Unit 1: Kinetics and Equilibrium ANSWER KEY Chem 1B Kinetics Name 1. What factors affect rates? Give examples of each. 1. Nature of the reactants- physical state, surface area-examples Li, Na, K react with water at different rates, liquid gasoline burns slower that gas vapor, powders react faster than larger chunks, dry solid reactants generally react slower than solutions of the same reactants. 2. The concentration/pressures of the reactants -Examples many times a reaction rate slows as the concentration of the reactants decrease. At any time an instantaneous rate can be calculated from the negative slope of the tangent to the curve of time (x axis) verses reactant concentration (y axis) or the positive tangent to the time verses product concentration. 3. The temperature at which the reaction occurs -Examples The rate increases as the temperature increases since more colliding particles will have the minimum energy to react. 4. The presence of a catalyst, enzymes -Examples Catalyst alter the mechanism/pathway of the reaction. The activation energy is lowered and the steps in the reaction vary from that without a catalyst. 2. Express the general rate of reaction in terms of the rate of change of each reactant and each product in the following reaction (this is an instantaneous rate used for relatively short time periods): 2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l) Rate = - [ C2H6]/2 t = - [ O2]/7 t= [ CO2]/4 t= [ H2O]/6 t 3 Write the balance equation and rate law expression for the combination reaction of NO (g) and O2 (g) to produce NO2 (g) if the reaction is second order in NO, first order in O2 and 3rd order overall. 2 NO (g) + O 2 (g) 2 NO 2 (g) Rate = k[no] 2 [O 2 ] 4 Rate data were collected for the following reaction at a particular temperature: A + 2B C + 2D Experiment Initial [A] Initial [B] Initial Rate form. C 1 0.20 M 0.10 M 4.0 x 10-4 M/min
2 0.20 M 0.30 M 1.2 x 10-3 M/min 3 0.60 M 0.30 M 3.6 x 10-3 M/min 4 0.40 M 0.20 M? a) What is the order of the reaction with respect to A, B, and overall? [A] = 1; [B] = 1; overall = 2 b) What is the rate law expression? Rate = k[a] [B] c) What is the value of k? k = 0.020 M -1 min -1 d) Calculate the rate for experiment 4. Rate = 1.6 x 10-3 M/min 5 Consider a graph where x is time in seconds and y is the concentration of reactant A in Molarity, [A]. The graph shows a straight line whose y intercept is 0.400 M and the slope is -0.0350 M/sec for the reaction: A Products a) What is the order of the reaction? Zero b) Write the rate law expression. Rate =k c) Determine the rate constant, k, for this reaction. k = 0.0350 M/sec 6 The reaction 2A products is second order in A and has the form Rate = k[a]2 where k has the value 6.0 x 10-4 M-1 min-1. The reaction is at an initial concentration of 0.100 M and 318K. a) How long is the first half-life? t 1/2 = 1.7 x 10 4 min; 280 hr b) How much time is required for the concentration to reach 0.085 M? 2.9 x 10 3 min 7 According to collision theory, to have an effective collision reactants must be in the proper ORIENTATION towards each other at the time of collision and have THE MINIMUM ENERGY REQUIRED, E a to rearrange outer electrons in breaking bonds and forming new ones. 8 The following is a proposed mechanism: 1) A + B <==> AB (fast, equilibrium) 2) AB + A A 2 B (slow) 3) A 2 B + B A 2 + B 2 (fast) a) What is the overall equation for this mechanism? 2 A + 2B A 2 + B 2 b) What is the rate law expression for the mechanism? Rate =k[a] 2 [B] c) What happens to the initial rate of formation of AB (step 1) if the concentrations of A and B are both doubled? Rate =k[2a][2b] 2 x 2 = 4 the rate is 4 times greater c) What happens to the rate of the reaction if the concentrations of A and B are both doubled? Rate =k[2a] 2 [2B] 2 2 x 2 = 8, the rate is 8 times greater d) Which letter(s) represents a reaction intermediate? AB, A 2 B
9 The activation energy for a given reaction is 95 kj/mol. If the rate constant is 0.042 sec-1 at 30 C, what is the value of the rate constant at 50 C? ln (k 1 /k 2 ) = (E a /R)(T 2-1 T 1-1 ); k = 0.43 s -1 10 How does a catalyst affect The activation energy? The rate? The equilibrium constant? The activation energy: Generally a catalyst lowers the activation energy The rate: A catalyst will speed up the rate The equilibrium constant: A catalyst has no effect on the equilibrium constant 11 The rate and therefore the specific rate constant of a particular reaction doubles when the temperature is increased from 25 C to 40 C. Calculate the activation energy, Ea, for this reaction. ln (k 1 /k 2 ) = (E a /R)(T 2-1 T 1-1 ); E a = 35.8 kj/mol 12 With regard to collision theory, (k = Zfp), a) What do each of the symbols, (k, Z, f, p) represent? Z = collision frequency, how often particles collide. This is directly proportional to the root mean squared speed, rms. From the kinetic molecular theory we learned rms = square root of 3RT/Molecular weight. As the temperature increases we know that the molecules move faster and will collide a bit more frequently. But, by doing the math one can find a 10 C rise has less than a 2% increase in collision frequency. Z does not noticeably affect the specific rate constant as the temperature changes. This is certainly far off from the tripling affect we saw in the NO and Cl 2 example. p = fraction of collisions with proper orientation (independent of Temp). f = fraction of collisions with the activation energy required for success. This energy allows the particles to create an activated complex at the height of the energy profile. This factor is greatly affected by small temperature changes. f = e -Ea/RT Using the same NO and Cl 2 example, f = e -Ea/RT and given that the activation energy is 85 kj/mol, one can solve for f. At 25 C, f = 1.3 x 10-15 and at 35 C, f = 3.9 x 10-15 which accounts for our k being over 3 times larger. b) What factor is most responsible for the fact that the rate of a chemical reaction generally increases sharply with a rise in temperature? f = fraction of collisions with the activation energy required for success.
13 Propose a plausible 3-step mechanism that fits with the information given. Overall reaction 2 Ce +4 + Tl +1 Mn+2 2 Ce +3 + Tl +3 Rate = k[ce +4 ][Mn +2 ] Step 1 is the rate determining slow step. Mn +3 is a reaction intermediate produced in step 1 and used up in step 2 Mn +4 is a reaction intermediate produced in step 2 and used up in step 3 Tl +1 is not used until step 3 Step 1) Ce +4 + Mn +2 Mn +3 + Ce +3 Slow Step 2) Ce +4 + Mn +3 Mn +4 + Ce +3 Fast Step 3) Tl +1 + Mn +4 Mn +2 + Tl +3 Fast 14 The experimentally observed rate law for the reaction: H 2 + 2 ICl 2 HCl + I 2 (all are gas) is Rate = k[icl][h 2 ]. a) (Circle) the following mechanism(s) that is (are) plausible? (can be more than one) b) For all the mechanisms identify, Reactants, Products, (Reaction Intermediates), and Catalysts c) Identify the molecular of each step in all the mechanisms (uni-, bi-, termolecular). Molecularity Mechanism 1 Step 1 H 2 + 2 ICl 2 HCl + I 2 Slow Mechanism 2 Step 1 2{ ICl I + Cl } Fast Step 2 I + Cl + H 2 HCl + HI Slow Step 3 HI + I + Cl HCl + I 2 Fast Mechanism 3 Step 1 2ICl Cl 2 + I 2 Slow Step 2 Cl 2 + H 2 2 HCl Fast Mechanism 4 Step 1 H 2 + ICl HCl + HI Slow Step 2 ICl + HI HCl + I 2 Fast Mechanism 1 Step 1 H 2 + 2 ICl 2 HCl + I 2 Slow Ter (Mechanism 2) Step 1 2{ ICl I + Cl } Fast Equilibrium Uni Step 2 I + Cl + H 2 HCl + HI Slow Ter Step 3 HI + I + Cl HCl + I 2 Fast Ter Mechanism 3 Step 1 2ICl Cl 2 + I 2 Slow Bi Step 2 Cl 2 + H 2 2 HCl Fast Bi (Mechanism 4) Step 1 H 2 + ICl HCl + HI Slow Bi Step 2 ICl + HI HCl + I 2 Fast Bi
15 Explain why a mixture of natural gas and air formed by leakage of gas from a pipe can stand for a long time without reacting but explodes if a spark is introduced. The reaction requires a spark in order to provide the needed activation energy to initiate the reaction. 16 Describe the characteristics of the specific rate constant, k. Include if and how k and its units are affected by temperature, time, concentration, overall order, catalyst, etc. Information on k, the specific rate constant 1. Its value is for a specific reaction represented by a balance equation 2. Its value will change if a catalyst is added 3. It s for a specific temperature and will change when Temp changes 4. Its units depend on the overall order. 1/(M (overall order 1 time) 5. It does not alter with the concentration changes 6. It does not change with time 7. It is determined through experiments 17 a) The table shows concentration changes with time as a reaction A products takes place. Fill in the table. time (hour) [A] (Molarity) ln[a] 1/[A] 0 0.200 M -1.61 5.00 M -1 1 0.100 M -2.30 10.0 M -1 3 0.050 M -3.00 20.0 M -1 4 0.040 M -3.22 25.0 M -1 7 0.025 M -3.69 40.0 M -1 b) What order is the reaction? Zero First Second Second c) Plot a graph labeling the axes appropriately for the determined order, which will give a straight line. 1/[A] vs hours d) Calculate the value of the specific rate constant, k. k = 5.00 M -1 hr -1 e) How long will it take [A] to reduce to 0.085 M? 1.35 hr f) How long is the first half-life of this reaction? 1.00 hr g) How long is the third half-life of this reaction? 4.00 hr
Equilibrium 18 Do most reactions go to completion (all reactants forming all products until one reactant is used up)? No 19 a) Write the equilibrium constant (Kc or Kp) for the following reaction: 2SO2 (g) + O2 (g) 2 SO3 (g) H = -198.4 kj/mol rxn K c = [SO 3 ] 2 /[SO 2 ] 2 [O 2 ] or K p = [P SO3 ] 2 /[P SO2 ] 2 [P O2 ] b) Does K remain constant as... 1) temperature varies? No, K changes If temperature increases how will the equilibrium shift? How does K change? As Temp increases this reaction will shift in the endothermic direction, for this example it is reversed. 2) initial concentrations vary? Yes, K is constant If you start with only products instead of reactants how does K change? There is no effect on K 3) reaction coefficients vary? No, K is changes If the equation is written as SO2 + 1/2 O2 SO3, how doe K change? One must take the effect and add it as a power to the value of K. For the example above the reaction is cut in half so the value of K is square rooted. c) if K << 1 then are products or reactants favored? REACTANTS d) How are Kc and Kp different? By what equation are they related? K p represents all activities came from partial pressures and K c represents all the activity values came from molarities. K p = K c (RT) n e) What is the difference between Q and K? Q is for any concentrations, while K must be for equilibrium conditions. 20 What is Le Chateliers Principle? Explain the meaning of STRESS added to a system at equilibrium and how the system reacts. (FROM THE NOTES) Le Chatelier s Principle and How to Manipulate the Reaction: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position as to counteract the effect of the disturbance. Changes to Consider: a) Concentration of any substance: Adding a reactant or removing a product will shift the reaction forward. Removing a reactant or adding a product will shift in reverse. b) Volume / Partial Pressures of any gas substance: Increasing the volume will decrease the partial pressures and the reaction will shift to create more pressure,
the direction that has more moles of gas. No effect occurs when an inert gas is added that does not participate in the reaction. c) Temperature: Increasing the temperature will favor the endothermic reaction to shift reduce the extra energy added. Decreasing the temperature will in the direction of the exothermic reaction. d) Catalyst: This has no affect on a system at equilibrium; it will just allow a system to reach the equilibrium quicker. 21 For the reaction Br2 (g) + F2 (g) 2 BrF (g) Kc = 54.7 a) What are the equilibrium concentrations of all the gases if the initial concentrations of bromine and fluorine were both 0.100-M with no BrF? x =0.079 M; 0.021 M Br 2, 0.021 M F 2 and 0.158 M BrF b) What are the equilibrium concentrations if the initial concentrations of all three gases (Br2, F2, & BrF) start at 0.100 M? x = 0.068 M; 0.032 M Br 2, 0.032 M F 2 and 0.236 M BrF 22 The system below is at equilibrium and has the following concentrations in a 1.00 L flask: [H2O2] and [CO] both are 0.37 M while [H2O] = [CO2] = 0.74 M. If 0.75 moles of ONLY hydrogen peroxide are added, calculate the new equilibrium concentrations. H2O2 (g) + CO(g) H2O (g) + CO2 (g) K c = 4.0, calculated from initial equilibrium values This needs the quadratic equation to solve. x = 0.16M; 0.96 M H 2 O 2, 0.21M CO, 0.90 M H 2 O, 0.90M CO 2 23 For the reaction at a 25 C temperature: N2O4 (g) 2 NO2 (g) If Kc = 250, what is Kp? K p = 6120 24 What is the equilibrium constant (Kc, Kp, Keq) expression for... a) MgCO3 (s) MgO (s) + CO2 (g) K c = [CO 2 ], K eq = K p = [P CO2 ] b) 2HCl (aq) + Mg (s) H 2 (g) + MgCl 2 (aq) K eq = [P H2 ][MgCl 2 ]/[HCl] 2, K c = [H 2 ][MgCl 2 ]/[HCl] 2 K p = not possible, c) Fe +3 (aq) + SCN -1 (aq) FeSCN +2 (aq) K c = K eq = [FeSCN +2 ]/[Fe +3 ] [SCN -1 ] K p = not possible,
25 What is the value of Keq when G = -150 kj/mol at 25 C? G = -RTlnK eq Use R = 8.314 J/mol K (this belongs in Unit II) K eq = 2 x 10 26 26 In the Haber process NH3 is manufactured from N2 and H2. At 25 C, Kp = 3.6 x 108. To increase the rate of reaction the reaction is run at a higher temperature. Use the van't Hoff equation to calculate Kp at 450 C N2 (g) + 3 H2 (g) 2NH2 (g) H = -92.2 kj Van't Hoff eq. ln(k2/k1) = H/R (1/T1-1/T2) Use R = 8.314 J/mol K (this belongs in Unit II) K p = 0.11 REVIEW 27 What volume of 18.0 M H2SO4 is required to prepare 1.50 L of 2.00 M H2SO4? 0.167 L 28 What is the difference between an equivalence point and an end point? Equivalence point is the stoichiometric amounts, End point is a visible sign to stop the titration. 29 What is the M of KOH if you have 38.7 ml of KOH solution, and 43.2 ml of 0.223 M HCl is required to react with all the KOH in a titration reaction? 0.249 M 30 What does it mean to be a strong electrolyte? Which types of substances will be strong electrolytes? What are the strong acids? What are the strong soluble bases? Strong electrolytes dissociate almost completely into ions. Strong electrolytes can be Strong Acids, Strong Bases and Soluble Salts. Strong Acids: HCl, HBr, HI, HNO 3, HClO 4, HClO 3, H 2 SO 4 Strong soluble Bases: Metal Hydroxides in groups 1A (excluding HOH) and IIA (excluding the first two; Be, Mg) 31 For the following double displacement reactions, predict the products and balance the reactions, include terms (s), (l), (g), & (aq): Use NR for no reaction. a) KNO3 (aq) + (NH4)3PO4 (aq) No Reaction-all spectators
b) HCl (aq) + K2SO3 (aq) 2HCl (aq) + K2SO3 (aq) 2KCl (aq) + H 2 O (l) + SO 2 (g) c) NaCl(aq) + AgNO3(aq) NaCl(aq) + AgNO3(aq) NaNO 3 (aq) + AgCl (s) Acids and Bases 32 Give a definition for an acid and a base for each of the 3 theories; Arrhenius, Bronsted-Lowry, and Lewis. Arrhenius theory Acid has H+ its willing to give up in aqueous solution Base has OH- its willing to give up in aqueous solution Bronsted-Lowry theory Acid is a proton donor Base is a proton acceptor Lewis theory Acid will accept an electron pair Base will donate an electron pair 33 What is the acid, base, conjugate base and acid in the equation below? NH3 + H2O NH4 + + OH- Base---------------------conj. acid acid-----------------------conj. base 34 a) What are some properties of an acid? a base? Properties of acids: Tastes sour, reacts with metals to liberate hydrogen, reacts with carbonates to liberate CO2, neutralizes bases, neutralizes metal oxides, change color of indicators, conduct electricity, react with a salt containing a conjugate base of a weaker acid Properties of bases: Tastes bitter, slimy to touch, neutralizes acids, conducts electricity, changes color of indicators. b ) List the strong acids and strong bases. Strong Acids: HCl, HBr, HI, HClO4, H2SO4, HNO3, HClO3 Strong Bases: LiOH, NaOH, KOH, RbOH, CsOH, FrOH, Ca(OH)2, Sr(OH)2, Ba(OH)2, Ra(OH)2
c) Describe the relationship between the strength of a strong acid and its conjugate base. Strong acids will always want to react to form weaker acids and vice versa. As an acid increases in strength its conjugate base decreases in strength. d) What is the relationship between the bond strength of a binary acid and the acid strength? As the bond strength increases the acid strength decreases since it takes more effort to break the bond and put the H+ into solution. e) What is the relationship between the oxidation state of an oxyacid with the same central atom and the acid strength? As the oxidation state increase so does the acid strength. As an example in H2SO4, S has an oxidation state of +6 and it is a stronger acid than H2SO3 in which sulfur has a +4 oxidation state. f) What other factors affect acid strength? (carboxylic acids, small highly charged metal ions) As more electronegative elements are attached to a carboxylic acid, the stronger the acid. The smaller a metal size and as the positive charge increases, the stronger the acid. 35 When would a soluble salt be basic? neutral? acidic? or unsure without knowing Kb or Ka values? Write the molecular equation, complete ionic equation, and net ionic equation for the following salts, indicate whether each salt is: basic, acidic, neutral, or unsure. NaCl, NH 4 NO 3, KF, CH 3 NH 3 NO 2 A salt that hydrolyzes to make a strong acid and a strong base is neutral an example is NaCl + H2O <==> NaOH + HCl Na +1 + Cl-1 + H2O <==> Na +1 + OH- + H++ Cl-1 H2O <==> OH- + H+ A salt that hydrolyzes to make a strong acid and a weak base is acidic an example is NH4NO 3 <==> NH3 + HNO 3 NH4 +1 + NO 3-1 <==> NH3 + H++ NO 3-1 NH4 +1 <==> NH3 + H+ A salt that hydrolyzes to make a weak acid and a strong base is basic an example is KF + H2O <==> KOH + HF K +1 + F- + H2O <==> K +1 + OH- + HF F- + H2O <==> OH- + HF
A salt that hydrolyzes to make a weak acid and a weak base is uncertain in acidity or basicity unless more information such as Kb and Ka values are known CH 3 NH 3 NO 2 + H 2 O CH 3 NH 2 + HNO 2 K b = 4.4 x 10-4 K a = 4.5 x 10-4 CH 3 NH +1-1 3 + NO 2 + H 2 O CH 3 NH 2 + HNO 2 This salt is neutral since the Ka is approximately equal to Kb. 36 What is the concentration of hydroxide ions in 0.25M Ba(OH) 2? 0.50 M [OH -1 ] 37 Water will autoionize, what does that mean? Kw is the equilibrium constant for water Kw = [H+][OH-] = 1 x 10-14 = KaKb at 25 C. ph+poh = 14. What is the ph range for acid? neutral? and base solutions? Autoionize means that an ionic substance will spontaneously ionize until it reaches its equilibrium. ph < 7 in acidic solution ; ph = 7 in neutral solution; ph > 7 in basic solution 38 For 0.020 M HCl calculate [H+], ph, poh, [OH-] [H+] = 0.020 M; ph = 1.70 ; poh = 12.30 ; [OH-] = 5.0 x 10-13 M 39 For cyanic acid write the ionization equation. What is the [HCN], [CN-], [H+], ph, % ionization in a 0.040 M HCN solution. Ka of HCN = 4.9 x 10-10 [HCN] = 0.040 M ; [CN-] = 4.4 x 10-6 M ; [H+] = 4.4x 10-6 M ; ph = 5.35 ; % ionized = 0.011% 40 Given that the ph of a NH3 solution is 10.95. Calculate the molarity of NH3 in this solution. Kb = 1.8 x 10-5 [NH3] = 4.5 x 10-2 M 41 Calculate the concentrations of [H+], [OH-], [H2S], [HS-], and [S-2] in 1.00 M H2S solution. Ka1 of H2S = 9.5 x 10-8, Ka2 of HS- = 1 x 10-19 [H+] = 3.1 x 10-4 M ; [OH-] = 3.2 x 10-11 M ; [H2S] = 1.00 M ; [HS-] = 3.1 x 10-4 M ; [S -2 ] = 1 x 10-19 M 42 What is the hydrolysis constant, Kb of CN- if Ka of HCN = 4.9 x 10-10. Calculate the [OH-], poh, ph and % hydrolysis of CN- in a solution that contains 0.22 M KCN. Kb for CN -1 = 2.0 x 10-5 ; [OH-1] = 2.1 x 10-3 M; poh = 2.67; ph = 11.33; % hydrolysis = 0.95 % 43 Look up the appropriate Ka and Kb constants. Would NH4CN be acidic, neutral, or basic? Basic