Equilibrium

Equilibrium (Chapter 15, pages: 629-661; chapter 18, pages: 799-803)

1. Be able to define the following concepts: a. Chemical Equilibrium A system of reversible reactions in which the forward and reverse reactions occur at equal rates, such that no net change in the concentrations occurs, even though both reactions continue. Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products. Equilibrium is a dynamic state. Equilibrium occurs in closed systems. b. **Reversible Reactions** Reversible reactions are the exact opposite of the forward reaction of a chemical equilibrium reaction. So, in a chemical equilibrium, not only can a forward reaction happen with reactant consumption and product formation, but a reversible reaction can also happen with the products being consumed and the reformation of the reactants. Also, about reversible reactions worth mentioning - the equilibrium concentration/pressure constant for a reversible reaction is the inverse of the equilibrium concentration/pressure constant of the forward reaction. A reversible reaction is one in which the reactants can be converted into products, but also the products can be converted back into the reactants. The reaction can run either direction, forward or backward. An example of this would be ice melting (forward direction) and then freezing again (backward direction). Theoretically, every reaction is reversible, however, experimentally, not enough energy can be harnessed to reverse some reactions. c. Equilibrium Constant (K) a number equal to the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants, with each concentrations raised to the power of its stoichiometric coefficient. A combination of various different constants combined to K. One of the things the equilibrium constant tells us is the extent to which a reaction proceeds at a particular temperature. The equilibrium constant= products/reactants (Equilibrium constant and rate constant are not the same. Keq is the equilibrium constant) " Reactions don't stop when they come to equilibrium. But the forward and reverse reactions are in balance at equilibrium, so there is no net change in the concentrations of the reactants or products, and the reaction appears to stop on the macroscopic scale."

A small K value is less than .01. This means that very few products will be created and most of the reactants will remain. An Intermediate value is .01100 This means that very few reactants will remain and most will be converted to products.

d. Equilibrium Expression the quotient of product concentrations and reactant concentrations, each raised to the power if its stoichiometric coefficient. This expression relates the equilibrium concentrations to the equilibrium constant. For Example: The Equilibrium expression for the reaction: 2SO3 (g) ↔ 2SO2 (g) + O2 (g) would be:

Product concentrations (along with their exponents) are multiplied together to form the numerator and reactants (along with their exponents) are multiplied together to form the denomenator. Solids and Liquids are not included in the expression because their concentration is constant throughout the whole reaction. e. Le Chatelier’s Principle can be used to predict the effect of change in conditions on a chemical equation. If a system at equilibrium changed, it responds by returning toward its original equilibrium position. It describes the phenomenon that if a stress is applied to a system at equilibrium, the reaction will shift to partially alleviate the stress and balance itself back out. .f. **Homogeneous Equilibrium** A homogeneous equilibrium is when the reactants and the products are both in the same physical state at equilibrium. "Has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is present in the same solution." Results when forward and reverse reactions are equal. g. **Heterogeneous Equilibrium** A heterogeneous equilibrium is when the reactants and the products are not all in the same physical state at equilibrium, but when the reactants and products are in different physical states at equilibrium. Solids and liquids are always thrown out when setting up the equilibrium constant. " has things present in more than one phase. The usual examples include reactions involving solids and gases, or solids and liquids." Results when the forward and reverse Reaction Quotients are not equal. l. Reaction Quotient (Q) a fraction with the product concentrations in the numerator and reactant concentrations in the denominator, each raised to its stoic hiometric coefficient The reaction quotient equals the equilibrium constant at equilibrium.Qc=Kc (equilibrium)

When Q > K the reaction moves in reverse. and if K > Q the reaction moves forward.

2. Be able to use the Le Chatelier’s Principle to predict the effects of the following changes to a reaction in equilibrium: (this principle states : If a system at equilibrium is changed it will respond by returning towards its original equilibrium position) a. Concentration of reactants Increasing the concentration of the reactants will not only increase the reactant rate, but it will also increase the amount of product. Likewise, decreasing the concentration of products will cause the reaction to produce more products to reach equilibrium. Decreasing [reactants] shifts to reactants, or left side. Also, increasing the products will do this. Increasing [reactants] shifts to products, or right side.Also, decreasing the products will do this. b. Concentration of products Increasing the concentration of the products will force a larger amount of products to result. Actually increasing the concentration of products will produce an increase in the reactants. If we remove product, the reactants will react more in order to balance out the effect of the product removal, in turn creating more product. Decreasing [products] shift to products, or right side. Increasing [products] shifts to reactants, or left side. c. Temperature The effect of changing the temperature in the equilibrium can be made clear by incorporating heat as either a reactant or product. When the reaction is exothermic (ΔH is negative, puts energy out), we include heat as a product, and when the reaction is endothermic (ΔH is positive, takes energy in), we include it as a reactant. In an endothermic (breaking bonds) reaction, if you increase the temperature the reaction shifts towards the products; if you decrease the temperature it shifts towards the reactants. In an exothermic (forming bonds) reaction, an increase in temperature causes a shift towards the reactants and a decrease in temperature shifts towards products. d. Pressure If the pressure of a system is changed ( at constant temperature and volume), it will respond in such a way as to return the pressure towards the original equilibrium vale. 1. An increase in pressure favors the side with fewer moles of gas. 2. An decrease in pressure favors the side with more moles of gas. e. Catalyst A catalyst only increases the rate of the reaction. It will not effect the equilibrium. For example, if uncatalyzed reaction's reactant A to product B ratio is 1:1, the catalyzed reaction's ratio is the same! The catalyst only increases the speed of which your products reach the equilibrium state. A substance that participates in a reaction, is not consumed, and modifies the mechanism of the reaction to provide a lower activation energy.

3. Be able to write the correct expression for the Equilibrium Constant for species in the following states: ( why and examples ) a. Solid (s) - will not be present it the equilibrium constant!ex: CaCO3(s) arrow CaO(s) + CO2(g) Ke= (CO2)/1 because the concentration of a solid is constant. b. Liquid (l) - will not be present it the equilibrium constant! because liquid only increased the volume but does not affect the concentration ie: Hg(l)+Hg2+(aq) arrow Hg2+/2 (aq)Kc = (Hg2+/2)/Hg2+ c. Aqueous (aq) - will be present in the equilibrium constant! ie HF(aq) arrow H+(aq) +F-(aq)Kc = (H+)(F-)/(HF) Because (aq) solutions influence the concentration of the chemicals involved and therefore the Keq d. Gas (g)- will be present in the equilibrium constant! ie; CO(g) + H2O(g) arrow CO2(g)+H2(g) Kc=(OC2)(H2)/(CO)(H2O) Because (g) solutions influence the concentration of the chemicals involved and therefore the Keq

4. Be able to calculate the equilibrium constant values using the concentration in equilibrium of the products and reactF K(eq) = K(1) / K(-1) = [B]/[A] = [products]/[reactants] This equation comes from dividing the rate law of the products by the rate law of the reactants.

5. Be able to predict the effects on changes in temperature on the equilibrium constant. Changes in concentration or volume may alter the position of the equilibrium but it will no change the value of the equilibrium constant. Only a change in temperature will alter the value of the equilibrium constant. If we treat heat as though it were a reactant, we can use Le Chateliers principle to predict what will happen if we add or remove heat. Increasing the temperature will shift an exothermic reaction in the forward direction because heat appears on the reactant side. Lowering the temperature will shift the exothermic reaction in the reverse direction. Naturally, for an endothermic reaction lowering the temperature causes a shift to the product side, and raising the temperature shifts it to the reactant side.

6. Be able to calculate the values of the reaction quotient (Q). For the general reaction, aA + bB <--> cC + dD at equilibrium, the reaction quotient Qc is equal to the equilibrium constant Kc Qc = [products]/[reactants] = Kc (at equilibrium)

7. Be able to predict the direction of the reaction given a reaction quotient value. If the reaction quotient is smaller than the equilibrium constant, the system proceeds in the forward direction. If the reaction quotient and the equilibrium constant are equal, there will be no net reaction in either direction. If the reaction quotient is larger than the equilibrium constant, then the system proceeds in the reverse direction.

If Q < K eq then the reaction moves forward to the products. The Q value directly correlates with the amount of products. If there are less products then the Keq dictates, then the reaction will move towards the products. If Q > K eq then the reaction moves backward to the reactants. The Q value here tells us that the products are greater than allowed by the Keq. Therefore, the reaction shifts towards the reactants to accommodate the overly abundant amount of product. If Q=K eq then the reaction is at equilibrium.

8. Correlate the equilibrium constant values with relative amounts of reactants and products The relationship between the equilibrium expression and the balanced chemical equation for the reaction: the numerator contains the product concentration raised to a power equal to its stoichiometric coefficient in the balanced chemical equation. This is in the equilibrium expression.

9. Calculate the following values (for small, intermediate and large Keq values) small Keq k<0.01 intermediate Keq 0.01100 a. Concentration in equilibrium of a reactant, given the initial concentration of a reactant and the equilibrium constant value b. Concentration in equilibrium of a product, given the initial concentration of a reactant and the equilibrium constant value c. Initial concentration of reactants, given a equilibrium concentration of a product and the equilibrium constant value

a. Sample Problem: What is the concentration of acetate in equilibrium in a solution prepared by adding 0.5 moles of acetic acid in 1L of water. Small K = [products]/[reactants] Keq less than or equal to ~10^-5 Ch3COOH <--> CH3COO- + H+ Keq = 1.8 x 10^-5 Keq = [CH3COOH-][H+] / [CH3COOH] Ch3COOH <--> CH3COO- + H+
 * = Initial (I) ||= 0.500 M ||= 0 ||= 0 ||
 * = Change (C) ||= -x ||= +x ||= +x ||
 * = Eq (E) ||= 0.500-x ||= x ||= x ||

Keq = [CH3COOH-][H+] / [CH3COOH] = x^2 / (0.5 -x) = 1.8 x 10^-5 x^2 / (0.5 -x) = 1.8 x 10^ -5 x = √((1.8 x 10^-5)(0.5M)) x = 3.0 x 10 ^ -3 or 0.003M
 * The "-x" in this problem can be canceled out because since k is small, this x is extremely close to 0 therefore the assumption is it is 0

10. Be able to solve equilibrium problems using partial pressures to solve the equilibrium with a partial pressures we will be using the ideal gas law PV=nRT If moles of gas are constant, then the equation can be used. 11. Be able to calculate Keq value for a reactions by: a. Using the equilibrium constant of the same reaction with different coefficients To calculate the Keq value for a reaction using the equilibrium constant we have to option we either divide or multiply the reaction to find the initial ( original equation) b. Manipulating the equilibrium constant from other reactions we need to know to work with reaction, mass- action, control experiment and equilibria. when given two or three reaction how do we go back to the initial equation?

12. Correlate equilibrium constant values to: a. Rate constant values The equilibrium constant is NOT the same thing as the rate constant. In the equilibrium constant, we can use the coefficients of the balanced chemical reaction equation, but for the rate constant k we cannot do this. b. Change in Gibbs free energy Where R is the universal gas constant, T is the temperature, DG is gibbs free energy, Ln is natural log, and Keq is the equilibrium constant.