Final 2012 Q6B pg. 247 in the course reader

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Yael Zoken 4H
Posts: 25
Joined: Fri Sep 26, 2014 2:02 pm

Final 2012 Q6B pg. 247 in the course reader

Postby Yael Zoken 4H » Wed Dec 03, 2014 5:55 pm

You are given number of moles and find the volume of the container by using the ideal gas law. When figuring out the invidiual molar concentration each variable (x) and (0.05-x) are dived by 1.072 liters. But in other equilibrium problems, only the initial value is divided by the volume and then everything else remains unchanged (ie. not dividing by the volume). Why is this? When do we divide everything by the volume and when do we only divide certain parts of the equation?

Erika Monasch 4I
Posts: 5
Joined: Fri Sep 26, 2014 2:02 pm

Re: Final 2012 Q6B pg. 247 in the course reader

Postby Erika Monasch 4I » Wed Dec 03, 2014 6:40 pm

In this case, you need to divide by the volume because you are not given the concentration of PCL5. Rather, you are given the number of moles that are placed in the vessel. In the equilibrium constant equation, you put in either the concentration (in moles per liter) or the partial pressures. Due to the fact that the question gave you the number of moles, it is easiest to proceed by using the concentrations. You need to divide by the volume to ensure that all parts of the equation are in the same state (M=mol/L). In addition, this allows you to get "x" in moles, so you can then easily divide by the initial number of moles and multiply by 100 to get the percent of dissociation. In a lot of other problems, you are already given the molar concentration, so you do not need to divide by the volume. Does this help?

Yael Zoken 4H
Posts: 25
Joined: Fri Sep 26, 2014 2:02 pm

Re: Final 2012 Q6B pg. 247 in the course reader

Postby Yael Zoken 4H » Wed Dec 03, 2014 8:40 pm

Yes, thank you very much Erica

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Final 2012 Q6B pg. 247 in the course reader

Postby Niharika Reddy 1D » Wed Dec 03, 2014 10:38 pm

I think you can do this problem like the other ones we have done as well, with a few variations to how it's shown in the solutions. If you convert just the initial amount of PCl5 to concentration and say the change is -x for the reactant and +x for the products, you can set up the equilibrium expression and set it equal to the given Kc value. This expression and the x value calculated from it will be different than the solution shows, but that's because the x value calculated this way is in terms of concentration whereas the course reader has it in moles. When calculating the percent dissociated, since the change, the x value, is a concentration, we must divide by the initial concentration of PCl5, not the initial number of moles (it's done this way in the solutions because x is caluclated in moles there). Both ways give the same final percent dissociation of PCl5: 61%.


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