants and the reaction would shift to the left to get the ratio just right so that it equaled the Keq again.

notice that (b) and (d) are really the same. Increasing the volume of the container will decrease the pressure.
 
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