Question

Can someone help me please?
If the volume of the vessel was decreased to 2 L, how would that affect the Kc and which way would
equilibrium shift and why.

Answer 1

Hi...Well if you decrease the volume of a vessel, that means that the concentration of your gas will increase...for example, if you had 2 moles of H2 gas in a 4 liter vessel, the Molarity would be 2/4 or 0.5 M H2...If you decrease that volume to say only 1 liter, then your ratio would be 2 moles H2 in a 1 liter vessel or 2/1= 2.0 M H2...So now you know by decreasing the volume, you increase the concentration of your gas which is the same as your Molarity...

So if they gave you a Kc value, take your new ratios of gas to volume, add them up, then just compare the values you added up with the original Kc value!

If the Kc value they gave you is "at equilibrium", then that simply means that that specific Kc value belongs to the product side of the equation...

If you add up the Molarity of your gases, just see if they are greater than or less than the other Kc value...if its greater than the Kc value, the reaction will shift in the opposite direction!

Again without the entire question, this is the best I can do...I hope this helps! :o) Good Luck!

Answer 2

Actually I made a mistake in the second to last paragraph...If your ratio addition is greater than the given Kc value, then the reaction is shifted left towards the reactants...if your ratio(Also known as Qc in your book) is less than the given Kc value, the reaction runs right towards the products! Sorry for the confusion.