# Chemistry equilibrium clarifications

1. Aug 8, 2010

### quantumtron

-My notes says:
1.Value of equilibrium constant (Kc) provides a measure of the position of equilibrium.
2.Position of equilibrium indicates whether the system contains a larger proportion of reactants or products.

Next,
it mentioned, changing the concentration of reactants doesnt change the Kc value but shift the position. Im confused....

-Another question is, the value of equilibrium constant is 4 for the reaction below
CH3CO2H(l) + C2H5OH (l) ->-< CH3CO2C2H5(l) + H2O (l)

in an experiment, 2 mol of ethyl ethanoate and 2 mol of water are mixed. Calculate the number of moles of each substance present when equilibrium is reached.

2. Aug 8, 2010

### quantumtron

just to add, a really big thanks if advance. I have an upcoming chem test and this is fretting all over my brain.

3. Aug 8, 2010

### Staff: Mentor

Take a solution containing some substances in the equilibrium - reaction quotient (Q) has a value of Kc. Now add one of the substances already present. Initially Q is different from Kc, but the reaction will proceed (shifting the equilibrium) till Q is again equal to Kc.

Write formula for Q (Kc) - there are four unknowns, however, using stoichiometry of the reaction you can easily express some of the unknowns using others. For example, number of moles of water and number of moles of ethyl ethanoate is identical, and number of moles of ethanol is 2 minus number of moles of ethyl ethanoate.

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methods

4. Aug 8, 2010

### quantumtron

im sorry but i still dont understand, since your Kc will remain constant, means your equilibrium position will not change right? But by LCP the position will change thus equilibrium constant Kc will change right? ( Im dam weak in chemistry :( )

5. Aug 8, 2010

### Staff: Mentor

No, equilibrium position will change, even if Kc stays identical.

Imagine we have a reaction

A + B <-> C

$$Q = \frac{[C]}{[A]}$$

(to simplify all stoichiometric coefficients are 1)

Let's assume Kc for this reaction is 1.

Imagine further we create a solution containing 1 mole of each A, B and C. Obviously

$$Q = \frac{[C]}{[A]} = \frac{1}{1\times 1} = 1 = K_c$$

and the solution is in equilibrium.

Now imagine we add additional 1 mole of A. Now we have 2 moles of A, 1 mole of B and 1 mole of C. Solution is no longer at equilibrium (Q = 2 <> Kc, so - according to the LeChatellier's Principle - it has to react. What happens is that equilibrium shifts right, till [A] = 1.7320, = 0.7320 and [C] = 1.2680 - then again

$$Q = \frac{[C]}{[A]} = \frac{1.2680}{1.7320\times 0.7320} = 1 = K_c$$

Where did I got these values from? I have calculated them using approach outlined in my previous post. Note how the amount by which concentration of C went up is identical to the amount by which concentrations of A and B went down. That's because of the reaction stoichiometry, for each mole of C produced one mole of both A and B are consumed.

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6. Aug 12, 2010