Le Chatelier TBR gen chem ex 3.11

[keikoblue2]

Hi, this is a simple question but I needed clarification ._.

"What is the observed result of increasing the total pressure under isothermal conditions in the following system initially at equilibrium?

PCL3(g) + Cl2(g) ---> <--- PCl5(g)"

C. A decrease in the ratio of of P(PCl5) to (P(PCl3) x (PCl2))
D. An increase in the P(PCl5) to P(PCl3) ratio

The correct answer is D and I understand it's because increasing the external pressure results in a shift to the right in order to decrease/counteract the increased pressure. Therefore the pressure of the product will increase while the pressures of the reactants will decrease, leading to answer D.

What I don't get is why can't C be correct? The book says C can't be correct because the Keq doesn't change unless temperature changes, but what about the reaction quotient? I thought C was referring to Q and not Keq.... does Q only mean CONCENTRATION of products over reactants? Therefore since C was referring to the pressures of products/reactants, it was referring to Keq, and not Q?

Also, for a dilution of a solvent, you're decreasing the solvent's concentration. Isn't this increasing the volume of the system, leading to a shift toward the side with more moles, since you want to increase it's concentration/collisions to balance the stress? I don't understand why the book equates "dilution of the solvent" with "Increasing the external pressure" since "increasing the external pressure" leads to a smaller volume and shifts to the side with less moles.

---

Members don't see this ad.

 

[gettheleadout]

Q and K are the same expression at different points in the equilibration of the reaction. This reaction starts at equilibrium, so Q equals K to begin with, and since the system change is isothermal K won't change and thus Q won't change.

 

[keikoblue2]

Q and K are the same expression at different points in the equilibration of the reaction. This reaction starts at equilibrium, so Q equals K to begin with, and since the system change is isothermal K won't change and thus Q won't change.

Thanks for your help! But I thought isothermal just means K won't change? After an external pressure is applied, the system is no longer in equilibrium so Q changes and does not equal K, right? Or are you saying Q won't change after the new equilibrium is established?

I guess I'm confused because in the book's previous example, they had this chart. Keq is assumed to be 1.00 and initially, the partial pressures of all components is 1.00 atm. After doubling the external pressure:

Reaction: A(g) + B(g) ---> <--- C(g) C/(AxB) State
Initially: 1.0 1.0 1.0 1.0 equilibrium
After stress: 2.0 2.0 2.0 0.5 not equilibrium
Shift: -x -x +x reacting
Final: 2-x 2-x 2+x 1.00 new equilibrium


After stress, the new C/(AxB) ratio is clearly less than K (not in equilibrium) so I thought it was the same for ex 3.11 and answer C was referring to Q and not Keq. OH but wait, did ex 3.11 mean what is observed after the new equilibrium is established from a result of increasing the pressure? Because that would rule out C (Since C/(AxB) is the same as initial after the new equilibrium is established) but the ratio of C to either A or B would increase.

I might have just answered my own question... if anyone can confirm this, please do. Thank you so much!

 

[enervate]

Also, for a dilution of a solvent, you're decreasing the solvent's concentration. Isn't this increasing the volume of the system, leading to a shift toward the side with more moles, since you want to increase it's concentration/collisions to balance the stress? I don't understand why the book equates "dilution of the solvent" with "Increasing the external pressure" since "increasing the external pressure" leads to a smaller volume and shifts to the side with less moles.

You're right about this part. I think TBR goofed on this one. Dilution is in line with decreasing the external pressure; the side with more molecules is favored. Evaporation "increases the pressure of the system" so the side with fewer molecules is favored.

 

[deh2ase]

Thanks for your help! But I thought isothermal just means K won't change? After an external pressure is applied, the system is no longer in equilibrium so Q changes and does not equal K, right? Or are you saying Q won't change after the new equilibrium is established?

I guess I'm confused because in the book's previous example, they had this chart. Keq is assumed to be 1.00 and initially, the partial pressures of all components is 1.00 atm. After doubling the external pressure:

Reaction: A(g) + B(g) ---> <--- C(g) C/(AxB) State
Initially: 1.0 1.0 1.0 1.0 equilibrium
After stress: 2.0 2.0 2.0 0.5 not equilibrium
Shift: -x -x +x reacting
Final: 2-x 2-x 2+x 1.00 new equilibrium


After stress, the new C/(AxB) ratio is clearly less than K (not in equilibrium) so I thought it was the same for ex 3.11 and answer C was referring to Q and not Keq. OH but wait, did ex 3.11 mean what is observed after the new equilibrium is established from a result of increasing the pressure? Because that would rule out C (Since C/(AxB) is the same as initial after the new equilibrium is established) but the ratio of C to either A or B would increase.

I might have just answered my own question... if anyone can confirm this, please do. Thank you so much!

Immediately after the stress, Q (C/(AxB) ratio) would change if the system has not reached equilibrium yet. But yes you are right, they allow the reaction to reach equilibrium after the stress, thus Q=K. The ratio of C/A or C/B increases because the same equilibrium has to be reestablished by synthesizing more of the products.