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Topic: Equilibrium  (Read 2448 times)

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Offline djohnny

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Equilibrium
« on: July 14, 2016, 07:40:48 PM »
Hello, I am having some issues with the following question. Thanks in advance!

1. [Ag(NH 3 ) 2 ] + is a type of ion called a complex ion. Consider the equilibrium:

AgCl (s) + 2 NH 3(aq)  :rarrow: [Ag(NH 3 ) 2 ] + (aq) + Cl¯ (aq)

If the equilibrium will shift (highlight):

Cl¯ is removed ← → no change

I 2 is added ← → no change

the pressure is increased ← → no change

[Ag(NH 3 ) 2 ] + is removed ← → no change

And this question, too:

2. Consider the equilibrium:

H 2(g) + I 2(g)  :rarrow: 2 HI (g) ∆H = +52 kJ

If the equilibrium will shift (highlight)

H 2 is removed ← → no change

HI is added ← → no change

the system is cooled ← → no change

the pressure is decreased ← → no change

Offline Borek

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Re: Equilibrium
« Reply #1 on: July 14, 2016, 11:48:42 PM »
Cl¯ is removed ← → no change

This is apparently about Le Chatelier's principle.

What does it say?
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Offline djohnny

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Re: Equilibrium
« Reply #2 on: July 15, 2016, 08:39:57 AM »
It asks what happens if Cl¯ is removed. Does the equilibrium shift right ( :rarrow:), left ( :larrow:), or doesn't change (no change).

Thanks!

Offline Borek

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Re: Equilibrium
« Reply #3 on: July 22, 2016, 03:46:49 PM »
I asked what does the principle say.
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Offline docnet

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Re: Equilibrium
« Reply #4 on: August 22, 2016, 01:59:48 PM »
According to the Le Chatelier's principle, in the complex ion formation, removing a product would shift the equilibrium towards the *product* side. It would be hard to say what kind of effect pressure would have in a reaction that takes place in an aqueous solution, but it would have very little effect.

In the reaction of the gases, Le Chatelier's principle still applies. Iodine gas and Chlorine gas are both diatomic homo nuclear molecules, and HI is a diatomic hetero nuclear molecule, therefore pressure alone would have no effect on the partial pressures of the gases at equilibrium. (In other words number of moles do not change from the reactant to the product side).

Since the reaction enthalpy is positive, which indicates that the formation of the product is endothermic, decreasing the system's temperature would mean less product, more reactant.

Edited: ...towards the *product side  :-[
« Last Edit: August 22, 2016, 05:25:35 PM by docnet »

Offline Borek

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Re: Equilibrium
« Reply #5 on: August 22, 2016, 03:50:07 PM »
According to the Le Chatelier's principle, in the complex ion formation, removing a product would shift the equilibrium towards the reactant side.

And not the opposite?
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