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Topic: Equilibrium Equations  (Read 3880 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?

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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?

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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.
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

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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