#### Transcript Equilibrium

Equilibrium Phase Solution Chemical Reversible Reactions a number of chemical reactions have a ΔH and ΔS that are both positive or both negative; one force favours the reaction, while the other opposes it: energy + H2O(l) ΔH = + H2O(g) ΔS = + for the opposite reaction the ΔH and ΔS are both negative: H2O(g) → energy + H2O(l) ΔH = ΔS = as a result this reaction is reversible Equilibrium occurs when a reaction can go in either direction. the majority of reactions can be equilibrium situations. there are 3 main types of equilibria. Phase Equilibrium occurs between the phases of a substance: energy + H2O(l) H2O(g) even at room temperature water evaporates until the air above is saturated. (relative humidity is the amount of water vapour in the air, in %, relative to the amount the air can hold at that temperature. The air is saturated if the relative humidity is 100%.) as a water molecule evaporates, another molecule of water vapour condenses. The rate of the evaporation reaction equals the rate of the condensation reaction; the reactions are at equilibrium: energy + H2O(l) H2O(g) Solution Equilibrium Occurs when a soluble non-volative substance is dissolved in a solvent. If the solution is saturated and there is still solid solute at the bottom of the beaker the system is at equilibrium. For example: NaCl(s) Na1+(aq) + Cl1-(aq) C12H22O11 (s) C12H22O11 (aq) Chemical Equilibrium occurs in a chemical reaction system (the other two were examples of physical reactions) Examples: 2 NO2 (g) N2O4 (g) PCl5 (g) PCl3 (g) + Cl2 (g) Equilibrium vs Steady State Equilibrium is a closed system; no inputs or outputs. Steady state is an open system; inputs and outputs. Dynamic Equilibrium 5 conditions are necessary for equilibrium to occur: • the reaction must be reversible • the system must be closed • both reactant and product must be present • concentrations of reactant and product do not change • conditions of temperature and pressure must be held constant Dynamic Equilibrium the forward reaction continues the reverse reaction continues the rate of the forward reaction equals the rate of the reverse reaction. Le Chatelier’s Principle reactions continue in both directions during equilibrium any factor which changes the rate of one or both reactions will alter the relative amounts of reactants and products. Le Chatelier’s Principle we can predict how an equilibrium system will respond to a given change by using Le Chatelier’s Principle: “When a system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimizes the stress.” Le Chatelier’s Principle in other words, for every action there is an equal and opposite reaction; whatever you do to a reaction system, the system will respond in the opposite direction. Effect of Changing Concentraton on a Chemical System Increasing the concentration of any component will cause the reaction system to try to decrease it, by favouring the opposite side. Effect of Changing Concentraton on a Chemical System 3 CH4 (g) + 6 H2O(g) + 4 N2 (g) 8 NH3 (g) + 3 CO2 (g) Increase [CH4]; system responds by trying decrease the [CH4] by making more product. The concentration of reactants will go down while the concentration of products goes up. In other words, increasing [CH4] in this reaction favours the creation of product. Effect of Changing Temperature on a Chemical System Increasing the temperature of a system will favour the endothermic direction. Decreasing the temperature will favour the exothermic direction. Effect of Changing Temperature on a Chemical System 3 H2(g) + N2 (g) 2 NH3 (g) + 92 kJ adding energy causes the system to try to absorb energy; the endothermic direction is favoured. More reactant is made and the amount of product decreases. taking energy away favours the exothermic reaction; in this case products are favoured. Effect of Changing Pressure on a Chemical System equal numbers of moles at the same temperature and pressure have the same volume. increasing pressure has the effect of favouring the side of the equation with fewer particles. Effect of Changing Pressure on a Chemical System 3 CH4 (g) + 6 H2O(g) + 4 N2 (g) 8 NH3 (g) + 3 CO2 (g) in this reaction the reactant side has 13 particles; the product side has 11. increasing the pressure will favour the products. decreasing the pressure will favour the reactants. Effect of Changing Volume on a Chemical System in a closed system reducing the volume has the effect of increasing the pressure. increasing the volume reduces the pressure. Catalysts and Inhibitors catalysts reduce activation energy in both directions; they increase the rate of reaction in both directions but does not change the equilibrium. inhibitors have the opposite effect, but also do not affect the equilibrium. Common Ion Effect is related to solubility equilibria. adding an ionic substance with an ion in common with an aqueous equilibrium will affect the equilibrium in the same way as changing concentration Common Ion Effect NaCl(s) Na1+(aq) + Cl1-(aq) if NaNO3(aq) is added to this system, it breaks into Na1+(aq) and NO31-(aq). It is just like adding Na1+ ions to the NaCl system. That system responds by favouring the formation of solid NaCl; the concentration of Cl1- ions decreases. Common Ion Effect NaCl(s) Na1+(aq) + Cl1-(aq) add AgNO3 Ag1+ precipitates chloride(Cl1-) removing it from solution. system responds by producing more chloride, also increases sodium ions and reduces the NaCl solid Equilibrium Constant For the reaction: aA(g) + bB(g) cC(g) + dD(g) the lower-case letters represent number of moles (the numbers used to balance the equation) the upper-case letters represent the chemical species. Equilibrium Constant aA(g) + bB(g) cC(g) + dD(g) At equilibrium a mathematical relationship exists where a constant results from the equation: Keq = [Products] = [Reactants] [C]c[D]d [A]a[B]b where [ ] represents the concentration. Equilibrium Constant this a useful mathematical relationship. no matter what concentrations of reactant or product you start with we know that the final ratio of products over reactants will equal the value of the Keq. This applies only if conditions of temperature and pressure stay constant. Writing Equilibrium Expressions Consider the reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) Keq = + 92 kJ [NH3]2 [N2] [H2]3 energy is not included in the equilibrium constant expression Writing Equilibrium Expressions 2 CaCO3 (S) 2 Ca(s) + 2 CO2 (g) + O2 (g) Keq = [CO2]2[O2] solids and liquids are not included in the equilibrium expression. Writing Equilibrium Expressions 2 H2 (g) + O2 (g) 2 H2O (l) Keq = 1 [H2]2[O2] since the liquid is not included in the equilibrium expression the numerator becomes a 1. Equilibrium Calculations the equilibrium expression is a mathematical relationship of several variables. given the values of all but one of the variables allows you to calculate the unknown, whether it be the Keq or any of the concentrations. Equilibrium Calculations consider the following case: N2 (g) + 3 H2 (g) 2 NH3 (g) + 92 kJ given this equation we can write the equilibrium expression: Keq = [NH3]2 [N2] [H2]3 Equilibrium Calculations at equilibrium the concentration of N2 = 0.142 mol/L, the [H2] = 0.341 mol/L and the [NH3] = 0.220 mol/L. calculate the value of the Keq Equilibrium Calculations [N2] = 0.142 mol/L [H2] = 0.341 mol/L [NH3] = 0.220 mol/L Keq = Keq = 8.60 [NH3]2 = (0.220 mol/L)2 [N2] [H2]3 (0.142 mol/L)(0.341 mol/L)3 This is slightly larger than 1; products are favoured over reactants ICE box problems This type of problem gives you information about the initial situation, before equilibrium is established. Then you get 1 or more pieces of information concerning the situation at equilibrium. You are required to fill in the blanks by figuring out the change. ICE box problems 2 A(g) + B(g) C(g) + 3 D(g) 0.100 mol/L each of A and B are placed in a flask and allowed to come to equilibrium. At equilibrium 0.022 mol/L of C is present. Calculate the value of the Keq. 2 A(g) + B(g) C(g) + 3 D(g) [Initial] 0.100 M 0.100 M 0.0 0.0 [Change] [Equilibrium] 0.022 M this is what you know. Nothing is said in the problem about the amount of C and D initially, so we assume they start at zero. 2 A(g) + B(g) C(g) + 3 D(g) [I] [C] [E] 0.100 M 0.100 M 0.0 0.0 +0.022 M 0.022 M since C started with nothing and ended with 0.022 mol/L we can see that it changed by +0.022 mol/L. 2 A(g) + B(g) C(g) + 3 D(g) [I] [C] [E] 0.100 M 0.100 M 0.0 0.0 +0.022 M 3/1(+0.022 M) 0.022 M if C goes up, so does D. the stoichiometry is 1 C : 3 D. For every 1 that C goes up, D goes up 3. 2 A(g) + B(g) C(g) + 3 D(g) [I] 0.100 M 0.100 M [C] 2/1(-0.022M) 1/1(-0.022M) [E] 0.0 0.0 +0.022 M 3/1(+0.022 M) 0.022 M the change to A and B are negative (teeter-totter) always take the stoichiometry into account. 2 A(g) + B(g) C(g) + 3 D(g) [I] 0.100 M 0.100 M [C] 2/1(-0.022M) 1/1(-0.022M) = -0.044 M [E] 0.056 M = -0.022 M 0.078 M 0.0 0.0 +0.022 M 3/1(+0.022 M) 0.022 M = +0.066 M 0.066 M adding or subtracting the changes from the initial concentration yields the equilibrium concentration of each species. all that remains is to write the equilibrium expression and solve for Keq Keq = [C][D]3 [A]2 [B] = (0.022 mol/L)(0.066 mol/L)3 (0.056 mol/L)2( 0.078 mol/L) Keq = 0.026 Another ICE box N2 (g) + 3 H2 (g) 2 NH3 (g) If 1.65 mol of ammonia is placed in a 3.00 L flask at 142 °C and allowed to come to equilibrium the concentration of hydrogen gas is 0.771 mol/L Calculate the value of the Keq N2 (g) [I] 0.00 mol/L [C] 1/3 [E] 3 H2 (g) 0.00 mol/L (+0.771 M) +0.771 mol/L = 0.257 mol/L 0.257 mol/L 0.771 mol/L 2 NH3 (g) 1.65 mol/3.00L = 0.550 mol/L 2/3 (-0.771 mol/L) = -0.514 mol/L 0.036 mol/L N2 (g) + 3 H2 (g) 2 NH3 (g) Keq = = [NH3]2 [N2] [H2]3 (0.036 mol/L)2 (0.257 mol/L)(0.771 mol/L)3 = 0.011 A(g) + B(g) 2 C(g) If the initial concentration of C is 0.350 mol/L, calculate the equilibrium concentrations of all three species if the value of the Keq is 145. A(g) + B(g) 2 C(g) [I] 0.0 0.0 0.350 mol/L [C] [E] this is what we know, other than the Keq value. where do we go from here ? A(g) + B(g) [I] 0.0 0.0 [C] +x +x 2 C(g) 0.350 mol/L - 2x [E] We don’t know the changes in concentration. When we don’t know something in math we call it x. A(g) + B(g) [I] 0.0 0.0 [C] +x +x [E] x x 2 C(g) 0.350 mol/L - 2x 0.350 mol/L - 2x The equilibrium concentrations are expressed in terms of x. Now we write the Keq expression Keq = [C]2 [A] [B] 145 = (0.350 mol/L – 2x)2 (x)(x) 145 = (0.350 mol/L – 2x)2 (x)2 √145 = 0.350 mol/L – 2x x x = 0.0249 mol/L A(g) + B(g) [I] 0.0 0.0 [C] +x +x [E] x x 2 C(g) 0.350 mol/L - 2x 0.350 mol/L - 2x [A] = [B] = x = 0.0249 mol/L [C] = 0.350 M – 2x = 0.350 M – 2(0.0249 M) = 0.300 mol/L A(g) + B(g) C(g) If the initial concentration of C is 0.350 mol/L, calculate the equilibrium concentrations of all three species if the value of the Keq is 145. A(g) + B(g) [I] 0.0 0.0 [C] +x +x [E] x x C(g) 0.350 mol/L -x 0.350 mol/L - x Keq = [C] [A] [B] 145 = (0.350 mol/L – x) (x)(x) 145 = 0.350 mol/L – x x2 145x2 = 0.350 mol/L – x 145x2 + x - 0.350 mol/L = 0 now we apply the quadratic equation: 145x2 + x - 0.350 mol/L = 0 a b c x = -b ± √ b2 – 4ac 2a x = 0.0458 mol/L A(g) + B(g) [I] 0.0 0.0 [C] +x +x [E] x x C(g) 0.350 mol/L -x 0.350 mol/L - x [A] = [B] = 0.0458 mol/L [C] = 0.350 M – x = 0.350 M – 0.0458 M = 0.304 mol/L Calculations using the Ksp A saturated solution of CuBr has a concentration of 2.0 x 10-4 mol/L. Calculate the Ksp. Write the dissociation equation and determine the concentration of each ion in solution: CuBr(s) Cu1+ (aq) + 2.0 x 10-4 mol/L 1(2.0 x 10-4 mol/L) = 2.0 x 10-4 mol/L Br1- (aq) 1(2.0 x 10-4 mol/L) = 2.0 x 10-4 mol/L Write the Ksp expression, insert the numbers and solve: Ksp = [Cu1+][Br1-] = (2.0 x 10-4 mol/L)( 2.0 x 10-4 mol/L) = 4.0 x 10-8 Calculate the Ksp for Bi2S3, which has a solubility of 1.36 x 10-15 mol/L at 25°C. Bi2S3 (s) 1.36 x 10-15 mol/L 2 Bi3+(aq) + 3S 2- (aq) 2(1.36 x 10-15 mol/L) 3(1.36 x 10-15 mol/L) = 2.72 x 10-15 mol/L = 4.08 x 10-15 mol/L Ksp = [Bi3+]2[S2-]3 = (2.72 x 10-15 mol/L)2(4.08 x 10-15 mol/L)3 = 5.02 x 10-73 The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at this temperature. Like the ICE box, express concentrations in terms of x: Cu(IO3)2 (s) x Cu2+(aq) + 1(x) Ksp 2 IO31-(aq) 2(x) = [Cu2+][IO31-]2 = (x)(2x)2 1.4 x 10-7 x = = 4x3 3√1.4 x 10-7 4 = 3.3 x 10-3 mol/L Predicting a Precipitate Ksp values can also be used to predict if mixing solutions will produce a precipitate. e.g. • A solution is prepared by adding 750.0 mL of 4.00 x 10-3 mol/L Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 mol/L KIO3. • The Ksp of Ce(IO3)3 is 1.9 x 10-10. • Will Ce(IO3)3 precipitate from this solution ? Step 1. Write a dissociation equation for each substance and determine the concentration of each ion in solution. Ce(NO3)3 4.00 x 10-3 mol/L KIO3 (aq) (aq) Ce3+(aq) + 1(4.00 x 10-3 mol/L) = 4.00 x 10-3 mol/L K1+(aq) 2.00 x 10-2 mol/L 1(2.00 x 10-2 mol/L) = 2.00 x 10-2 mol/L 3 NO31-(aq) 3(4.00 x 10-3 mol/L) = 1.20 x 10-2 mol/L + IO31-(aq) 1(2.00 x 10-2 mol/L) = 2.00 x 10-2 mol/L Step 2. Use the dilution equation to determine the concentration of each of the significant species. In this case they are Ce3+(aq) and IO31-(aq) [Ce3+] = csvs vd = (4.00 x 10-3 mol/L)(750.0 mL) 750.0 mL + 300.0 mL = 2.86 x 10-3 mol/L [IO31-] = csvs vd = (2.00 x 10-2 mol/L)(300.0 mL) 750.0 mL + 300.0 mL = 5.71 x 10-3 mol/L Step 3. Write the Ksp expression for the substance you are testing and solve for the Ksp using the numbers you just calculated. This is a trial Ksp Substance: Ce(IO3)3 Ce(IO3)3 (s) Ksp Ce3+(aq) + 3 IO31-(aq) = [Ce3+][IO31-]3 Trial Ksp = (2.86 x 10-3 mol/L)( 5.71 x 10-3 mol/L)3 = 5.52 x 10-10 Step 4. Compare the trial Ksp with the actual Ksp from the question: if the trial Ksp is larger than the actual Ksp, there is more solute than can dissolve; there will be a precipitate. if the trial Ksp is smaller than the actual Ksp, there is less solute than can dissolve; there will not be a precipitate. In this case 5.52 x 10-10 is larger than 1.9 x 10-10 there will be a precipitate