Equilibrium Notes AP Chemistry What is chemical equilibrium

Equilibrium Notes AP Chemistry

What is chemical equilibrium? n A dynamic state of a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction in a closed system, but no net effect is observable No change in macroscopic properties such as pressure, color, or concentration at constant temperature n But at the atomic level, atoms, ions, or molecules are in constant reaction n

Representing Equilibrium Reactions

The concept of equilibrium n Consider Forward reaction: A B Rate = kf[A] Reverse reaction: B A Rate = kr[B] n At equilibrium kf[A] = kr[B]. n Therefore, as the reaction progresses § § § [A] decreases to a constant, [B] increases from zero to a constant. When [A] and [B] are constant (but not necessarily equal), equilibrium is achieved

n Alternatively: § kf[A] decreases to a constant, § kr[B] increases from zero to a constant. § When kf[A] = kr[B] equilibrium is achieved.

Equilibrium Constant Expressions n Mathematical expression of the equilibrium condition n Results in a quantitative value for an equilibrium constant under specified conditions n Based on the law of mass action § § § The equilibrium constant (Keq) is computed as the ratio of the mathematical product of the concentrations of the products to the mathematical product of the concentrations of the reactants. When molarities are used, the equilibrium constant is termed Kc When partial gas pressures are used to represent concentrations, the constant is termed Kp

Equilibrium constant expressions using the law of mass action n For the reaction a. A(aq) + b. B(aq) c. C(aq) + d. D(aq) Kc= [C]c [D]d [A]a [B]b n If the reaction was a. A(g) + b. B(g) c. C(g) + d. D(g) K p= P Cc P Dd P Aa P Bb

Important things to remember about equilibrium constant expressions n The reaction must be reversible n The coefficients in the balanced equation are equal to the exponents n Products over reactants n The concentrations must be the equilibrium concentrations n Equilibrium constants n n are unitless Are temperature dependent, since temperature changes affect the forward and reverse reaction rates by different amounts, thus affecting the ratio of concentrations at equilibrium and thus the value of the equilibrium constant n Pure solids (s) and pure liquids (l) are excluded from the expression

Names and symbols for equilibrium constants n Special names and symbols are used for different n n n types of chemical reactions Kc (most general form with molar concentrations) Kp (can be used with partial pressures when working with a gas phase reaction) Ka (used for the dissociation of weak acids in water) Kb (used for the dissociation of weak bases in water) Kw (the equilibrium expression for the dissociation of water into its ions) Ksp (used for the dissociation into ions of sparingly soluble solids in water)

Kp vs. Kc n Mathematical relationship between Kp and Kc for gas phase reactions where either molar concentrations or partial pressures can be used Kp= Kc(RT)Δn(gas) R=the gas law constant (. 0821 L atm/mol K) T=temperature in K Δn(gas)= moles of gaseous products - moles of gaseous reactants (from coeffiecients) **This is the “politically correct” (pc) equation- to remember the order of the K’s in the eqn

Size of Keq : The Equilibrium Position n If we say equilibrium lies far to the left or favors the reactants The reactant concentrations are much larger than the product concentrations at equilibrium n K<1 n n If we say equilibrium lies far to the right or favors the products The product concentrations are much larger than the reactant concentrations at equilibrium n K>1 n

Size of Keq : The Equilibrium Position n Whether a reaction lies far to the right or to the left depends on three main factors Initial concentrations (more collisions- faster reaction) n Relative energies of reactants and products (nature goes to minimum energy) n Degree of organization of reactants and products (nature goes to maximum disorder) n

Size of Keq : The Equilibrium Position n If Keq > 10 n The reaction goes nearly to completion; products are favored; equilibrium lies to the right n The concentration of the products are much greater than the concentration of the reactants, which simplifies calculations n If Keq < 0. 1 n The reaction doesn’t go very far to completion; reactants are favored; equilibrium lies to the left n The concentrations of the reactants are much greater than the concentrations of the products, which simplifies calculations n If Keq =1 n There are substantial amounts of both product and reactant present n No approximations may be made

Other rules of interest n Coefficient rule n If the coefficients of the reaction are increased or decreased by a factor of n, then the K is raised to that power n Reciprocal rule n If you reverse a reaction, you reciprocate (invert) the K n Multiple equilibria rule n When two or more reaction equations are added together, the equilibrium constant for the sum reaction is the product of the equilibrium constants of the added reactions

n Coefficient rule: 1) 2 A(aq) + B(aq) ↔ 4 C(aq) K 1 = [C]4. [A]2[B] 2) A(aq) + ½ B(aq) ↔ 2 C(aq) K 2 = [C]2. [A][B]1/2 SO K 2 = (K 1)1/2

n Reciprocal Rule 1) 2 A(aq) + B(aq) ↔ 4 C(aq) K 1 = [C]4. [A]2[B] 2) 4 C(aq) ↔ 2 A(aq) + B(aq) K 2 = [A]2[B] [C]4 SO K 2 = 1/K 1

n Multiple Equilibria Rule 2 A(aq) + B(aq) ↔ 4 C(aq) K 1=[C]4/ [A]2[B] 4 C(aq) + E(aq) ↔ 2 F(aq) K 2=[F]2/ [C]4[E] 2 A(aq) + B(aq) + E(aq) ↔ 2 F(aq) K 3= K 1 x K 2 =[F]2/ [A]2[B][E]

Example Questions n Write the equilibrium constant expression for the following reaction (the Haber process) N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) Kc= [NH 3]2 [N 2][H 2]3

n Write the equilibrium constant expression for the following reaction Ca. CO 3(s) ↔ Ca. O(s) + CO 2(g) Kc = [CO 2] *remember pure solids are pure liquids are left out of the equilibrium constant expression

n In the synthesis of ammonia from nitrogen and hydrogen, Kc= 9. 60 at 300◦C. N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) Calculate Kp for this reaction at this temperature. Kp= Kc(RT)Δn(gas) = (9. 60)(0. 0821 x 573)(2 -4) =4. 34 x 10 -3

n The equilibrium constant for the reaction H 2(g) + I 2(g) ↔ 2 HI(g) varies with temperature in the following way: Kc=794 at 298 K n Kc=54 at 700 K n Is the formation of HI favored more at the higher or lower temperature? At the lower temperature because of the larger Kc value (the larger Kc, the more products)

n The value for the equilibrium constant for a reaction is 1. 25 x 10 -3 at a certain temperature. What is the value of the equilibrium constant for the reverse reaction at this temperature? 1/ 1. 25 x 10 -3=800 (K of the reverse reaction is reciprocal of the K of the forward reaction)

n For the reaction N 2 O 4(g) ↔ 2 NO 2(g) the equilibrium concentration of N 2 O 4(g) is 0. 00452 M and that of NO 2(g) is 0. 0310 M. Calculate the value of the equilibrium constant Kc. Kc= [NO 2]2 = (0. 0310)2 = 0. 213 [N 2 O 4] (0. 00452)

When a reaction is NOT at equilibrium. Predicting the direction of a reaction: the reaction quotient, Q n For the general reaction a. A + b. B c. C + d. D Reaction quotient= Qc = [C]c[D]d [A]a[B]b n Q has the appearance of K, but the concentrations aren’t necessarily equilibrium concentrations n The reaction quotient is found by substituting in the initial concentrations into the expression.

n By finding the value of Q, we can predict the direction of a reaction n If Q < K, the system is not at equilibrium. n To reach equilibrium, the reaction proceeds in the forward direction- “shifts right” -(the reactant concentrations will decrease, and the product concentrations will increase until Q=K) If Q=K, then the reaction is at equilibrium If Q > K, the system is not at equilibrium n To reach equilibrium, the reaction proceeds in the reverse direction- “shifts left”- (the product concentrations will decrease, and the reactant concentrations will increase until Q=K)

n Timesaver… n If your initial concentration of any reactant is zero, the reaction will proceed in the reverse direction (to the left) n If your initial concentration of any product is zero, the reaction will proceed in the forward direction (to the right)

Q vs K

example n For the synthesis of ammonia at 500◦C, the equilibrium constant is 6. 0 x 10 -2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: a)[NH 3] = 1. 0 x 10 -3 M, [N 2]=1. 0 x 10 -5 M, [H 2]= 2. 0 x 10 -3 M b) [NH 3] = 2. 00 x 10 -4 M, [N 2]=1. 5 x 10 -5 M, [H 2]= 3. 54 x 10 -1 M c) [NH 3] = 1. 0 x 10 -4 M, [N 2]=5. 0 M, [H 2]= 1. 0 x 10 -2 M (Zumdahl 5 th ed page 628)

Determining equilibrium concentrations using the RICE box n To determine equilibrium concentration given an initial set of conditions: 1. Set up the RICE box n n 2. R=balanced reaction I=initial concentrations C=change in concentrations as reaction proceeds E=Equilibrium concentrations Underneath each formula, fill in the given information about concentrations

4. 5. 6. Calculate Q to determine which direction the reaction proceeds by substituting in initial concentrations in the equilibrium constant expression Fill in the change line in the ICE box n If Q<K, the reactant concentrations decrease by a value of (coefficient)x and the initial concentration of each product is increased by a value of (coefficient)x Or n If Q>K, the reactant concentrations increase by a value of (coefficient)x and the initial concentration of each product is decreased by a value of (coefficient)x Fill in the missing information in the equilibrium line in terms of x (Sum of the “I” and “C” lines)

7. 8. Substitute the equilibrium concentrations into the equilibrium expression Solve for x n May require use of quadratic formula, which will give 2 solutions, choose the logical one! n Sometimes approximations can be made to make solving for x easier § If K <<1, x represents a relatively small number, and can be treated as negligible when subtracting it out from an initial concentration; check the validity of the approximation after solving for x; a difference of less than 5% of the initial concentration is considered a valid assumption 9. Determine the equilibrium concentrations of the reactants and products by substituting in the value of x and solving

Examples n Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5. 10. Calculate the equilibrium concentrations of all species if 1. 000 mol of each component is mixed in a 1. 000 L flask. (Zumdahl 5 th ed. P 631)

n Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1. 15 x 102 at a certain temperature. In a particular experiment, 3. 000 mol of each component was added to a 1. 500 L flask. Calculate the equilibrium concentrations of all species. (Zumdahl 5 th ed. Page 633)

n For a synthesis of hydrogen fluoride from hydrogen and fluorine, 3. 000 mol H 2 and 6. 000 mol F 2 are mixed in a 3. 000 L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1. 15 x 102. Calculate the equilibrium concentration of each component. (Zumdahl 5 th ed page 635)

Determining K given initial concentrations and at least one equilibrium concentration n To determine K given initial and one equilibrium concentration: 1. 2. Write the balanced equation Set up the ICE box n n n 3. I=initial concentrations C=change in concentrations as reaction proceeds E=Equilibrium concentrations Underneath each formula, fill in the given information about initial and equilibrium concentrations

4. 5. 6. 7. Use the given equilibrium and initial concentration of one substance to determine the change in concentration that has occurred. The change in concentration equals (coefficient)x, so solve for x. Use the value of x along with the stoichiometry to determine the concentration changes and then the equilibrium concentrations of all other species in the reaction. Substitute the equilibrium concentrations into the equilibrium constant expression and solve for K.

Example n A 1. 00 L flask was filled with 2. 00 mol gaseous SO 2 and 2. 00 mol gaseous NO 2 and heated. After equilibrium was reached, it was found that 1. 30 mol gaseous NO was present. Assume that the reaction SO 2(g) + NO 2(g) SO 3(g) + NO(g) occurs under these conditions. Calculate the value of the equilibrium constant, K, for this reaction.

Le Chatelier’s Principle n If a closed system at equilibrium is subjected to a stress, a shift will occur to counteract the stress and establish a new equilibrium. n The most common “stresses” are changes in n Temperature n Concentration or partial pressure n Volume n Note: Addition of a catalyst or an inert gas does not result in an equilibrium shift

n A stress initially results in a nonequilibrium condition n Because the rates of the forward and reverse reactions are no longer equal n A new equilibrium is then established over time as the forward and reverse rates once again become equal.

Temperature Stresses n Increasing the temperature of an equilibrium system provides more available energy for the reaction n increasing the rate of both the forward and reverse reactions, but by different amounts n Because the Ea of the forward reaction and reverse reactions are not equal

Temperature increase n Endothermic (think of heat as a reactant) n The forward rate will be increased more than the reverse rate n Resulting in a net shift of the reaction toward the products n Exothermic (think of heat as a product) n The reverse rate will be increased more than the forward rate n Resulting in a net shift of the reaction toward the reactants n NOTE: the reverse will happen with a temperature decrease

n A temperature change is the ONLY change that will result in a new value for the equilibrium constant

Concentration or partial pressure stresses n Note: there is no change in the value of the equilibrium constant with changes in concentrations n The addition of a reactant in an equilibrium system will produce a shift toward the products (in the forward direction) n The addition of a product in an equilibrium system will produce a shift toward the reactants (in the reverse direction)

n Increasing the concentration, or partial pressure, of a reactant in an equilibrium system will result in an increase in the rate of the forward reaction (assuming the reactant appears in the rate law) n Initially, the system is no longer at equilibrium n Eventually, the forward and reverse reactions will again be at the same rates, and equilibrium will be reestablished.

n Similarly, the removal of a reactant in an equilibrium will cause a shift toward the reactants (in the reverse direction) n Removal of a product will cause a shift toward the products (in the forward direction)

n 3 common ways of removing species from equilibrium systems n n n Neutralization n The addition of a strong acid to an equilibrium containing a base or the addition of a strong base to an equilibrium containing an acid Precipitation n The addition of an ion having low solubility with an ion present in an equilibrium system Complex ion formation n The addition of a ligand that may bind some ion present in an equilibrium system

Volume change stresses n NOTE- volume changes do not change the value of the equilibrium constant n If the volume is decreased, the pressure will increase (Boyle’s law), if the volume is increased, the pressure will decrease. n An increase in pressure means the gas particles will be closer together, and an increase in collision frequency. Both the forward and reverse rates increase

n Whether the forward or reverse reaction rate will be increased more than the other depends on the STOICHIOMETRY of the reaction n In other words, the coefficients of the gaseous reactants and products

n If reactants and products have an equal number of moles, then changing the volume will have NO EFFECT on equilibrium n If the volume decreases, and the pressure increases, the system will attempt to decrease the pressure by shifting to the side with the fewest moles of gas n If the volume increases, and the pressure decreases, the system will shift to the side with more moles of gas. n Note- only the gas moles are considered, liquid or solid moles are ignored