Question 5.39 in Textbook; How to go about half-ing the volume?
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Question 5.39 in Textbook; How to go about half-ing the volume?
Hi! I was doing question 5.39 in the textbook, which essentially asks you to calculate equilirbium quantities based on the initial concentration and Kc. However, it has a part two that states "the volume of the flask is reduced to half its original value. Calculate the new equilibrium concentrations of the gases". In this example, the book just simply divides the volume in half, and uses the new concentration values to calculate. I was wondering how to do this if it was pressures, rather than concentration given?
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Re: Question 5.39 in Textbook; How to go about half-ing the volume?
Hi Meghna,
It's difficult to say how much the partial pressures would change, but if we halve the volume then the pressures of the gasses are going to increase as the gasses are going to collide more often. At that point we would recalculate the equilibrium concentrations using the ICE table.
I hope this helps Meghna!
It's difficult to say how much the partial pressures would change, but if we halve the volume then the pressures of the gasses are going to increase as the gasses are going to collide more often. At that point we would recalculate the equilibrium concentrations using the ICE table.
I hope this helps Meghna!
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Re: Question 5.39 in Textbook; How to go about half-ing the volume?
If the pressures were given rather than the concentration, assuming we know the volume, temperature, and moles of the reactants/products, we could use the ideal gas law (PV=nRT) in order to solve for what the new partial pressures would be. It would be a similar methodology as what was done for the concentrations, instead, we would just half the value of V in PV=nRT.
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