Chemistry equilibrium clarifications

Click For Summary

Discussion Overview

The discussion revolves around clarifications regarding chemical equilibrium, specifically the equilibrium constant (Kc) and its implications on the position of equilibrium in reactions. Participants explore concepts related to changes in concentration and their effects on equilibrium, as well as calculations involving equilibrium states in a specific reaction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • Some participants note that the value of the equilibrium constant (Kc) indicates the position of equilibrium, which reflects the relative proportions of reactants and products.
  • There is confusion expressed about how changing the concentration of reactants affects the position of equilibrium without altering the Kc value.
  • One participant describes a scenario where adding a substance to an equilibrium mixture causes the reaction quotient (Q) to differ from Kc, prompting a shift in equilibrium until Q equals Kc again.
  • Another participant presents a specific reaction (CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H2O) and asks for help calculating the number of moles of each substance at equilibrium, given an initial mixture of 2 moles of ethyl ethanoate and 2 moles of water.
  • There are discussions about using stoichiometry to express unknown quantities in the equilibrium calculation, with some participants suggesting relationships between the amounts of different substances.
  • One participant expresses uncertainty about the implications of Le Chatelier's Principle (LCP) and whether the equilibrium position can change while Kc remains constant.
  • Another participant provides a detailed example illustrating how the addition of a reactant affects the equilibrium position and the calculation of concentrations at equilibrium.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between changes in concentration, equilibrium position, and the constancy of Kc. There is no consensus on the implications of these concepts, and confusion remains regarding the application of LCP in this context.

Contextual Notes

Some participants highlight the need for clarity regarding the definitions of equilibrium and the assumptions underlying the calculations, particularly in relation to stoichiometry and the behavior of reaction quotients.

quantumtron
Messages
8
Reaction score
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.
 
Chemistry news on Phys.org
just to add, a really big thanks if advance. I have an upcoming chem test and this is fretting all over my brain.
 
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.

--
methods
 
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 :( )
 
No, equilibrium position will change, even if Kc stays identical.

Imagine we have a reaction

A + B <-> C

Q = \frac{[C]}{[A]<b>}</b>

(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]<b>} = \frac{1}{1\times 1} = 1 = K_c</b>

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]<b>} = \frac{1.2680}{1.7320\times 0.7320} = 1 = K_c</b>

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.

--
 

Similar threads

Activity - Equilibrium Expressions
  • · Replies 2 ·
Replies
2
Views
2K
Does the Value of K Depend on Initial Reactant and Product Amounts?
  • · Replies 3 ·
Replies
3
Views
2K
Question about equilibrium position
  • · Replies 5 ·
Replies
5
Views
2K
Iodine Adsorption and Titration with Na2S2O3
  • · Replies 4 ·
Replies
4
Views
3K
Trouble with Gibbs Free Energy & Equilibrium Constant Calc.
  • · Replies 3 ·
Replies
3
Views
3K
Equilibrium constant expression; why exponents?
  • · Replies 8 ·
Replies
8
Views
6K
Chemistry Decomposition of C5H6O3 equilibrium
  • · Replies 1 ·
Replies
1
Views
3K
Effects of Crushing a Solid on Equilibrium Position
  • · Replies 9 ·
Replies
9
Views
3K
Approximations in Chemical Equilibrium (add a weak acid HA into pure water)
  • · Replies 3 ·
Replies
3
Views
2K
Calculating time to reduce alcohol in wine using heating method
  • · Replies 131 ·
5
Replies
131
Views
11K