Equilibrium+Overview

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=**__ Equilibrium Overview __**= "Order is not pressure which is imposed on society from wihout, but an equlibrium which is set up from within." - Jose Ortega y Gasset A gaseious chemical system can only achieve equilibrium when the chemical reaction is carried out in a closed container. For all systems trying to achieve equilibrium, the forward reaction occurs at a faster rate than the reverse reaction until equilibrium is reached. Once at equilibrium, the forward and reverse reactions occur at the same rate, and the concentrations of all species present no longer change. [1] For the above diagram, the forward reaction first occurs at a fast rate. As the forward reaction continues, it has a decreasing amount of reactants to undergo the reaction, therefore the rate slows. However, when an increasing amount of products are formed, the reverse reaction can occur at an increasing rate. This process continues until the rates of the forward and reverse reactions become equal (at equilibrium). To help illustrate this concept, let's take a look at the N 2 O 4 - NO 2 Equilibrium System. When N 2 O 4 is placed in a closed container at 100ºC, NO 2 forms. After this NO 2 forms, it can, in turn, create N 2 O 4. The partial pressure of N 2 O 4 and rate of the forward reaction decrease as more NO 2 is formed, and the partial pressure of NO 2 and rate of the backward reaction increase as more NO 2 is formed and some N 2 O 4 is reformed. This process continues until the rates of the forward and backward recations are equal. At this point in time, the system is in equilibrium, and the partial pressure of N 2 O 4 and NO 2 remain constant (as demonstrated in the diagram below). [2] The equilibrium constant, K, is a numerical value that represents an equilibrium system at a certain temperature. This number is independent of the original composition, the volume of the container, or the total pressure and changes with temperature only.
 * __Equilibrium Basics__**

An equilibrium constant expression can be written for every gaseous chemical system. aA(g) + bB(g) ‹—› cC(g) + dD(g) - A and B = reactants - C and D = products - a, b, c, and d = corresponding coefficients in the balanced equation This expression, otherwise known as K p, involves specifically the partial pressures of the gases at equilibrium, which must be expressed in atmospheres. Other equilibrium expressions involve equilibrium concetrations in moles per liter. Put simply, the equilibrium constant expression, K, is always expressed by equilibrium product values divided by equilibrium reactant values with the coefficients of each equation acting as exponents. __Examples__ Write the equilibrium constant expression for the following reactions: 4NH 3 (g) + 5O 2 (g) ‹—› 4NO(g) + 6H 2 0(g) HF(aq) ‹—› H^+(aq) + F^-(aq) The two constants K p and K c are simply related by the equation - delta ng = change in the number of moles of gas in the equation - R = gas law constant [0.0821 (L x atm) / (mol x K)] - T = Kelvin temperature [|Kp/Kc Relationship] (example)
 * __The Equilibrium Constant Expression__**

"The expression for K depends on the form of the chemical equation written to describe the equilibrium system." [2] The coefficient rule, K' = K^n, states that "if the coefficients in a balanced equation are multiplied by a factor n, the equilibrium constant is raised to the nth power." [2] The reciprocal rule, K" = 1/K, states that "the equilibrium constants for forward and reverse reactions are the reciprocals of each other." [|Quotient and Reciprocal Rules] (examples) The rule of multiple equilibria states "if a reaction can be expressed as the sum of two or more reactions, K for the overall reaction is the product of the equilibrium constants of the individual reactions. If reaction 3 = reaction 1 + reaction 2, then K(reaction 3) = K(reaction 1) + K(reaction 2)." [|Rule of Multiple Equilibria] (example)

Hetergenous equilibria are systems in which more than one phase of matter is present in the reaction. If solids and/or liquids are present in these systems, their equilibrium positions do not depend on the amounts of such solids and/or liquids. Therefore, solids and liquids can be left out of all equilibrium constant expressions. __Example__ Given the following description of reversible reactions, write a balanced net ionic equation and the equilibrium constant expression. Hydrogen sulfide gas (H 2 S) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions. H 2 S(g) + Pb^2+(aq) ‹—› PbS(s) + 2H^+(aq)

For a better understanding or a different interpretation of the above information, please view the following video. [|Introduction of Chemical Equilibrium] [3]

In order to calculate the value of an equilibrium constant for a system, you must know the partial pressures of the products and reactants at equilibrium. When determining K, you may be given the partial pressures of the products and reactants at equilibrium. [|K with Equilibrium Partial Pressures] (example)
 * __Determination of K__**

Sometimes, you may be given the initial partial pressures and the equilibrium partial pressure of one species. To calculate K in this case, you must deterime the partial pressures of all the species at equilibrium, which can be quite challenging. For problems like this, we use an I.C.E. Chart, which visually represents the initial partial pressures, the change in those partial pressures, and the equilibrium partial pressures of a system in a table. In correlation with the I.C.E. Chart, we follow the principle that the changes in the partial pressures of products and reactants must agree with the reaction stoichiometry. This means that the changes are related through the coefficients in the balanced equation. [|Determine_K_with_I.C.E._Chart.jpg] [2] (example)

In a system at equilibrium, if... - K < 1: the denominator is larger than the numerator; the partial pressures of the reactants are bigger than those of the products; the reactants are favored; the forward reaction basically does not go - K > 1: the numerator is larger than the denominator; the partial pressures of the products are bigger than those of the reactants; the products are favored; the forward reaction basically goes to completion.
 * __Applications of the Equilibrium Constant__**

The reaction quotient, Q, takes the same form as the equilibrium constant, K (products / reactants). However, it involves the partial pressures of reactants and products in a system at any one moment in time, not necessarily at equilibrium. "The reaction quotient can be used to determine in which direction a system must shift in order to reach equilibrium." [4] If... - Q < K: The amount of reactants is too high, and the amount of products is too low. The system must shift to the right to form a larger amount of products from the already large amount of reactants. This decreases the denominator, increases the numerator, and raises the value of Q until it is equal to K. - Q > K: The amount of reactants is too low, and the amount of products is too high. The system must shift to the left to form a larger amount of reactants from the already large amount of products. This increases the denominator, increases the numerator, and lowers the value of Q until it is equal to K. - Q = K: The system is at equilibrium. Therefore, the system does not need to shift. [|Q v. K] (example)

You can use the equilibrium constant to calculate the equilibrium partial pressures of the reactants and products present in a system. Sometimes, you will be given K and the equilibrium partial pressures of all species except one, and you will be asked to find the unknown equilibrium partial pressure. However, usually, you will be given the inital partial pressures of the reactants and products in the system and be asked to find their equilibrium partial pressures. To solve problems like this, follow these steps: 1. Write: Write the equilibrium constant expression using the balanced equation for the reaction. 2. Set up: Set up an I.C.E. Chart. Place the initial partial pressures on the top row in the appropriate columns. Fill in the appropriate amount of change (±#X) on the middle row. Write the equilibrium terms on the bottom row. 3. Substitute: Substitute the equilibrium terms into the equilibrium constant expression. 4. Solve: Solve the equation for x. This can be quite challenging at times. You may decide to use the quadratic formula (if applicable) or simply plug it into a graphing calculator (Y 1 = numerical K value, Y 2 = equilibrium constant expression, graph, intersection x-coordinate is x). However, most of the time, the amount of change is so small that when added or subtracted to an initial partial pressure, we assume that it is negligable and solve for x as if it weren't there. To make sure the assumption is valid, use the 5% rule. Divide the value you found to be x by the initial partial pressure and multiply that number by 100. If the result is less than 5%, the asumption is valid. 5. Calculate: Calculate the equilibrium partial pressure of all species by plugging your obtained x value into the equilibrium terms of each species. [|Determine Equilibrium Partial Pressures with I.C.E Chart by Shannon R.]

For a better understanding or different interpretation of the applications of the equilibrium constant with a focus on I.C.E. Chart examples, please visit the website linked below. This website also includes an introduction to the rate unit, which you should have learned about already. [|Gas-Phase Equilibrium] [4] For a more in depth discussion about the mechanics of the I.C.E. Chart accompanied by examples, please view the following video. [|Calculating Magnitude of Equilibrium Shifts] [5]

If you are having issues with any of the information above, need a more in depth explanation, or want more examples to reference. Please visit the website linked below and scroll to the "Chemical Equilibrium" section. It covers everything you might need to know about solving chemical equilibrium problems. [|How Do I Solve It?] [6] This website also discusses Acid and Base Equilibria and Precipitation Equilibria, which involve major types of equilibrium systems that you will learn about and manipulate once you have become an expert on the basics of equilibrium. The website even introduces other topics that you will need to know for the AP Exam, so it may be a good idea to bookmark it.

Le Chatelier's Principle states that "if a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the system will, if possible, shift to partially counteract the change." [2]
 * __Le Chatelier's Principle__**

Adding or Removing a Gaseous Species: "If a chemical system at equilibrium is disturbed by adding a gaseious species (reactant or product), the reaction will proceed in such a direction as to consume part of the added species. Conversely, if a gaseous species is removed, the system shifts to restore part of that species." [2] Put more simply, if the partial pressure of a gas or concentration of an aqueous ion is increased, the reaction will shift to the opposite side to resetablish equilibrium. If the partial pressure of a gas or concentration of an aqueous ion is decreased, the reaction will shift to the same side (the side from which the species is being removed) to reestablish equilibrium.

Compression or Expansion: "When the system is compressed, thereby increasing the total pressure, reaction takes place in the direction that decreases the total number of moles of gas. When the system is expanded, thereby decreasing the total pressure, reaction takes place in the direction that increases the total number of moles of gas."[2] Put more simply, if the pressure increases (compression), the reaction shifts toward the side with less moles of gas. If the pressure decreases (expansion), the reaction shifts toward the side with more moles of gas. This is illustrated by the diagram below. [2] Change in Temperature: "An increase in termperature causes the endothermic reaction to occur." [2] This portion of Le Chatelier's Principle involves a concept of thermodynamics - delta H or the change in heat or energy. If dela H is positive, the system is endothermic, and this change in heat serves as a reactant. If delta H is negative, the system is exothermic, and this change in heat serves as a product. Once it is determined whether delta H is a product or reactant, the disturbance to equilibrium by a change in temperature becomes very easy to deal with. If the reaction is endothermic (delta H is a reactant) and the temperature increases, the reaction shifts to the right. If the temperature decreases, the reaction shifts to the left. If the reaction is exothermic (delta H is a product) and the temperature increases, the reaction shifts to the left. if the temperature decreases, the reaction shifts to the right. it is also important to know that a change in temperature is the only disturbance that can change the value of the equilibrium constant. [|Le Chatelier's Principle] (example provided by Kyla)

For a better understanding or different interpretaion of Le Chatelier's Principle, please visit the website linked below. [|Le Chatelier's Principle] [7] For a more in depth discussion about how to determine K and Le Chatelier's Principle, please view the following video. [|Using and Calculating Keq, and Le Chatelier] [8] For a funny way of showing Le Chatelier's Principle check out this video. [|Le Chatelier's Principle; the fun way.] (Added by Robert Hudson)

**The Common Ion effect is another example of Le Chatelier's principle** The common ion effect tells us that the solubility of an ionic compound is //**decreased**// by the addition to the solution of another ionic compound that contains one of the ions involved in the solution equilibrium. AN EXAMPLE: If we have a barium sulfate solution, the solid salt is in equilibrium with its ions:
 * BaSO4 (s) || [[image:http://www.saskschools.ca/curr_content/chem30_05/graphics/site_wide/darrow.gif width="20" height="10" caption="in equilibrium with"]] || Ba2+(aq) + SO42-(aq) ||

If we then add solid barium chloride to this solution, which dissolves to produce Ba2+ and Cl- ions, we are increasing the concentration of Ba2+ ions in our solution. (The new Cl- ions will remain in solution as spectator ions). Ba2+ is the ion common to both solutions. Le Châtelier's Principle tells us that if the concentration of one of the reaction participants is increased, then equilibrium will shift to use up the additional substance. So adding more Ba2+ will force the equilibrium to shift to the left (the reverse direction) in order to use up the added Ba2+ ions, producing more solid BaSO4. The concentration of SO42- will decrease, indicating that solubility has decreased. (JENNIFER SAMUELS)

 

There will be an example problem using the Nernst equation here! Santo-

[1] [] [2] __Chemistry: Principles and Reactions, Fifth Edition__ by Masterton and Hurley, pages 320 - 349 [3] [] [4] [] [5] [] [6] [] [7] [] [8] []