Study Guide Exam 3 - Chemistry II | LB 172, Study notes of Education Planning And Management

Download Study Guide Exam 3 - Chemistry II | LB 172 and more Study notes Education Planning And Management in PDF only on Docsity! Exam 3 Review Guide If you have any questions about any of the material, or feel as if I have missed anything please feel free to email me or call me Email: fadelrae@msu.edu Phone: 1-3132057485 Chapter 13: Chemical Kinetics  A Reaction Mechanism is a series of individual steps in an overall reaction o The sum of the individual steps gives the overall equation o Compile the individual steps by adding together or canceling out to find overall reaction  Compounds on similar side of equation add up  Compounds on different sides cancel out  Ex. NO2 + CO  CO2 + NO (overall equation) 1st step: 2NO2  NO3 +NO 2nd step: CO + NO3  NO2 + CO2 o Elementary Steps  Elementary steps are simple reactions that include one or more similar or different compounds to form product  The amount and type of product dictates rate law  A  product Rate = k[A]1  A + A  product Rate = k[A]2  A + B  product Rate = k[A]1[B]1 o Rate Determining Step  Within each reaction mechanism, there is one intermediate step which is the slow step  This slow step, or the “Rate Determining Step” dictates what the rate law of the overall reaction will be  Ex. 2NO + 2H2  N2 + 2H2O 1st step: 2NO  N2O2 (Fast) 2nd step: N2O2 + H2  N2O + H2O (Slow) 3rd step: N2O + H2  N2 + H2O (Fast) In this case the 2nd step would be the Rate Determining Step. Therefore: Rate = k[Reactants of Slow Step] Rate = k[N2O2][H2] However, N2O2 is an intermediate, and is consumed and used up in the reaction. Therefore, we cannot determine its concentration. In order to replace an intermediate, kf[NO]2 = kr[N2O2] *Set kr[intermediate] = kf[reactant in overall] equation (kf/kr)[NO]2 = [N2O2] *Now replace [N2O2] with new concentration Rate = k (kf/kr)[NO]2 [H2] or Rate = k’ [NO]2 [H2] * k’= k(kf/kr)  Energy Diagrams o Energy Diagrams are diagrams illustrating the overall behavior of a reaction o Energy Diagrams tell us  1) Amount of Elementary Steps (# of humps in the diagram) The 2NO2 and NO2 cancel to form NO2 and the NO3’s cancel out o Before we discussed when Keq is large, there is more product, and when Keq is small, there is more reactant o However, when Q<Keq, the reaction will move toward the products (to the right) because there are more reactants than products, and the reaction will attempt to restore equilibrium o Conversely, when Q>Keq, the reaction will move toward the reactants (to the left) because there are more products than reactants, and the reaction will attempt to restore equilibrium o Ex. The following reaction takes place in a 10 L vessel @ 214 K: N2 + 3H2  2NH3 Kc = 2.37x10-3 The partial pressures were measured: N2 = 1.2 atm ; H2 = 15.5 atm ; NH3 = 1.84 atm. What is the Qc of the reaction? Which direction will the reaction go in order to reach equilibria? Given: N2 = 1.2 atm H2 = 15.5 atm NH3 = 1.84 atm T = 214 K KC = 2.37x10-3  Le Chatelier’s Principle o Le Chatelier’s Principle states: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance. o In other words, there are several changes that a system may experience which could throw the equilibrium of that system off. When such changes are experienced, the system will shift right (to favor products) or left (to favor reactants). Therefore, if a change in the system increases the concentration of products, the system will shift to the left to create more reactants to combat the change in concentration of product, and vice versa. o Some of the “stressors” that can affect a systems equilibrium include:  Concentration  If the concentration of product increases, the reaction will shift to the left to create more reactants, and vice versa.  Pressure/Volume  An increase in pressure is directly related to an increase in concentration, so an increase in the pressure of products (increase in concentration of products) results in the system shifting to the left to create more reactants to restore equilibrium, and vice versa.  An increase in Volume is indirectly related to an increase in concentration, so if the volume of a system is increased and the concentration of products decreases, the system will shift to create more product to restore equilibrium.  ONLY TRUE FOR GASSES!!!!  Temperature  An increase in heat results in a system shift toward the products [N2]= 1.2 atm/(.08206x214K) = .06M [H2]= 15.5 atm/(.08206x214K) = .88M [NH3]= 1.84atm/(.08206x214K) = .105M QC = [.105]2 / [.88]3[.06] = .234 QC > KC or Q > Keq meaning the reaction will go to the left P=MRT  A decrease in heat (decrease in temperature) results in a system shift toward the reactants.  Catalysts  Catalysts do not affect the reaction’s equilibrium  Catalysts only increase the rate of the overall reaction, increasing both the forward and reverse reactions, favoring no side.  ICE Tables o ICE tables are tables that include the Initial concentration, Change in concentration, and Equilibrium concentration o If initial concentrations are given, ICE tables can help solve for individual concentrations at equilibrium o Solving for KC using ICE tables  Ex. .100M H2 and .100M I2 are in a system. At equilibrium, [H2] = .020M and [I2] = .020M. What is the concentration of HI at equilibrium? What is the Kc of this system? H2 + I2  2HI Initial .100M .100M 0 Change .100-x .100-x +2x Equilibrium .020M .020M .160M o What if the concentration of reactants/products at equilibrium are not given?  Plug initial concentrations into ICE table  Find the change (-x or +x) depending on if Q is greater than or less than Keq  If Q is less than Keq the change for products will be +x and the change for reactants will be –x  Conversely, if Q is greater than Keq the change for products will be –x and the change for reactants will be +x  Ex. Solve for the concentrations at Equilibrium CO(g) + H2O(l)  CO2(g) + H2(g) I .100M DC 0 0 C -x DC +x +x E (.100-x) DC x x o o **To check your x value and your Kc = ? [HI]eq = ? Given [H2] = .100M [I2] = .100M [H2]eq = .020M [I2]eq = .020MX is calculated out by finding the difference between the initial and equilibrium concentrations of either H2 or I2 Once X is solved for, plug it in to +2x to Solve for the [HI] at equilibrium Kc = [HI]2 / [H2][I2] = (.160)2 / (.020)(.020) = 64 Given: KC= 4.06 [CO]Initial = .100M [CO2]initial = 0 [H2]Initial = 0Tips: As soon as you approach this problem, you notice H2O is in (l) form, meaning we DC (don’t care) about it. Also, in order to solve for the concentrations at EQ, x must first be solved for. X can then be plugged into EQ to find concentrations KC = 4.06 = [CO2][H2] / [CO] = (x)(x) / (.100-x) = x2 / (.100-x) After using quadratic formula, x= -4.16 ; .0977. The x value cannot be negative, so the -4.16 value is thrown out. Therefore, x = .098 [CO] = .100 – x = .100 - .098 = .002M [CO2] = x = .098M [H2] = x = .098M concentrations at equilibrium, plug the concentrations into the Kc expression and see if you get the same Kc as the one given to you at the beginning of the problem. o What if the initial concentrations given to you are not equal?  Ex. Solve for the concentrations at Equilibrium H2 + I2  2HI I .006M .004M .022M C -x -x +2x E (.006-x) (.004-x) (.022+2x)  Solving Equilibrium Problems: Steps to Solving Equilibrium Problems o 1) Write Balanced Equation, Write Keq Expression o 2) Determine Direction of Reaction  Solve for Q  If Q>K: products are (-x) reactants are (+x)  If Q<K: products are (+x) reactants are (-x) o 3) Fill in ICE Table o 4) Use equilibrium values ([A]initial(+/-)x) and plug into Keq expression o 5) Set Keq expression equal to given K value and solve for ‘x’ o 6) Use ‘x’ value to determine concentration of reactants/products at Equilibrium  **NEGATING X** o One of the most promising and convenient short-cuts learned in this section is the ‘negate x’ short-cut o Most of the time, the change in x in the denominator is so small compared to the overall equation that we could just negate it.  This is the case when reactions have very small Ka meaning the overall reaction is mostly reactants  The change in reactants is small o To negate x, simple remove it from the denominator, and go about solving the equation without the quadratic formula, saving you tons of time. o In order to determine if removing x from denominator is valid, divide x by the initial reactant concentration, then multiply it by 100. The answer must be less than 5.00% If it is, then you can negate x o **Simple tip. I always believed that by the time you test x and do all that, you could have completed quadratic formula. This shortcut, if not tested, does have HUGE risks, so be cautious about using it. Chapter 15: Acids and Bases  Definitions of Acids and Bases o Arrhenius Definition Given: Kc = 54.3 [H2]Initial = .006M [I2]Initial = .004M [HI]Initial = .022M KC = 54.3 = [HI]2 / [H2][I2] = (.022+2x)2 / (.006-x)(.004-x) = 4x2+.088x+4.84x10-4 / x2+.010x+2.4x10-5 Using quadratic formula, x = -.011 ; .0015, and since x cannot be negative, x = .0015 [HI] = .022+2x = .025M [I2] = .004-x = .0025M[H2] = .006-x = .0045M  In a Kb expression, a weak base will take a H+ from H2O to form an OH- as a product  If a problem involves an equation that produces OH-, but the Ka is given, you must calculate for the Kb  Using Kw to solve for either Kb or Ka depending on which one is given o Kw is the ion product constant for water (dissociation constant for water) o Kw = [H3O+][OH-] = [H+][OH-] = Ka x Kb o Kw = 1.0x10-14 @ 25 degrees Celsius (Dr. Cass will assume each reaction in which she intends us to use Kw to be at this temp so assume 1.0x10-14 = Kw always) o So, if a problem has an equation that produces H3O+, you know you must use Ka However if the Kb is given, you must calculate for the Ka  Kw = Ka x Kb  Ka = Kb / Kw and you know Kw is always 1.0x10-14 so just plug whatever Kb value Dr. Cass gives you to solve for Ka o If a problem has an equation that produces OH- you know you must use Kb However if the Ka is given, you must calculate for the Kb (approach this problem exactly how you would approach the previous one, except youre solving for Kb instead of Ka)  Ex. A solution has [OH-] = 3.2x10-4; What is the [H3O+]? Kw = [H3O+][OH-] 1.0x10-14 = [H3O+][3.2x10-4] [H3O+] = 3.125x10-11  pH Scale, pOH and Acidity o If something is added to a solution, and the concentration of [H+] becomes larger than [OH-], the solution is Acidic o If something is added to a solution, and the concentration of [OH-] becomes larger than [H+], the solution is Basic  Ex. Is the previous problem acidic or basic? [H+] = 3.125x10-11 [OH] = 3.2x10-4 [H+] < [OH-] ; Solution is Basic!!! o pH is a compact way to specify the acidity of a solution  pH = -log[H3O+] *[H3O+] is the same as [H+]  pH = 14 - pOH  pH<7 Solution is acidic  pH=7 Solution is neutral  pH>7 Solution is basic o pOH is a compact way to specify the basicity of a solution  pOH = -log[OH-]  pOH = 14 - pH o pOH + pH = 14  Calculating the pH and pOH from strong acids and bases o When you are dealing with strong acids and strong bases, it is easy to solve for the pH or pOH o Ex. A solution has .200M HCl. What is the pH? pH = -log[H+] pH = - log[.200] pH = .699 o Ex. A solution has .050M [HI]. What is the pOH? pOH = -log[OH] = 14 - pH pH = - log[H+] = -log[.050] = 1.30 pOH = 14 – 1.30 = 12.699 o In the two above examples, solutions contained strong acids/bases and asked to find the pH or pOH. In these situations, when the concentration of a strong acid or base is given, and element/ion connected to the H+ in the acid or OH- in the base you know is whats called a spectator ion. This means that it is not significant, so the [OH] in a solution of NaOH is equal to the [NaOH].  Determination of [H3O+] and [OH-] in weak acid/base solutions o BACK TO ICE TABLES o Ex. What is the pH of .120M solution of Acetic Acid? (Tip: pH = -log[H3O] … knowing this, you know you want to solve for the concentration of H3O at equilibrium) CH3CO2H + H2O  CH3CO2- + H3O+ I .120M DC 0 0 C -x DC +x +x E .120M-x DC x x  **In the above problem, it asked you to solve for the pH of the solution. As soon as the problem asks for the pH, you must understand it is asking to first find the [H3O] and then take the –log of that concentration to satisfy the equation pH = -log[H+]. Conversely, as soon as the question asks for the pOH of the solution, you must note that it is asking for the concentration of [OH] first, then you must plug in [OH] into pOH = -log[OH]. Using ICE tables allows you to find the concentrations of OH and H+ at equilibrium so you could use them to solve for pH or pOH  **Again, finding [OH] and [H+] using ICE tables only holds true for weak bases and weak acids that do not completely dissociate in solution. The equation must have a  arrow to symbolize equilibrium. (For strong acids and strong bases, see previous section on how to calculate pH and pOH)  Effect of Salts on pH o Salts in which neither the cation nor the anion acts as an acid or a base form pH- neutral solutions o Salts in which the cation does ot act as an acid and the anion acts as a base form basic solutions o Salts in which the cation acts as an acid and the anion does not act as a base for acidic solutions o Salts in which the cation acts as an acid and the anion acts as a base form solutions in which the pH depends on the relative strengths of the acid and the base o This table represents the effects of salts on pH. Given: Ka = 1.8x10-5 [CH3CO2H] = .12M Ka = 1.8x10-5 = [CH3CO2][H3O+] / [CH3CO2H] Ka = 1.8x10-5 = x2 / .12-x Using quadratic formula, x = .001461 ; -.00148 but x cannot be negative so x = .00146 [H3O] = x = .00146 pH = -log[H3O] = 2.836  Polyprotic acids & Determination of pH in these solutions o Polyprotic acids ionize in successive steps, each step with its own Ka value  Ex. The ionization of sulfurous acid: 1st step: H2SO3(aq)  H+(aq) + HSO3-(aq) Ka1 = 1.6x10-2 2nd step: HSO3+(aq)  H+(aq) + SO32-(aq) Ka2 = 6.4x10-8  In this example, the ionization takes place in 2 separate steps. The Ka is smaller for the second step than the first, which will be the case for all polyprotic acids.  Ex. Find the pH of .01M Sulfuric Acid solution 1st step: H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) Strong 2nd step: HSO4(aq) + H2O(l)  H3O+(aq) + SO42-(aq) Ka2 = 1.2x10-2 *The first step is a strong acid completely ionizing in water, so the [H3O] is .1M *Again, H2O is in (l) form, so we do not count it in our ICE table! HSO4  H3O+ + SO42- I .010M .010M 0 C -x +x +x E .010M-x .010M+x x Chapter 16: Aqueous Ionic Equilibrium  Buffer: resists pH change by neutralizing added acid or added base. Ka2 = 1.2x10-2 = [H3O][SO4-] / [HSO4-] [H3O+] = .010M + x = 1.2x10-2 = .010x + x2 / .010 – x = .010M = x2 + .022x - .00012 = .0145M After quadratic formula, x = -.027; .0045 and pH = -log[H3O+] = 1.84 since x cannot be negative, x = .0045 Anion Catio n Conjugate base of strong acid Conjugate base of weak acid Conjugate acid of weak base Acidic Depends on relative strengths Small, highly charged metal ion Acidic Depens on relative strengths Counteranion of strong base Neutral Basic