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Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs.
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Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs. At equilibrium: rateforward= ratereverse A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.
CHEMICAL EQUILIBRIUM - RATES OF REACTION kF Reactants products kB Chemical reactions are a dynamic process, that is, reactions involve both forward and reverse processes. Chemical Equilibrium is reached by as reaction mixture when the rates of forward and reverse reactions becomes equal. kF = kB NO net change appears obvious although the system is still in constant motion.
Rate Forward Rate Reverse Reaction Dynamics Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Rate Time
N2O4(g)2NO2(g) Reaching equilibrium on the macroscopic and molecular levels.
H2(g) + I2(g) 2 HI(g) at time 0, there are only reactants in the mixture, so only the forward reaction can take place [H2] = 8, [I2] = 8, [HI] = 0 at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H2] = 6, [I2] = 6, [HI] = 4 Tro, Chemistry: A Molecular Approach
H2(g) + I2(g) 2 HI(g) at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated [H2] = 4, [I2] = 4, [HI] = 8 at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium
LAW OF MASS ACTION K equilibrium constant aA + bB cC + dD K = Products Reactants K = [C]c [D]d = Q reaction quotient [B]b [A]a Write the equilibrium equation for: a. HC2H3O2 H+ + C2H3O2- b. H2O + H2O H3O+ + OH- c. 4NH3(g) + 302(g) 2N2(g) + 6H2O(g) d. N2(g) + 3H2(g) 2NH3(g)
[C]c[D]d Q = [A]a[B]b Q - The Reaction Quotient At any time, t, the system can be sampled to determine the amounts of reactants and products present. A ratio of products to reactants, calculated in the same manner as K tells us whether the system has come to equilibrium (Q = K) or whether the reaction has to proceed further from reactants to products (Q < K) or in the reverse direction from products to reactants (Q > K). We use the molar concentrations of the substances in the reaction. This is symbolized by using square brackets - [ ]. For a general reaction aA + bB cC + dD where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as
Predicting the direction of reaction: Q > K forms more reactants Q = K equilibrium Q < K forms more products Note: 1. Kf = 1 Kr 2. K = Kn If the balanced equation is multiplied by a factor then the K (& Q) is multiplied by the exponent.
small K large K intermediate K The range of equilibrium constants
Reaction Progress Reaction Progress reactants products Equilibrium: no net change reactants products Reaction direction and the relative sizes of Q and K.
K as either Kc or Kp Kc = the equilibrium constant using concentrations. Kp = the equilibrium constant using pressure Kp = Kc(RT) n gas Dn is the difference between the number of moles of reactants and moles of products
Deriving the RelationshipBetween Kp and Kc for aA(g) + bB(g) cC(g) + dD(g) substituting
workshop on Kc vs. Kp Kp = Kc(RT) n gas Write Kp the Kc for: 1. N2(g) + 3H2(g) 2NH3(g) 2. N2O4(g) 2NO2(g) 3. Calculate kp if kc = 0.105 for #1 4. 2SO3(g) 2SO2(g) + O2(g) if Kc = 4.07 x 10-3, what is kp?
DIRECTION OF REACTIONS AND Keg 1. The following reaction is a means of “fixing” nitrogen: N2(g) + O2(g) 2 NO(g) A. If the value for Q at 25°C is 1 x 10-30, describe the feasibility of this reaction for Nitrogen fixation. B. Write the equilibrium expression, Kc C. Write the equilibrium expression for 2NO(g) N2(g) + O2(g) D. Determine the Kc for “C”
HOMOGENEOUS EQUILIBRIA H2(g) + I2(g) 2HI(g) Kp = (PHI)2 (PI2)(PH2) Kc = ? 2O3(g) 3O2(g) Kp = (PO2)3 (PO3)2 Kc = ? HETEROGENEOUS EQUILIBRIA 2H2O(l) H3O+(aq)_ + OH-(aq) Kc = [H3O+][OH-] C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) Kc = ?
HETEROGENEOUS EQUILIBRIA Substance in more then 1 phase 1. CaCO3(s) CaO(s) + CO2(g) Kc = [CaO][CO2] [CaCO3] How do the [ ] of solid express? A. D = g/cm3 = mol MW g/mol cm3 Pure solids & liquid have constant [ ] Kc = constant [CO2] constant Kc = Kccon = [CO2] con
The reaction quotient for a heterogeneous system. solids do not change their concentrations
Q. Each of the mixtures listed below was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equilibrium, expressed by 1 a) pure CaCO3 b) CaO & PCO2 > Kp c) solid CaCO3 & PCO2 > Kp d) CaCO3 & CaO
Calculating Variations on Q and K [C]c[D]d Qc = aA + bB cC + dD [A]a[B]b Q’ = 1/Qc cC + dD aA + bB Qc’ = (Qc)n n aA + bB cC + dD For a sequence of equilibria, Koverall = K1 x K2 x K3 x …
CALCULATING THE Keq 1. In one experiment, Haber introduced a mixture of H2 & N2 into a reaction vessel and allowed the system to attain chemical equilibrium at 472°C. The equilibrium mixture of gases were analyzed and found to contain 0.1207M H 2, 0.0402M N2, and 0.00272M NH3. Calculate Keq. 2. Nitryl Chloride, NO2Cl, is in equilibrium in a closed container with NO2 and Cl2. 2 NO2Cl(g) 2 NO2(g) + Cl2(g) Calculate Keq if [NO2Cl] = 0.00106M [NO2] = 0.0108M & [Cl2] = 0.00538M
3. For the Haber process N2(g) + 3H2(g) 2NH3(g) Kp = 1.45 x 10-5 at 500°C If an equilibrium mixture of the three gas started with partial pressures of 0.928 atm for H2 and 0.432 atm for N2, what is the partial pressure of NH3? 4. A 1.00 L flask is filled with 1.00 mol of H2 and 2.00 mol I2 at 448°C Kc is 50.5. What are the equilibrium concentrations of H2, I2 & HI?
Workshop CALCULATING Keq 1. A mixture of 5.0 x 10-3 mol of H2 and 1.0 x 10-2 mol of I2 is placed in a 5.0L container at 448°C and allowed to come to equilibrium. Analysis of this equilibrium mixture shows that the [HI] is 1.87 x 10-3 M. Calculate Kc: H2(g) + I2(g) 2HI(g) 2. At 448°C the equilibrium constant Kc for the reaction below is 50.5. H2(g) + I2(g) 2HI(g) Predict how the reaction will proceed to reach equilibrium if the initial amount of HI is 2.0 x 10-2 mol, H2 is 1.0 x 10-2 mol, and I2 is 3.0 x 10-2 mol in a 2.00 L container.
APPLICATION OF Keq 1. Predicting the direction of reaction Q = reaction quotient at equil Q = K Q > K species on Rt (prod) (no net Rx) react to form left K = [Equil] Q = [Non Equil] Q < K forms more products Goal: Calculate Q to determine state of Rx, equil, more product or more reaction
Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures • first decide which direction the reaction will proceed • compare Q to K • define the changes of all materials in terms of x • use the coefficient from the chemical equation for the coefficient of x • the x change is + for materials on the side the reaction is proceeding toward • the x change is for materials on the side the reaction is proceeding away from • solve for x • for 2nd order equations, take square roots of both sides or use the quadratic formula • may be able to simplify and approximate answer for very large or small equilibrium constants
1. At 448°C the equilibrium constant Kc for the reaction is 50.5. H2(g) + I2(g) 2HI(g) Predict how the Rx will proceed to reach equil at 448°C if the initial amount of HI is 2.0 x 10-2 mol, 1.0 x 10-2 mol, H2 and 3.0 x 10-2 mol I2 in at 2.0 L container. [HI]° = 2 x 10-2 mol/2.0L = 1.0 x 10-2M [H2]° = 1.0 x 10-2mol/2.0L = 5.0 x 10-3M [I2]° = 3.0 x 10-2mol/2.0L = 1.5 x 10-2M Q = Prod = [HI]2 = (1.0 x 10-2)2 = 1.3 React [H2][I2] (5 x 10-3)(1.5 x 10-2) since K = 50.5, Q = 1.3, Q < K [HI] will need to increase and [H2][I2] will decrease to reach equilibrium.
2. At 1000K the value of Kc for the reaction 2S03(g) 2SO2(g) + O2(g) is 4.07 x 10-3. Calculate the value for Q and predict the direction in which the reaction will proceed towards equil if the initial concentration of reactants are: [SO3] = 2 x 10-3 M [SO2] = 5 x 10-3 M [O2] = 3 x 10-2 M Q = 0.2 reaction will proceed from Rt to left forming SO3.
CALCULATING Keq 1. Sulfur Trioxide decomposes at High temperature in a sealed container. 2 SO3(g) 2 SO2(g) + O2 (g) Initially the vessel is filled at 1000K with SO3(g) at a concentration of 6.09 x 10-3M. At equilibrium, the [SO3] is 2.44 x 10-3M. Calculate Kc. 2. Calculate the value for Q and predict the direction in which the reaction will proceed towards equilibrium if the initial concentrations are: [SO3] = 2.0 x 10-3M [SO2] = 5.0 x 10-3M [O2] = 3.0 x 10-2M
LE CHATELIER’S PRINICIPLE “If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of one of its components, that system will shift it’s equilibrium position as to ‘counteract’ the effect of the disturbance.” Equilibrium can be disturbed by: - adding or removing components - a change in pressure - a change in volume - a change in temperature
PREDICTING THE DIRECTION OF THE SHIFT I. CATALYST A catalyst increases the rate at which equilibrium is achieved but not the composition of the equilibrium mixture. II. THE REACTION QUOTIENT Q < K: the reaction shifts to the products Q > K: the reaction shifts to the reactants III. CHANGES IN VOLUME Reducing the volume of a gas at equilibrium causes the system to shift in the direction that reduces the number of moles of gas.
+ lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas The effect of pressure (volume) on an equilibrium system.
When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. The Effect of Volume Changes on Equilibrium
PROBLEM: How would you change the volume of each of the following reactions to increase the yield of the products. (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) PLAN: When gases are present a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. Sample Problem Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position SOLUTION: (a) CO2 is the only gas present. To increase its yield, we should increase the volume (decrease the pressure). (b) There are more moles of gaseous reactants than products, so we should decrease the volume (increase the pressure) to shift the reaction to the right. (c) There are an equal number of moles of gases on both sides of the reaction, therefore a change in volume will have no effect.
IV. CHANGES IN TEMPERATURE When heat is added to a system, the equilibrium shifts in the direction that absorbs heat. Cooling has the opposite effect and shifts the equilibrium towards the side which produces heat. Endothermic Reactions: heat + Reactants Products An increase in temperature leads to a shift towards the products, a decrease leads to a shift towards the reactants. (Keq increases) Exothermic Reactions: Reactants Products + heat An increase in temperature leads to a shift towards reactants. (Keq decreases)
PROBLEM: How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions? (a) CaO(s) + H2O(l) Ca(OH)2(aq) DH0 = -82kJ (b) CaCO3(s) CaO(s) + CO2(g) DH0 = 178kJ (c) SO2(g) S(s) + O2(g) DH0 = 297kJ PLAN: Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc. (a) CaO(s) + H2O(l) Ca(OH)2(aq) heat (b) CaCO3(s) + heat CaO(s) + CO2(g) (c) SO2(g) + heat S(s) + O2(g) Sample Problem Predicting the Effect of a Change in Temperature on the Equilibrium Position SOLUTION: An increase in temperature will shift the reaction to the left, decrease [Ca(OH)2], and decrease Kc. The reaction will shift right resulting in an increase in [CO2] and increase in Kc. The reaction will shift right resulting in an decrease in [SO2] and increase in Kc.
Practice - Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? • adding more O2 to the container • condensing and removing SO3 • compressing the gases • cooling the container • doubling the volume of the container • warming the mixture • adding the inert gas helium to the container • adding a catalyst to the mixture
Practice - Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? • adding more O2 to the container shift to SO3 • condensing and removing SO3 shift to SO3 • compressing the gases shift to SO3 • cooling the container shift to SO3 • doubling the volume of the container shift to SO2 • warming the mixture shift to SO2 • adding helium to the container no effect • adding a catalyst to the mixture no effect
Example 1: N2O4(g) 2 NO2(g) H = 58 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add N2O4 B) remove NO2 C) increase the total pressure by adding N2 D) increase the volume of the container E) decrease the temperature
Answer: N2O4 2NO2 A) The system will adjust so to decrease [N2O4] shifts to products B) Shifts to products (NO2 removed) C) N2 will increase total pressure but since it is not involved in Rx the partial pressures of N2O4 and NO2 are unchanged, no shift. D) volume shifts to more moles of gas E) Temp - Rx is endo heat + N2O4 2 NO2 Temp shifts so more heat produced K is affected
Example 2: PCl5(g) PCl3(g) + Cl2(g) H = 88 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add Cl2 B) temperature is increased C) the volume of the reaction system is decreased D) PCl5 is added E) a catalyst is added
1. The equilibrium constant for the Haber process at 472°C is Kc = 0.105. A 2.00 L flask is filled with 0.500 mol of ammonia and is then allowed to reach equilibrium at 472°C. What are the equilibrium concentrations? N2(g) + 3 H2(g) 2 NH3(g) 2. For the reaction PCl5(g) PCl3(g) + Cl2(g) at a certain temperature Kc equals 450. What will happen when 0.10 mol of PCl5, 1.0 mol of PCl3, and l.5 mol of Cl2 are added to a 2.0-L container and the system is brought to the temperature at which Kc=450. What are the equilibrium concentrations?
EFFECT OF VARIOUS DISTURBANCES ON AN EQUILIBRIUM SYSTEM DISTURBANCE NET DIRECTION OF REACTION EFFECT ON VALUE OF K Concentration Increase (reactant) Toward formation of product None Decrease (reactant) Toward formation of reactant None Pressure (volume) Increase P Toward formation of lower amount (mol) of gas1 None Decrease P Toward formation of higher amount (mol) of gas None Temperature Increase T Toward absorption of heat Increases H°rxn>0 Decreases if H°rxn<0 Decrease T Toward release of heat Increases H°rxn<0 Decreases H°rxn>0 Catalyst added None; rates of forward and reverse reactions increase equally None
VAN’T HOFF EQUATION Changes in K due to T In K2 = H°rxn ( 1 - 1) K1 R T1 T2 R = 8.314 J/mol K T = Kelvin The formation of methanol is an important industrial reaction in the processing of new fuels. At 298K, Kp = 2.25 x 104 for the reaction CO(g) + 2 H2(g) CH3OH(l) If H°rxn = -128 kJ/mol CH3OH, calculate Kp at 0°C.
H0rxn H0rxn Hvap K2 1 1 1 1 1 1 1 1 = - - ln R R R T1 T2 T1 T1 T2 T1 T2 T2 K1 P2 Ea k2 - = - - ln = - ln P1 R k1 K2 - = - ln K1 The van’t Hoff Equation The Effect of T on K R = universal gas constant = 8.314 J/mol*K K1 is the equilibrium constant at T1 Temperature Dependence
Q 2. CH4(g) + CO2(g) 2CO(g) + 2 H2(g) A. What is the percent yield of H2 when equimolar mixture of CH4 and CO2 with a total pressure of 20.0 atm reaches equilibrium at 1200K at which Kp = 3.548 x 106? B. What is the percent yield of H2 for this system at 1300K, at which Kp = 2.626 x 107? C. Use the Van’t Hoff equation to find H°rxn.
