UNIT 2: KINETICS RATES of Chemical Reactions (TEXT: Chap 13-pg 573)

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1 UNIT 2: KINETICS RATES of Chemical Reactions (TEXT: Chap 13-pg 573) UNIT 2: LAB 1. A Brief Introductory Kinetics Investigation A) Set up 4 test tubes containing about 5 ml of 0.1 M sodium oxalate sol n. Acidify each by adding 1 dropperful of 2 M H 2 SO 4. Add a "pinch" of manganous sulfate ( MnSO 4 ) to tubes # 1 and # 3 1) Remove tubes 1 and 2 from the rack and simultaneous add 1 dropperful of 0.1 M potassium permanganate (KMnO 4 ). Invert once to mix and note the time it takes for the reaction to occur. time for tube # 1 = s; time for tube # 2 = s 2) Remove tubes 3 & 4 from the rack and place in a warm water bath (~50 C ) for 3-4 min. Remove and simultaneous add 1 dropperful of 0.1 M KMnO 4. Invert once to mix and replace in the water bath and note the time it takes for the rx to occur. time for tube # 3 = s; time for tube # 4 = s B) Arrange another 5 test tubes in your rack. Into # 1 and # 2 place 5 ml of 6 M HCl, into #3 place 5 ml of 6 M CH 3 COOH, into # 4 place 5 ml of 1.2 M HCl and into # 5 place 5 ml of 0.12 M HCl. Now add 2-3 small pieces of limestone (CaCO 3 ) into #2 thru 5 and a small scoop of powdered limestone into # 1 and rank the relative rates of the reactions. (1st, 2nd etc) Conclusion: List 5 factors which affect the rates of chemical reactions UNIT 2: LAB 2. Quantitative Relationship between Concentration and Rx Rates - Rx rate law expressions The reaction we will investigate is the IODINE CLOCK REACTION IO 3 (sol n A) + 3 HSO 3 (sol n B) I SO H then 5 I + 6 H + + IO 3 3 I 2 (turns starch blue) + 3 H 2 O We will determine the rate at which the "HSO 3 " is used up. When all the HSO3 is used up the I will react with the excess IO 3 producing I2 which will turn the starch a bluish colour. (Rx 2) * [ Sol'n A =4.3 g KIO 3 / L [ IO 3 ] = 0.02 M; Sol'n B = 0.2 g NaHSO3 /L [ HSO 3 ] = 1.92 x 10 3 M] Part 1: Effect on the rate of changing [IO 3 ] ** Keep the [solution B] constant as well as the temp but change [IO ] according to the data chart below. Measure "t" 3 (appearance of blue colour ) and calculate "Rate" in moles / s *( Rate = [HSO 3 ] / t ) Tube # 1 # ml 0.02 M IO # ml of distilled water Tube #2 #ml of sol n B [ IO 3 ] your reaction time (s) average reaction time (s) total # moles HSO 3 used in rx 1.92 x 10 5 held constant REACTION RATE moles HSO 3 used /second From the data above plot the following graphs: I) [IO 3 ] vs time (a curve) & II) rate vs [IO3 ] (straight line)

2 Factors that affect the rate of a reaction (13.2 pg 574) 2:1. ATTEMPT QUESTIONS a) 13.5 b) 13.9 c) d) e) f) & g) on Page 616 2:2. Aluminum metal is used on the exterior of buildings and doors etc. yet a bottle of powdered aluminum in our stock room is labeled "danger flammable" EXPLAIN 2:3. Discuss some advantages of catalyzing a reaction to speed it up rather than increasing the temperature. Include an example. 2:4. ATTEMPT QUESTIONS a) b) c) on Page 618 Concentration & RATE, Rate Laws & Reaction Mechanisms 2:5. ATTEMPT QUESTIONS a) b) c) d) on Page :6. ATTEMPT QUESTIONS a) b) on Page 619 2:7. A reaction proceeds with the following 2 step mechanism: I) 2 A A 2 (slow) a) Identify any activated complexes? II) A 2 + B A 2 B (fast ) b) Identify any catalysts? c) Write the eq'n for the overall rx d) Write the Rate Law expression e) What is the effect on the rate if i) [A] is tripled ii) [B] is doubled iii) [A] & [B] are both quadroupled 2:8. ATTEMPT QUESTIONS a) b) c) on Page 617 2:9. Another reaction proceeds by the following 3 step mechanism I) A + C AC (fast) II) AC + B ABC (fast) a) Write the eq'n for the overall rx III) ABC + B AB 2 + C (slow) b) Write the Rate Law expression c) What is the effect on the rate if i) [A] x 5 ii) [B] x 0.5 iii) [A] x 0.5, [B] x 4, [C] x :10. A reaction occurs between A, B, and C. i) doubling [A] doubles the rate ii) doubling [B] has no effect iii) doubling [C] increases the rate x 4 * Write the Rate Law Expression for this rx. 2:11 ATTEMPT QUESTIONS a) b) c) d) e) on Page A reaction occurs between P, Q, & R. Experiments to measure the rate produced the following results: Initial [P] Initial [Q] Initial [R] RATE( mol/l.s) x x x x 10 3 * Write the rate law expression for this reaction and calculate the rate law constant. 2:13. A reaction occurs between X, Y, & Z. The rate law expression is RATE = k [X] 2 [Z] 1/2. * Calculate numerical values for the rates (a), (b) & (c) in the chart below. Initial [X] Initial [Y] Initial [Z] RATE ( mol/l.s ) x x a b c 2:14. The reaction between nitric oxide and oxygen takes place in two steps: I) NO + O 2 NO 3 a) Write the rate law expression you would expect for this reaction if II) NO 3 + NO 2 NO 2 i) the first step (I) is slower & ii) the second step (II) is slower b) Design a rate studies experimental procedure that would verify which step in the above mechanism is the slowest!

3 2:15. In the stratosphere there is a very small concentration of oxygen atoms (O). There is also a fairly large amount of ozone (O 3 ), which absorbs all but a fraction of the u-v radiation from the sun. Nitric oxide introduced into the stratosphere by man reacts with the ozone in a 2-step rx I) O 3 + NO NO 2 + O 2 (fast) II) NO 2 + O NO + O 2 (slow) a) Write the equation for the overall reaction b) Explain how it is that a small amount of NO can decompose a large amount of O 3 c) Write the rate law expression for this reaction 2:16. In basic sol'ns, alkyl bromides (alkyl =R = a hydrocarbon group) are hydrolyzed into alcohols (R-OH) thusly RBr + OH ROH + Br ( R represents any alkyl group ) Two mechanisms below have been proposed for this rx: 1) RBr R + + Br (slow) 2) RBr + OH ( BrROH) (slow) R + + OH ROH (fast) (BrROH) ROH + Br (fast) a) When methyl bromide (CH 3 Br) is the reactant the rate varies with [ CH 3 Br] & [OH ]. i) Which mechanism is this rate consistant with? ii) What would be the formula of the activated complex? b) When tertiary butyl bromide ((CH 3 ) 3 CBr) is used the rate varies with the concentration of it only. i) Which mechanism is this rate consistant with? ii) What would be the formula of the activated complex? 2:17. Under certain conditions this rx NO 2 (g) + CO (g) CO 2 (g) + NO (g) is found to follow the rate : RATE = k[no 2 ] 2 Work up a mechanism that is consistant with this data. Include the RDS. Concentration & time (pg 587) 2:18 ATTEMPT QUESTIONS a) b) 75 c) 76 d) 77 *e) 78 f) 80 g) 81 h) 84 on Page 620 Temperature effects, Maxwell Boltzmann Curves, Energy Profile Diagrams (pg ) 2:19 Which of the letters on the Maxwell-Boltzmann curve and total energy profile diagram label the following. a) The rx co-ordinate axis =. a) The rx co-ordinate axis b) b) The The kinetic kinetic energy energy axis axis =. c) c) The Potential potential energy energy axis axis =..... d) d) The % molecules % molecules axis axis =. e) e) The activation energy =. f) The ²H for H the for reaction the rx =. A axis G E F B axis C axis L K J D axis H Three(3) different reactions are done at 300 C and then at 600 C. The Maxwell-Boltzmann curves are shown at the right a) Which reaction(s) will be very 300 C. b) Which reaction(s) will be very 600 C. c) Which reaction(s) will be 300 C. d) Which reaction(s) will be 600 C. e) What could you do to increase the rate of rx #3 300 C 600 C Ea (rx#1) Ea (rx#2) Ea (rx#3) KE

4 2.21 These 2 Energy Profile diagrams represent the combustion in air of 2 elements. a) Which would be the best Fuel?. b) Which has the lower ignition temperature?. c) Which one is carbon?. Which is phosphorous... Explain why! PE A B 2.22 a) Is the forward rx endo or exothermic.. PE (kj/mol) b) What is H rx forward?. With the catalyst? c) Is the reverse rx endo or exothermic.. d) What is H rx reverse?. With the catalyst? e) What is Ea (forward) With the catalyst?.... f) What is Ea (reverse) With the catalyst?.... g) Which reaction (forward or reverse) do you think would be -40 faster?.. Why? On the diagrams below, indicate any change by B). On the diagrams below, indicate any change raising the temperatureof the system. by adding a catalyst to the system. PE PE KE KE 2:24 a) On the 1 st set of axes below, draw a PE diagram showing b) On the 2 nd set of axes below draw a PE diagram for a reaction a reaction with a Hrx=-150kJ/Ea(no catalyst)=120kj/ with a Hrx=50kJ/Ea(no catalyst)=200kj/ea (catalyst)=60kj Ea (catalyst)=80kj - Be sure it is fully labelled 2.25 A given reaction mechanism below is known to have 3 separate steps: i) A + B 2 AB 2 Ea = 4.0 kj H 1 = 10.0 kj ii) AB 2 + C 2 ABC + BC Ea = 30.0 kj H 2 = +5.2 kj iii) ABC + B 2 AB 2 + BC Ea = 10.0 kj H 3 = 7.8 kj a) Write the overall equation for the reaction. Include Hrx b) Draw a complete energy profile diagram for the entire rx. Roughly draw it to scale and label it fully with all Ea's and H's. c) Which step is the rate-determining step? Why? d) Write the rate law expression for the overall reaction. Calculations involving the Activation Energy (pg ) ATTEMPT QUESTIONS **a) *b) *c) on Page 621

5 A Rate Law Studies Lab: DETERMINATION OF THE ORDER OF A REACTION Introduction: In this experiment you will measure the time required for magnesium metal to react with hydrochloric acid solutions of various concentrations. From these data, you can calculate the rate of reaction for each experiment and graphically determine the order of reaction with respect to the hydrogen ion (from the hydrochloric acid). You will clean a 28 cm length of magnesium ribbon (to remove any oxide) and then cut it into 7 equal pieces, 4 cm long. Since the ribbon is fairly uniform, the length is proportional to the surface area, and hence the area for reaction will be the same for each piece ie the concentration of the Mg is constant for each trial. The variations in rate then will be solely attributable to the variations in acid concentration. You will make 7 aliquots of HCl, 50 ml each of 0.5 M, 1.0 M, 1.5 M, 2.0 M, 2.5 M, 3.0 M, 3.5 M using 6.0 M stock HCl solution. The reaction involved is Mg (s) + 2 H + (aq) H 2 (g) + Mg 2+ (aq) ( the Cl is a spectator ion only) The general Rate Law expression for this rx is rate = k [H + ] n *where n is the order of the reaction **the rate can be expressed in terms of reciprocal time(ie rate = 1/t) as [Mg] is constant Then if the natural log of both sides is taken ln (1/t) = n * ln [H+] + ln k which now has the form of a linear equation y = m x + b * where the slope(m) is equivalent to the order of the reaction n **and the y-intercept is the natural log of the rate law constant k. Procedure: 1. Prepare the 7 aliquots of HCl as described in the introduction using your dilution equation - [stock sol n] * # ml stock sol n = [diluted sol n] * # ml dil sol n Use graduated cylinders and pipets for accuracy. 2. Prepare the 6 pieces of magnesium metal. 3. Drop the pc of Mg into the acid solution and time the reaction carefully. Be sure the metal does not stick to the sides of the beaker. Repeat with the other 6 sol ns. Trial [HCl] ln e [H + ] time 1/time ln e (1/t) Rate (use the rate expression) M M M M M M M Plot the graph then determine the order of the reaction as well as the rate law constant k then complete the Rate Law Expression. Now use the Rate Law Expression to complete the chart above ( show all your work)

6 A Rate Law Studies Lab-2: Graphical Determination of a Complex Rate Law Introduction: The rate of a chemical reaction can be measured in terms of the rate of disappearance of one of the reactants or in terms of the rate of the appearance of one of the products. For example in the hypothetical reaction, A + 2B AB 2 the rate could be measured in terms of the disappearance of either reactant A or B, or in terms of the appearance of the product AB 2. The rate is nearly always proportional to each reactants concentration raised to some power, where the power is the order of that reactant in the Rate Law expression For the example above this means we would write the following relationship: rate = k*[a] p *[B] q This expression is called the Rate Law expression for the reaction. The value of k, called the rate constant, is unique for a specific reaction and is dependent only on the temperature. A study of the kinetics of any reaction involves determining the values of k, p, and q. In this experiment you will determine the Rate Law expression for the reaction of potassium persulfate, K 2 S 2 O 8, with potassium iodide. The reaction is: S 2 O 8 + 2I 2SO 4 + I2. 2 p q The Rate Law expression for this reaction will be rate = k * [ S 2 O 8 ] * [I ] In logarithmic form this equation becomes: ln e (rate) = ln e (k) + p*ln[s 2 O 8 2 ] + q*ln[i ] In order to determine the value of p you will measure the rate of a series of reactions for a constant iodide ion concentration as the persulfate ion concentration is changed and plot : ln(rate) vs ln[s 2 O 8 2 ]. The value of p is the slope of this plot because at a constant iodide ion concentration the logarithmic form of the rate 2 equation reduces to: ln(rate)= p*ln[s 2 O 8 ] + C ; where C=k+q*ln[I ]. As can be seen this equation now takes the form of the algebraic " slope intercept equation" (Y=mx + b) where p and m are equivalent. The value of q can be determined in a similar manner. The reaction rate constant (k) will be determined by substituting the values of p, and q back into the rate equation, calculating k for each trial and averaging the results. The rate of the reaction will be the time it takes to produce a given quantity, 0.1 mmols, of iodine. To aid in determining when 0.1 mmols of iodine has been produced we will react the iodine produced with sodium thiosulfate, Na 2 S 2 O 3 This reaction is: 2S 2 O I2 ----> 2I 1- + S 4 O 6 2- The reaction of Na 2 S 2 O 3 with I 2 is many orders of magnitude faster than the reaction we are studying so its presence will have no effect on our investigation. Exactly 0.2 mmols of Na 2 S 2 O 3 will be added to the reaction mixture for each trial. Therefore, excess iodine will begin to accumulate, and react with a starch indicator, only after all of the sodium thiosulfate has reacted. Since 0.2 mmols Na 2 S 2 O 3 will consume 0.1 mmols of I 2, the starch indicator will turn blue when 0.1 mmols of iodine have been produced by the reaction we are studying. The rate of the reaction can be found by dividing 0.1 mmols iodine by the time in seconds it takes for the blue color to appear.

7 Procedure The table below summarizes the preparation of the test solutions. Measure the volumes of each as accurately as possible using either a pipet or a graduated cylinders. All volumes are in ml. Test Solution A Solution B time rate ln e k # water 0.3 M KI starch 0.02M Na 2 S 2 O 3 [I ] in ln e [I ] 0.1 M K 2 S 2 O 8 [S 2 O 8 2 ] in ln e [S 2 O 8 2 ] (s) mmols/s rate Prepare solution A in a 250 ml beaker or 250 ml Erlenmeyer flask. Stir the solution thoroughly and record its temperature. Prepare sol n B in a separate beaker. 2. The reaction begins when solution B is poured into solution A, therefore, be prepared to start timing the reaction in seconds with a watch or clock. Place the reaction vessel on a white sheet of paper so that the color change is more easily detected. Calculations 1. Determine the initial concentration, after mixing, of both persulfate and iodide ions for all trials using [sol n ] = n/l or [I ] = [I ] * (# L used) / total # L. 2. Determine the rate for each trial in mmols iodine/sec. 3. Using trials 1,2, & 3 plot ln e (rate) vs. ln e ([ I ]). Determine q from this plot. 4. Using trials 3, 4, & 5 plot ln e (rate) vs. ln e ([S 2 O 8 2- ]). Determine p from this plot. 5. Using the p and q determined in steps 3 and 4 and the measured rate, substitute into the rate equation for each trial and determine k. Calculate the average k. 6. Write the rate law for the reaction investigated in this experiment with the determined values of p, q, and k. The rate equation for this reaction will be: rate =...*[S 2 O 8 2- ]... * [I 1- ] Hand-in your graphs as well as the chart above with all calculations shown

8 A Rate Law Studies Lab-2: Determining the Energy of Activation Discussion The activation energy of a reaction is the amount of energy needed to start the reaction. It represents the minimum energy needed to form an activated complex during a collision between reactants. In slow reactions the fraction of molecules in the system moving fast enough to form an activated complex when a collision occurs is low so that most collisions do not produce a reaction. However, in a fast reaction the fraction is high so that most collisions produce a reaction. For a given reaction the rate constant, k, is related to the temperature of the system by what is known as the Arrhenius equation: k = A*e (-Ea/R*T) see page in your text (Brady & Holum) where R = the ideal gas constant [8.314 J/(mol- K)], T = the temperature in degrees Kelvin, Ea = the activation energy in joules per mol, A = a constant called the frequency factor; which is related to the fraction of collisions between reactants having the proper orientation to form an activated complex. ln e k = ln e A + (Ea/R) * (1/T) y = b + m x In this experiment you will measure the rate constant (k) for a chemical reaction, the same one investigated in " Determination of a Rate Law", at several temperatures, plot ln e k vs 1/T and determine the activation energy of the reaction from the slope. (It is left to the experimenter to determine how to calculate the activation energy.) This lab will be done after the lab entitled " Determination of a Complex Rate Law". Review the rate Law expression calculated in that lab to calculate the rate in this lab and then the value of k the rate constant at your temperatures. Then use test solution 3 as the base data then collect four data points: room temperature, one below room temperature(~7-10 C), and two above room temperature(~ 35 C & 50 C). This will require you to prepare water baths and carefully monitor the temperature of the baths and the solutions in the baths. Test Solution A Solution B temp 1/temp time rate k ln e k # water 0.3 M KI starch 0.02M Na 2 S 2 O 3 [I ] in 0.1 M K 2 S 2 O 8 [S 2 O 8 2 ]in ( C) K 1 (s) mmols/s Hand-in your graph as well as all your calculations showing how you calculated the values required to find the Activation Energy.

9 Kinetics Assignment 1. 2 A + 2 B C + D The following data about the reaction above were obtained from three experiments: Exp [A] [B] Initial Rate of Formation of C (mole. liter 1 min 1 ) x x x 10-4 (a) What is the rate Law expression for the reaction? (b) What is the numerical value of the rate constant k? What are its units? (c) Propose a reaction mechanism for this reaction. Indicate activated complexes & which step is the RDS 2. For a hypothetical chemical reaction that has the stoichiometry 2 X + Y Z, the following initial rate data were obtained. All measurements were made at the same temperature. Initial Rate of Formation of Z, (mol. L 1. sec 1 ) Initial [X] o, (mol. L 1 ) Initial [Y] o, (mol. L 1 ) 7.0x x x x (a) Give the rate law expression for this reaction from the data above. (b) Calculate the specific rate constant for this reaction and specify its units. (c) How long must the reaction proceed to produce a concentration of Z equal to 0.20 M, if the initial reaction concentrations are [X] 0 = 0.80 M, [Y] 0 = 0.60 M and [Z] 0 = 0 M? (d) Select from the mechanisms below the one most consistent with the observed data, and explain your choice. In these mechanisms M and N are reaction intermediates. (1) X + Y M (slow) (2) X + X M (fast) (3) Y M (slow) X + M Z (fast) Y + M Z (slow) M + X N (fast) N + X Z (fast) 3. Radioactive strontium-90 has a 1/2 life of 28 a. If the amount of 90 Sr remaining in a sample is only 1/6 of the amount originally present, how many years has the sample been decaying? * The following question requires you to read section 13.7-page in your text on Calculating the Activation Energy and go through examples & 13:13 4. Organic compounds that contain large proportions of nitrogen and oxygen tend to be unstable and are easily decomposed. Hexanitroethane, C 2 (NO 2 ) 6, decomposes according to this equation: C 2 (NO 2 ) 6 2NO 2 + 4NO + 2CO 2 The reaction is 1 st order with respect to C 2 (NO 2 ) 6. At 70 C, k=2.41 x 10-6 s -1 & at 100 C k=2.22 x 10-4 s -1 a) What is the 1/2 life of this reaction( in minutes & hours) of C 2 (NO 2 ) i) 70 C and ii) 100 C b) If 0.10 mol of C 2 (NO 2 ) 6 is dissolved in 1.0 L of CCl 70 C what will the concentration of C 2 (NO 2 ) 6 be in 8.33 h? c) Calculate the value of the activation energy required for this reaction in kj. d) What is this reactions rate constant at 120 C? What would its 1/2 life be at this temperature? *5. ATTEMPT QUESTION on Page 622 ( note: Use appendix E-pg A-46 to determine BP 355 torr)

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