Chemistry equilibrium clarifications

In summary, the equilibrium constant measures the position of equilibrium for a reaction and indicates whether the system contains a larger proportion of reactants or products. Changing the concentration of reactants does not change the equilibrium constant, but instead shifts the position of equilibrium.
  • #1
quantumtron
8
0
-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 doesn't change the Kc value but shift the position. I am 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.
 
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  • #2
just to add, a really big thanks if advance. I have an upcoming chem test and this is fretting all over my brain.
 
  • #3
quantumtron said:
changing the concentration of reactants doesn't change the Kc value but shift the position. I am confused...

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.

-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.

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
Borek said:
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.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

im sorry but i still don't 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? ( I am dam weak in chemistry :( )
 
  • #5
No, equilibrium position will change, even if Kc stays identical.

Imagine we have a reaction

A + B <-> C

[tex]Q = \frac{[C]}{[A]}[/tex]

(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

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

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

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

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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Related to Chemistry equilibrium clarifications

1. What is equilibrium in chemistry?

Equilibrium in chemistry refers to the state where the rate of a forward reaction is equal to the rate of its reverse reaction. This means that the concentrations of reactants and products do not change over time.

2. How is equilibrium achieved in a chemical reaction?

Equilibrium is achieved when the forward and reverse reactions occur at the same rate, meaning that the concentrations of reactants and products no longer change. This can happen when the reactants are fully consumed, the products are fully formed, or when the reaction is reversible.

3. What factors affect equilibrium in a chemical reaction?

The factors that affect equilibrium in a chemical reaction include temperature, pressure, and concentration of reactants and products. Changes in these factors can shift the equilibrium in either the forward or reverse direction.

4. How does Le Chatelier's principle explain equilibrium?

Le Chatelier's principle states that when a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the system will shift in a direction that minimizes the effect of that change. This helps to maintain the equilibrium state.

5. How can equilibrium be manipulated in a chemical reaction?

Equilibrium can be manipulated by changing the temperature, pressure, or concentration of reactants and products. By altering these factors, the equilibrium can shift in either the forward or reverse direction, resulting in a change in the concentrations of reactants and products.

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