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Topic: Equilibrium Equations  (Read 3072 times)

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Offline 1QWK96GT

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Equilibrium Equations
« on: November 07, 2013, 02:56:02 PM »
Hey everybody. I am trying to solve some of these equations my teacher gave us to do.

1) The initial concentrations of A, B, and C are 1.1 M, 2.2 M, and 3.3 M respectively and the value of K is 1.3x10^3 at 25celsius.
a. Is this system at equilibrium?
b. If not, which way will it shift?
A(aq) + B(aq)  ::equil:: C(aq) + D(l)

Now for a.) I think they are equal because both arrows show the reaction going both ways?

For b.) I believe it is equal so this does not apply?

We briefly went over these in class there must be something I am missing.

Thanks,

Offline Corribus

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Re: Equilibrium Equations
« Reply #1 on: November 07, 2013, 03:06:46 PM »
How the reaction is written doesn't make a bit of difference.  Are you familiar with the reaction quotient, Q?
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Offline 1QWK96GT

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Re: Equilibrium Equations
« Reply #2 on: November 07, 2013, 03:58:00 PM »
I just looked it up its the reaction of products/reaction of reactants?

Also now I think that it is in fact not at equilibrium because of the liquid on one side. Which will cause it to shift to the left?

Offline Borek

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Re: Equilibrium Equations
« Reply #3 on: November 07, 2013, 04:13:06 PM »
I just looked it up its the reaction of products/reaction of reactants?

Not "reaction of products" - product of concentrations of products in correct powers over product of reactants in correct powers.

Quote
Also now I think that it is in fact not at equilibrium because of the liquid on one side. Which will cause it to shift to the left?

If something is in a separate phase (as solid or liquid) its concentration (more precisely activity) is assumed to equal 1.

Try to write the quotient for this particular reaction.
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Offline 1QWK96GT

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Re: Equilibrium Equations
« Reply #4 on: November 07, 2013, 04:32:46 PM »
Okay,
A=1.1M
B=2.2M
C=3.3M

It does not give me D for some reason.

1.1M+2.2M=3.3M Thats for A+B

3.3 + ? For d= 3.3M

Divide the two 3.3/3.3=1

Compare Q to K
K=1.3x10^3
Q=1

So K>Q

So it will shift to the right?

Offline Borek

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Re: Equilibrium Equations
« Reply #5 on: November 07, 2013, 04:47:15 PM »
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Offline Corribus

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Re: Equilibrium Equations
« Reply #6 on: November 07, 2013, 05:06:27 PM »
Here's a quick remedial lesson:

The equilibrium constant for a reaction xA + yB  ::equil:: zC + rD is:

[tex]K = \frac{[C]^z[D]^n}{[A]^x[ B]^y}[/tex]

Where here, the concentrations are concentrations once equilibrium has been reached.  In reality, we should use chemical activities, but a good approximation is that the activities are equal to the molar concentrations, except for pure liquids and solids, which are assigned activities (and hence "concentrations") = 1.  This is why they didn't give you a value for D.

Q, the reaciton quotient, is obtained by using the real concentrations for a system at a given moment of time.  By computing Q and comparing it to K, you can then determine which direction the reaction will proceed from the given starting point.
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Offline Corribus

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Re: Equilibrium Equations
« Reply #7 on: November 07, 2013, 11:45:00 PM »
Sorry, typo: the reaction should be xA + yB  ::equil:: zC + nD
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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