Reversible reactions and equilibrium

Reversible reactions and equilibriumSection 14.1Completion reactions and reversible reactionsCompletion Reactions: a chemical reaction in which all of the reactants are changed into products.

Reversible Reactions: a chemical reaction in which the products re-form the original reactants. Reversible Reaction ExampleCaCl2(aq) + Na2SO4(aq) CaSO4(s) + 2NaCl(s)Because Cl- and Na+ are spectator ions, the net ionic equation better describes what is happeningCa2+(aq) + SO42-(aq) CaSO4(s)At the same time, CaSO4 is breaking down to re-form ionsCaSO4(s) Ca2+(aq) + SO42-(aq)These forward and reverse reactions take place at the same time. When the forward and reverse reaction rates are equal, we have chemical equilibriumCa2+(aq) + SO42-(aq) CaSO4(s)

As HI is formed, fewer collisions occur between H2 and I2 molecules and the rate of forward reaction dropsAs HI molecules are made, more collisions occur and the rate of re-formation of H2 and I2 increases.When the forward and reverse rate are equal, the system is at chemical equilibrium Chemical Equilibria are DynamicStatic equilibrium: a state in which nothing changesDynamic equilibrium: two opposite changes occur at the same time in which there is no net change in the systemhttps://www.youtube.com/watch?v=wlD_ImYQAgQ https://www.youtube.com/watch?v=JsoawKguU6A Chemical Equilibria are Dynamic

Higher concentration of products than reactants. Products are favored. Higher concentration of reactants than products. Reactants are favored.Equilibria Involving Complex IonsComplex Ion: any metal or ion that is bonded to more than one atom or molecule a.k.a coordination compoundMany complex ions consist of a metal ion surrounded by a number of ligandsLigand: a molecule, such as NH3+ or H2O, or anions such as cyanide, CH- that readily bond to metal ions.

Equilibria Involving Complex IonsComplex ions form cations (+) or anions (-)The charge on a complex ion is the sum of the charges on the species from which the complex ion forms[CoCl4]2-Co2+ bonds with four Cl- ligandsTotal charge = (+2) +4(-1) = -2Complex ions often form in systems that reach equilibriumhttps://www.youtube.com/watch?v=cWr3UDo-WeU

The protein haemoglobin has an iron atom that is co-ordinately bonded to four nitrogen atoms that are part of larger molecules. Oxygen co-ordinates with the Fe2+ ion and can be transported. 11Systems at equilibriumChapter 14: Section 2ObjectivesWrite Keq expressions for reactions in equilibrium, and perform calculations with them.Write Keq expressions for the solubility of slightly soluble slats, and perform calculations with them.The Equilibrium Constant, KeqThere is a mathematical relationship between product and reactant concentrations at equilibrium. This relationship can be expressed using Keq, Equilibrium ConstantThe value of Keq is unitlessWe do not include solids or liquids in the expression

Equilibrium ConstantA general equation can be written for a reversible reactionaA + bB cC + dDThe ratio of products to reactants is expressed as:

KeqEquilibrium ConstantThe value of Keq gives us an indication of whether the reaction favours the products or reactantsIf K > 1 then equilibrium favours products i.e. there will be more products than reactants If K < 1 then equilibrium favours the reactants i.e. there will be more reactants than products

Keq small: numerator (products) is larger than the denominator (reactants). Concentrations of products will be greater than those of reactants. This is a favorable reaction. Keq big: numerator (products) is smaller than the denominator (reactants). Concentrations of reactants will be greater than those of products. This is a unfavorable reaction. Calculating Keq from Concentrations of Reactants and ProductsAn aqueous solution of carbonic acid reacts to reach equilibrium as described below: H2CO3 + H2O HCO3- + H3O+The solution contains the following solute concentrations: carbonic acid 3.3 x 10-2 mol/L; bicarbonate ion 1.19 x 10-4 mol/L; hydronium ion 1.19 x 10-4 mol/L. Determine the Keq. Calculating Keq from Concentrations of Reactants and ProductsKeq for the equilibrium below is 1.8 x 10-5 at a temperature of 25C. Calculate [NH4+] when [NH3] = 6.82 x 10-3NH3 + H2O NH4+ + OH-Point of SolubilitySolubility Equilibrium is dynamic as ions crystallize at the same rate as ions re-enter solution in a saturated solution.

The Solubility Product Constant, KspSolubility Product Constant, Ksp: the equilibrium constant for a solid that is in equilibrium with the solids dissolved ionsThe dissolution of slightly soluble salts are given a special name/symbol - KspCaF2(s) Ca2+(aq) + 2F-(aq)Ksp = [Ca2+][F-]2 = 1.6 x 10-10The reverse is also true when the solid precipitates back out of solution. Calculating Ksp from SolubilityMost parts of the oceans are nearly saturated with CaF2. The mineral fluorite may precipitate when ocean water evaporates. A saturated solution of CaF2 at 25C has a solubility of 3.4 x 10-4 M. Calculate the solubility product constant for CaF2. CaF2(s) Ca2+(aq) + F-(aq)Calculating Ionic Concentrations Using KspCopper (I) chloride has a solubility product constant of 1.2 x 10-6 and dissolves according to the equation below. Calculate the solubility of this salt in ocean water in which the [Cl-] = 0.55CuCl(s) Cu+(aq) + Cl-(aq)Using Ksp for Hydroxides We use the concept of Ksp to extract metals as hydroxides are slightly soluble.Mg(OH)2(s) Mg2+(aq) + 2OH-(aq)[Mg2+][OH-]2 = Ksp = 1.8 x 10-11Because the Ksp is below 1, the reactants are favored. Equilibrium systems and stressChapter 14: Section 3 ObjectivesState Le Chteliers principleApply Le Chteliers principle to determine whether the forward or reverse reaction is favored when stresses are applied to the system TemperatureConcentrationPressure Discuss the common-ion effect in the context of Le Chteliers principleDiscuss the practical uses of Le Chteliers principle

Le Chteliers principleThe principle that states that a system in equilibrium will oppose a change in a way that helps eliminate the changeif a chemical system is at equilibrium and we add a substance (either a reactant or a product), the reaction will shift so as to reestablish equilibrium by consuming part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance.Change in ConcentrationAdd product = increases the concentration of product The system responds by decreasing this new concentration and creating more reactant. An increase in the reverse reaction occurs the equilibrium shifts leftAdd reactant = the exact reverse occurs

Addition of H2 would cause the system to shift, so as to reduce the concentration of H2 toward its original value. This can occur only if the equilibrium shifts to form more NH3. At the same time, the quantity of N2 would also be reduced slightly. This situation is illustrated in Figure 15.12. Addition of more N2 to the equilibrium would likewise cause a shift in the direction of forming more NH3. However, if we add NH3 to the system at equilibrium, the concentrations will adjust to reduce the NH3 concentration toward its original value; that is, some of the added ammonia will decompose to form N2 and H2.30Change in TemperatureRecall: exothermic reactions release energy & endothermic reactions absorb energy

Increase in Temperature: The reaction will always shift in the direction that will absorb heat.Decrease in Temperature: The reaction will always shift in the direction that will release heat.

Forward reaction absorbs heatHeat addedMoves to side to consume added heatHeat removedMoves to side to add heat removedChange in PressureIn an equilibrium, a pressure increase favors the reaction that produces fewer gas molecules. A pressure decrease favors the reaction the produces more gas molecules.Changes in pressure only affect gases in equilibriumIf there are equal gas molecules on both sides of the equation, pressure changes will not affect equilibrium

2NO2(g) N2O4(g)

N2 + 3H2 2NH3ChangeWhat OccursEffect on EquilibriumAddition of ReactantAdded reactant consumedShift to right (produce more products)Addition of ProductAdded product consumedShift to left (produce more reactants)Increase in PressurePressure decreasedShift to fewer gas moleculesDecrease in PressurePressure increasedShift to more gas moleculesIncrease Temp.Heat is consumedShift in endo. directionDecrease Temp.Heat is producesShift in exo. directionThe Common-Ion EffectThe phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduces ionizationThe reduction of the solubility of salt due to the addition of a common ion.If a common ion is added to a saturated solution of a salt, then the salt will be precipitated.The equilibrium will shift to consume excess ion and as a result produce more reactant (typically, an insoluble salt)

The Common-Ion EffectCuCl(s) Cu+(aq) + Cl-(aq)[Cu+][Cl-] = Ksp = 1.2 x 10-6The addition of sea water, which is rich in Cl-, will result in an increase of [Cl-]. The equilibrium will shift left to consume this excess and produce a precipitate of CuCl(s).The ion Cl- is the common-ion in this case as the Cl- comes from two sourcesThe Common-Ion EffectBaSO4(s) Ba2+(aq) + SO42-(aq) Ksp < 1Na2SO4(s) 2Na+(aq) + SO42-(aq).

BaSO4 is not very soluble but there is still some Ba2+ produced. This is poisonous to the human body so Na2SO4 is added so that [Ba2+] is decreased to a safe level.

Practical Uses of Le Chateliers Principlehttps://www.youtube.com/watch?v=NWhZ77Qm5y4