Equilibrium AP Chem Mr. Nelson

The Concept of Equilibrium The term reversible reaction is used to describe reactions that can go in either the forward or the reverse direction Equilibrium is used to specify that a reversible reaction has reached an equal rate for both directions

The Concept of Equilibrium Chemical equilibrium occurs when opposing reactions are proceeding at equal rates Concentrations of the reactants/products remains constant

The Concept of Equilibrium Its important to note that forward and reverse reactions still occur, although outwardly no apparent change occurs

The Concept of Equilibrium Note that the equilibrium concentrations are independent of initial concentrations A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached

The Concept of Equilibrium Fish Tank Example

Connection between Equilibrium & Kinetics For generalized reaction: A⇋B Forward reaction: AB Forward Rate Law: Ratef = kf[A] Reverse reaction: BA Reverse Rate Law: Rater = kr[B]

Connection between Equilibrium & Kinetics By definition, at equilibrium forward and reverse rates must be equal, so: Ratef = Rater kf[A] = kr[B] Also, concentrations remain constant,so:

The Equilibrium Constant The equilibrium constant (Keq) can be determined at constant temperature for any reaction at equilibrium: aA+ bB  cC + dD Keq = [C]c [D]d [A]a [B]b Keq is unitless

The Equilibrium Constant Note that only equilibrium concentrations can be placed into an equilibrium constant expression! Also, no solids or liquids are included in an equilibrium constant expression The reason for this is their concentrations remain fairly constant in equilibrium Note that they are still required for the reaction to proceed however

The Equilibrium Constant The Many Types of Equilibrium Constants, Keq Kc Kp Ksp Ka/Kb Kw

Concentration vs. Pressure Equilibria Equilibrium expressions can be written in terms of concentration (Kc) measured in molarity or pressure (Kp) measured in atm Example: or

Concentration vs. Pressure Equilibria In general, Kc is not equal to Kp The following equation converts between the two: n = moles of gas products – moles of gas reactants Note: When no change in moles, Kp=Kc

Homogeneous vs. Heterogeneous Equilibria The term homogeneous equilibrium is used to describe reactions where all the species are in the same phase Example: The term heterogeneous equilibrium is used to describe reactions where species are in different phases H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq)

Multiple Equilibria A + B  C + D C + D  E + F K1 A + B  E + F K2 Kc When the products of one reaction are the reactants of another Example: A + B  C + D C + D  E + F K1 K2 A + B  E + F Kc

Multiple Equilibria Using their equilibrium expressions, we can derive a mathematical relationship:

Manipulating Equilibria Expressions If an equilibrium is written in the reverse, the following relationship exists: PCl3 + Cl2  PCl5 PCl5  PCl3 + Cl2

Manipulating Equilibria Expressions If an equilibrium is multiplied by a factor, the following relationship exists: PCl3 + Cl2  PCl5 2PCl3 + 2Cl2  2PCl5

The Magnititude of Equilibrium Constants What does Keq tell us? If Keq is very large, products will be favored This implies the numerator is large in the Keq expression compared to denominator If Keq is very small, reactants will be favored This implies the denominator is large in the Keq expression compared to numerator If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium

The Reaction Quotient (Q) Recall: In electrochemistry, we utilized “Q” which is similar in all respects to “Keq” except in definition Q is utilized when given initial concentrations, not concentrations at equilibrium

Predicting the Direction of Reversible Reactions If given an initial set of conditions for a reversible reaction, utilizing Q and Keq, a prediction of the direction of the reaction (towards reactants/towards products) can be made

Comparing Kc and Qc Ratio of prds to rcts is too small Qc<Kc Ratio of prds to rcts is too small To reach equilibrium, rcts must go towards prds Shifts to the right Qc=Kc Initial concentrations are at equilibrium concentrations System is at equilibrium Qc>Kc Ratio of prds to rcts is too large To reach equilibrium, prds must go towards rcts Shifts to the left

Comparison of Kc and Qc Qc Kc Qc Kc Kc Qc

Calculating Equilibrium Concentrations Tabulate all known initial and equilibrium concentrations When both are known, calculate the change Using coefficients, calculate changes in concentration for all participants in a rxn Calculate equilibrium concentrations

Initial molarity Change in molarity Equilibrium molarity The ICE Method This is the process used for determining equilibrium concentrations Initial molarity Change in molarity Equilibrium molarity

Example #1: Calculating Equilibrium Concentrations A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00 L stainless-steel flask at 430 °C. The equilibrium constant Kc for the reaction: H2 (g) + I2 (g)  2 HI (g) is 54.3 at this temp. Calculate the concentrations of H2, I2, and HI at equilibrium.

H2 (g) + I2 (g)  2 HI (g) 0.500 -x +2x 0.500 – x 2x Initial (M) Change (M) -x +2x Equilibrium (M) 0.500 – x 2x

Taking the square root of both sides: Substituting, we get: Taking the square root of both sides: x = 0.393 M

So, the concentrations at equilibrium would be the following: [H2] = (0.500 – 0.393) M = 0.107 M [I2] = (0.500 – 0.393) M = 0.107 M [HI] = 2 x 0.393 M = 0.786 M

Example #2: Calculating Equilibrium Concentrations When 3.0 mol of I2 and 4.0 mol of Br2 are placed in a 2.0 L reactor at 150oC, the following reaction occurs until equilibrium is reached: I2(g) + Br2(g)  2IBr(g) Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?

I2 (g) + Br2 (g)  2 IBr (g) 1.5 2 -x +2x 1.5 – x 2 – x 1.6 Initial (M) 1.5 2 Change (M) -x +2x Equilibrium (M) 1.5 – x 2 – x 1.6

When initial and final concentrations are known, calculate the change: 0 + 2x = 1.6 x = 0.8 Then determine all equilibrium concentrations: [I2] = 1.5 – 0.8 = 0.7 M [Br2] = 2 – 0.8 = 1.2 M [IBr] = 1.6 M

Plug equilibrium concentrations into equation and solve:

Calculating Equilibrium Concentrations when Keq is small When solving equilibrium concentrations and Keq is small, assume the effect of the change (“x”) is negligible (for rcts only)

Example: Calculating Equilibrium Concentrations when Keq is small At 100 °C the equilibrium constant (Kc) for the reaction: COCl2 (g)  2 CO (g) + Cl2 (g) is 2.19 x 10-10. If the initial concentration of [COCl2] = 1.5 M, calculate the concentrations of all species at equilibrium.

COCl2 (g)  2 CO (g) + Cl2 (g) 1.5 -x +2x +x 1.5 – x Initial (M) Change (M) -x +2x +x Equilibrium (M) 1.5 – x

Le Chatelier’s Principle If a stress is applied to a system in equilibrium, the equilibrium shifts to relieve that stress Forms of “stress”: Concentration Pressure Temperature

Change in Reactant or Product Concentration Adding a substance shifts a system at equilibrium away from the added substance System must consume added item Removing a substance from a system at equilibrium shifts the reaction toward the removed substance System must replace removed item

Changes in Pressure Increase in pressure forces an equilibrium to shift in the direction that reduces the number of moles of gas in the system Example: An increase in pressure causes a shift to the left N2O4 (g)  2 NO2 (g)

Changes in Pressure Recall: Changes in volume can also affect changes in pressure Increase volume = decrease pressure Decrease volume = increase pressure Also the addition of an inert gas (one not in the reaction) can cause an increase in pressure as well

Changes in Temperature Enthalpy of the reaction plays a major roll in how heat effects the equilibrium Recall: Endothermic (+∆H) means heat is a reactant, while exothermic (- ∆H) means it is a product Treat changes in heat in the same manner that you treat changes in concentration

Effects of Catalysts on Equilibrium Recall: Catalysts lower energy barrier between reactants and products Ea for both forward and reverse are lowered to the same extent Catalysts increase the rate at which equilibrium is reached However, it does NOT change the composition of the equilibrium mixture.

Solubility Equilibria Saturated solutions of salts are another type of chemical equilibria Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution

The Solubility Product, Ksp The equilibrium constant, the Ksp, is no more than the product of the ions in solution (Recall: solids do not appear) For a saturated solution of AgCl, the equation would be: The solubility product expression would be:

Sequence of Steps for Solubility Equilibria Solubility of compound Molar solubility of compound Concentrations of cations and anions Ksp of compound Ksp of compound Concentrations of cations and anions Molar solubility of compound Solubility of compound

Solubility Solubility can be expressed two ways: Molar solubility is the moles of solute per 1L of saturated solution (mol/L) Solubility is the number of grams of solute per 1 L of saturated solution (g/L) Ksp expressions require molar solubility

Determining Ksp by Experimental Measurements Use ICE table, but with “s” for change (instead of “x”) “s” is not only the change, but the molar solubility

Determining Ksp: Example #1 The solubility of CaSO4 is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. Solubility of CaSO4 in g/L Molar solubility of CaSO4 [Ca2+] and [SO42-] Ksp of CaSO4

Determining Ksp: Example #1 Initial (M) Change (M) -s +s Equilibrium (M) s

Determining Ksp: Example #1 As CaSO4 dissolves, we get 1 mole Ca2+, and 1 mole SO42-, so at equilibrium:

Determining Ksp: Example #2 Lead (II) chloride dissolves to a slight extent in water according to the equation: Calculate the Ksp if the solubility of lead (II) chloride has been found to be 1.62 x 10-2 M

Determining Ksp: Example #2 Molar solubility of PbCl2 [Pb2+] and [Cl-] Ksp of PbCl2 Initial (M) Change (M) -s +s +2s Equilibrium (M) s 2s

Determining Ksp: Example #2

Calculating Solubility from Ksp: Example #1 The Ksp for CaCO3 is 3.8 x 10-9 at 25 °C. Calculate the solubility (in g/L) of calcium carbonate in pure water. Ksp of CaCO3 Concentrations of [Ca2+] and [CO32-] Molar solubility of CaCO3 Solubility of CaCO3

Calculating Solubility from Ksp: Example #1 Initial (M) Change (M) -s +s Equilibrium (M) s

Calculating Solubility from Ksp: Example #1

Predicting Precipitation Reactions using Q Using the reaction quotient, Q, we can Decide whether a ppt will form Calculate ion concentration required to begin the ppt. of an insoluble salt Q < Ksp System has too much reactant No ppt Q = Ksp System at equilibrium Q > Ksp System has too much product Forms ppt!

Predicting Precipitation Reactions Steps Determine insoluble product using solubility rules Calculate cation and anion concentration of insoluble product (if mixed) Calculate Q Compare to Ksp

Example: Predicting Precipitation Exactly 200 mL of 0.0040 M BaCl2 are mixed with exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? (Ksp of insoluble salt is 1.1 x 10-10)

The Common Ion Effect Le Chatelier’s Principle can describe the effect of an added common ion For example, if NaCl were added to a saturated solution of PbCl2, what effect would occur? Increase product concentration shifts equilibrium towards reactants, forming more ppt.

Solubility and the Common Ion Effect The presence of a common ion will cause the equilibrium to shift so that even less of the substance (with smaller Ksp value) will dissolve

The Common Ion Effect Be aware: pH can affect solubility The addition of an acid or base could have a side reaction, decreasing concentration of an ion If an acid were added to the following, what effect would be observed? H+ combines with OH-, so [OH-] is decreased, forcing equilibrium to the right

Example: Common Ion Effect Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10-3 M silver nitrate solution. (Ksp = 1.6 x 10-10) Since silver nitrate is a soluble, strong electrolyte: 6.5 x 10-3 M 6.5 x 10-3 M

Example: Common Ion Effect Initial (M) 6.5 x 10-3 0.00 Change (M) -s +s Equilibrium (M) (6.5 x 10-3 + s) s Because Ksp is so small, s must be very small in comparison to 6.5 x 10-3, so we approximate that 6.5 x 10-3 + s  6.5 x 10-3

Selective Precipitation In certain problems, you will have to determine which of two possible precipitates will form The salt with the lower Ksp will precipitate first To determine the amount of an ion needed to precipitate, recall that Ksp concentrations are saturated, so anything greater than this will form a ppt.

Example: Selective Precipitation A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt