Molarity- molarity is defined as moles of solute per liter of solution.

Transcription

1 Solubility Solubility is defined as the amount of solute that will dissolve in a given amount of solution at a certain temperature. Our main measurement of solubility is molarity. Molarity- molarity is defined as moles of solute per liter of solution. Molarity (M) = moles solute L solution Example What is the molarity of 1 liter of a solution that contains 0.5 moles of NaCl. Answer To solve we divide the number of moles by the liters of solution. Example Molarity (M) = 0.5 moles 1 L = 0.5 M What is the molarity of a solution that contains grams of NaCl in 2 L of solution. Answer The first thing we must do is determine how many moles of solute g 1 mole 58.4 g = 1 mole We then divide the number of moles by the volume of the solution. 1 mole 2 L = 0.5 M Exercise 1 What is the molarity of a solution made by dissolving 4.2 grams of CaCl2 in enough water to make 125 ml of solution? 0.30 M 2. What is the molarity of the chloride ion in the above solution 0.60 M 72

2 Equilibrium involving insoluble compounds When certain salts are placed in water very little of the salt dissolves to come to equilibrium. A small amont of solid dissovles and then dissociates into ions. The following is a general equilbrium equation for this process. MxAy(s) < > x M +y (aq) + y A -x (aq) Example Write the equilbrium equation for the addition of solid AgBr to water. AgBr(s) Ag + (aq) + Br- (aq) Exercise Write the equilibrium equations for the following compounds. PbI2 PbI 2 (s) Pb 2+ (aq) + 2I- (aq) Ag3PO4 Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3- (aq) PbCO3 PbCO 3 (s) Pb 2+ (aq) + CO 3 2- (aq) Al(OH)3 Al(OH) 3 (s) Al 3+ (aq) + 3 OH- (aq) 73

3 Ksp calculations. Ksp is another example of an equilibrium constant. It can be determined like any other equilibrium constant if the concentration of the products are known. We only need to know the product concentrations because the reactants are solids. Example AgCl(s) is an insoluble compound and when placed in water the chloride concentration was found to be 1.26 x 10-4 M. What is the Ksp of AgCl(s)? Answer AgCl (s) Ag + (aq) + Cl- (aq) [Ag + ] = [Cl - ] = 1.26 x 10-4 K sp = [Ag + ] [Cl - ] = (1.26 x 10-4 ) 2 = 1.58 x 10-8 Exercise 1. What is the solubilty of a compoumd with MW of 128 if after coming to equilibrium we find 0.23 g/l in the solution. 1.8 x 10-3 M 2. A saturated solution of Al(OH)3 had a ph = 9.0 what is the solubility of the compound at this temperature. 3.3 x A solution of PbCl2 has a chloride concentration of 1.6 x 10-2 M. What is the Ksp of Lead(II)chloride? 2.0 x

4 Determination of solubility given the Ksp When excess solid is placed in water an equilibrium reaction takes place in which the solid dissociates into ions. Solid (s) Cation (aq) + Anion (aq) The following equilibrium is what would happen when excess solid AgCl is added to water. AgCl (s) Ag + (aq) + Cl- (aq) The solubility of the silver chloride is equal to the concentration of the silver cation or the chloride anion. The solubility is defined as the concentration of either the silver cation or the chloride anion. Solubility(s) = [Ag + ] = [Cl - ] If more than two ions are produced the solubility of the compound must take into account the multple ions. C x A y(s) x C + (aq) + ya- (aq) Solubility (s) = (1/x[)C + ] = (1/y)[A - ] The solubility of lead(ii) chloride is defined as the following. PbCl 2(s) Pb 2+ (aq) + 2Cl- (aq) Solubility(s) = [Pb 2+ ] = 1/2 [Cl - ] We calculate the concentration of the aqueous species just as we would with any other equilibrium using an ICE table and the equilibrium expression. 75

5 Example What is the solubility of BaCO3? Ksp = 5.1 x 10-9 Answer We set up the ice table and plug the values into the equilibrium expression. Note that we us s instead of x, but everything else is the same as before. BaCO 3(s) Ba 2+ (aq) + 2+ CO 3 (aq). initial N 0 0 change NA +s + s equilibrium NA s s [Ba 2+ ][CO3 2- ]= s 2 = Ksp = 5.1 x 10-9 s = 7.1 x 10-5 Exercise Determine the solubility of the following compounds using the Ksp values. CaSO4 Ksp = 9.1 x x 10-3 Al(OH)3 Ksp = 9.1 x x 10-2 HgCl 2 Ksp = 9.1 x x 10-2 CaCO3 Ksp = 9.1 x x

6 Common Ion Effect. (LeChatlier s) When a solid is allowed to come to equilibrium in a solution of one of it constituent ions we call this the common ion effect. As we would expect from LeChatlier s principle addition of a product ( the common ion) would shift the equilibrim to the reactant side. An example would be letting solid silver chloride come to equilibrium in a solution of sodium chloride as seen in the following example. AgCl (s) Ag + (aq) + Cl- (aq) initial NA 0 1M change NA +s + s equilibrium NA s 1M + s The solution has a chloride ion which is in common with the chloride from the silver chloride. The net effect is to decrease the solulbility of the silver chloride which is also what we would predict from LeChatlier s principle. We solve for the solubility, using the techniques we learned in the earlier equilibrium section. Example What is the solubility of PbCl2 in 0.1M solution of sodium chloride? PbCl 2(s) Pb 2+ (aq) + 2Cl - (aq) initial NA 0 0.1M change NA +s + s equilibrium NA s 0.1M + s We substitute the variables into the equilibrium expression and solve for s. K sp = [Pb +2 ] [Cl - ] 2 = [s] [0.1 + s]2 = 1.58 x 10-8 assume s small s = 1.58 x 10-7 the asumption that s is small was correct Precipitation When two solutions of ions are combined that make an insoluble precipitate it is usefull to be able to predict if a precipitate will form. To do this we calculate Q as we did earlier and compare it to Ksp. The following relationships are used to predict precipitation. 77

7 Q > K precipitate will form Q = K no precipitate saturated solution Q < K no precipitate unsaturated solution Example 25 ml of 2.3 x 10-6 M Pb(NO3)2 was combnined with 25 ml of 1M NaCl. Will a precipitate form? Ksp = Answer Write the solubility equilibrium for the insoluble compund. PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq). Ksp = 1.6 x 10-5 Write the Qsp expression. Q sp = [Pb +2 ] [Cl - ] 2 Calculate the concentration of the Ions in the solubility expression using the diultion equation. M1V1 = M2V2 [Pb 2+ ] = [(2.3 x10-6 M) (25 ml)]/ 50 ml = 1.15 x 10-6 [Cl - ] = [(1M) (25 ml)]/ 50 ml = 0.5M Plug the diluted values into the Q expression. Q sp = [Pb +2 ] [Cl - ] 2 = [(1.5 x 10-6 ) (0.5) = 7.5 x 10-7 Compare Qsp to Ksp Qsp < Ksp so a precipitate will not form. 78

8 Exercise Determine if a precipitate will form when the following solutions are mixed together. Assume all concentrations are 1M ml NaCl and 50 ml Pb(NO3)2 PbCl2 Ksp = 1.6 x 10-5 ppt will form ml NaOH and 5 ml Mg(NO3)2 MgOH2 Ksp = 1.8 x ppt will form ml AgNO3 and 20 ml Na2(CO3) Ag2CO3 Ksp = 8.1 x ppt will form ml Na2(CO3) and 50 ml PbCO3 PbCl2 Ksp = 1.6 x 10-5 ppt will form 79

9 Seperation of Ions We are going to seperate solutions of metal ions based on the differnce in solubility. There are two types of seperations that will interest us. They are seperation of an ion to a certain concentration and maximum seperation of an ion. This is done by adding an anion that will cause the metal cations to precipitate. The less soluble cation will precipitate first and the more soluble cation will remain in solution. If the concentration of anion is too large the more soluble ion will precipitate and we will not be able to seperate the cations. The following are the rules we use for the seperation of the metal cations. Ion seperation 1. Determine which ion is the most souble by calculating solubility for each ion. 2. Find the concentration of anion that will make Qsp = Ksp for the most soluble ion. This will be the maximum amount of anion that can be added to give us a maximum seperation of the less soluble compound. Example What concentration of chloride will give the maximum seperation of a 1M solution of Ag + and 1M Pb 2+? AgCl Ksp = 1.8 x PbCl2 Ksp = 1.6 x 10-5 Answer We first find that the AgCl is the least soluble and will be the one we can seperate. We then find the concentration of Cl- that will give Q=Ksp for the PbCl2. [Pb + ] [Cl - ] 2 = 1.6 x 10-5 [1M] [Cl - ] 2 = 1.6 x 10-5 [Cl-] = 4.0 x 10-3 this is the maximum amount that can be added before the lead will precipitate. We can check and see how much silver will be left after precipitation. [Ag + ] [Cl - ] = 1.8 x we can use the chloride values calculated above and solve for the silver concentration remaining. Ag + = 1.8 x Cl - = 1.8 x x 10-3 = 4.5 x 10-8 That is the amount of Ag + that is left along with all the lead cation. 80

10 Exercise 1. What is the concentration of hydroxide that will give the maximum seperation of 0.5M Fe 3+ and 0.5M Cu 2+? Fe(OH)3 Ksp = 1.0 x Cu(OH)2 Ksp = 2.2 x [OH-] = 2.1 x What is the concentration of sulfate that will give the maximum seperation of 0.1M Ba 2+ and 0.1M Ag +? BaSO4 Ksp = 1.1 x Ag2SO4 Ksp = 1.4 x 10-5 [SO4 2- ] = 1.4 x 10-3 Percent Seperation If we want less than maximum seperation we can determine the amount of the less soluble cation and then calculate the amount of anion required to do the seperation. To determine the concetration of anion needed to give a certain seperation we do the following. i. Determine the concentration of the less soluble cation that will remain after the desired precipitation is completed. 81

11 ii. Determine the concentration of anion that will allow this concentration to remain in solution. Qsp = Ksp iii. Make sure that this concentration of anion is less than the concentration of anion that will give the maximum seperation. Example What concentration of chloride will give a 90% 1M solution of Ag + and 1M Pb 2+? AgCl Ksp = 1.8 x PbCl2 Ksp = 1.6 x Determine concetration of Ag + that will remain after the seperation x 1M = 0.9 M Amount remaining = = Determine concentration of chloride that will give a final silver ion concentration of 0.1. [Ag + ] [Cl - ] = 1.8 x [0.1M] [Cl - ] = 1.8 x [Cl-] = 1.8 x 10-9 this is the maximum amount that can be added to leave the silver at a concentration of 0.1M 3. We will now check and make sure that the lead will not precipitate. Qsp < Ksp Exercise Q = [Pb + ] [Cl - ] 2 = [1M] [1.89 x 10-9 ] 2 = 3.6 x x < Ksp = 1.6 x 10-5 so no precipitate 1. What concentration of sulfate will give a 80 % seperation of a 1M solution of Ba 2+ from a 1M Solution of Ag +. Ag2SO4 Ksp = 1.4 x 10-5 BaSO4 Ksp = 1.1 x [SO42-] = 5.5 x

12 2. What concentration of carbonate will give a 99 % seperation of a 1M solution of Mg 2+ from a 1M Solution of Mn 2+. MgCO3 Ksp = 1.4 x 10-5 MnCO3 Ksp = 1.1 x [CO3 2- ] =1.4 x What concentration of sulfate will give a 80 % seperation of a 1M solution of Ba 2+ from a 1M Solution of Ag +. Ag2SO4 Ksp = 1.4 x 10-5 BaSO4 Ksp = 1.1 x [SO4 2- ] = 5.5 x

13 Kinetics Kinetics- The study of rates of reactions. What we are going to study is the rate of chemical reactions and things that are experimentally and theoretically related to rates. Rate- a rate is defined as an experimentally determined positive quantity, that relates the change in concentration, of a reactant, or product as a function of time. This is represented mathematically below. Rate of change of a reactant rate = - 1 change in concentration of reactant ( n change in time Example ) = - 1 final concentration - initial concentration ( n final time - initial time In the following chemical reaction the initial concentration of CH3Br was found to be 1M if after 1.0 min the concentration of CH3Br is 0.45 M what is the rate. CH3Br + CH3OH > CH3OCH3 + HBr rate = - 1 final concentration - initial concentration n final time - initial time = - (0.45M - 1.0M) 1 min - 0 min = 0.55 M min This will give us a average rate because we will find that often the rate will change as the reaction proceeds. Also you might observe that the negative sign in front of the equation makes the value positive, as required by the definitition of rate. The rate is the experimentally determined change in concentration versus the change in time. ) Rate of change of a product rate = 1 n (change in concentration of product change in time ) = 1 final concentration - initial concentration ( n final time - initial time ) Notice that it is not necessary to put a negative sign in front, because the quantity will be positive. Example In the reaction above after 2 minutes the final concentration of HBr was found to be 0.7M what is the rate. rate = 1 concentration - initial concentration (final n final time - initial time ) = 0.7M - OM 2 min - 0 min = 0.35 M min 84

14 As was stated before the rates determined by these methods are the average rates. We are often interested in the instantaneous rate which can be determined from a graph or by a computer program. The graphical method is shown below. Procedure for determining instantaneous rates. Step 1 An experiment is done in which the change in concentration of the species of interest is followed as a function of time. We look at the concentration as time increases. Data table from a concentration vs time experiment for the decomposition of N2O5. N2O5(aq) > 4NO2(aq) + O2(aq) concentration N2O5 (M) time(s) Step 2 The data is graphed as illustrated on the next page. 85

15 conc. N 2 O 5 (M) Conc. N 2 O 5 vs Time Time (s) We can find the instantaneous rate by taking a tangent to the point at the desired concentration, as is listed below for two points. concentration N2O5 Rate (M/s) 0.90M 5.4 x M 2.7 x 10-4 Excercises Using the data for the decomposition of N 2 O 5 answer the following questions. 1 What is the average rate between seconds? 3.3 x 10-4 M/s 2. What is the average rate between seconds? 3.25 x 10-4 M/s 3. From examination of the instantaneous rate how does rate change with concentration? Decreases 86

16 Rates and rate laws We are interested in the experimental results but what is of greater interest is how does the rate vary as a function of concentration. We can graph our experimental data as instantaneous rate vs concentration and come up with the following sort of graph. I have not used numbers as I am discussing this in a qualitative manner. Rate vs concentration Rate (M/min) Concentration of reactant (M) This is a line and we know that the general equation of a line is the following. y = mx + b y= vertical axis x= horizontal axis m= slope of line ( change in x / change in y) b= y intercept ( where the line crosses the y axis) With this in hand we can write an equation that relates the rate of a reaction with the reactant concentration. This equation that goes with the graph above takes the following form. rate = k[reactant] [ ] = concentration brackets k= rate constant (slope of the line) 87

17 General Rate Law Not all rate vs concentration graphs are linear, and to account for this we can write a more general rate law that has the following form. rate = k [ ] m m = order of the reaction. We will only investigate reactions in which m is equal to 0, 1 or 2; which are refered to as zero, first or second order repectively. We can also write a general rate law for reactions that have fractional or negative orders, but that is a topic that would be dealt with in more advanced classes. The following graphs show how rate changes with concentration for each of the orders. 0 order 1st order 2nd order Rate Rate Rate Concentration Concentration Concentation Experimental Determination of rate laws. The rate law must be determined experimentally from data collected in the lab. Using the rate law we can theoretically determine what happen to the rate when the concentration of reactant is doubled. We can then do experiments in which the concentration of a reactant is doubled and then determine the order from the change in rate as compared to the theoretical expectation. The following is a summary of how rate change versus a doubling in concentration. Order Concentration Change in rate 0 x2 no change 1 x2 x2 2 x2 x2 88

18 Example Using the relationships above and the experimental data, determine the order of the reaction. Use the data to determine the rate constant for the reaction. N2O4(g) > 2NO2(g) Experiment Conc. N2O4(M) rate(m/s) The concentration was doubled in experiment 2 compared to experiment 1. The data shows that the rate doubled in experiment 2. This shows that the reaction is 1st order with respect to N2O4. The rate law can be written as the following. rate = k[n2o4] To determine the rate constant we can use the data from one of the experiments and plug the values into the rate law and solve for k. Exercise k = rate / [N2O4] = 0.34 / 0.12 = 2.8 s Using the experimental data in the following table write the rate law for the decomposition of Ozone. 2O3(g) > 3O2(g) Experiment Conc. O3(M) rate(m/s) Rate = k[o3] 2 k = 110 1/M s 89

19 2. For the reaction in which A decomposes to B the following data was collected. A(aq) > B(aq) Experiment Conc. A (M) rate(m/s) Determine the order, rate constant and write the rate law for the reaction. rate = k k = 0.94 M/s Determination of rate laws for reactions with 2 reactants Like before the rate law of reactions with 2 reactants must be determined by experiment. To do this we will double the concentration of one reactant and hold the other reactant constant and determine the order by the change in rate. Example The following experimental data was collected for the reaction of hydrogen and nitrogen monoxide. Determine the order for each reactant and write the rate law. Use the data to determine the rate constant. 2H2(g) + 2NO(g) > N2(g) + 2H2O(g) Experiment Conc. H2 (M) Conc. NO (M) rate(m/s)

20 Answer We will determine the order of each reactant in the manner described above. If we compare experiment 1 to experiment 2 we held the concentration of NO constant and doubled the concentration of H2, and we saw a doubling in the rate. This would indicate that the reaction is 1st order with respect to H2.. rate = k [H2] [NO] 2 We can use the data from experiment 1 to determine the rate constant for the rate law. k = rate [H 2 ][NO] = 0.10 M/S [0.1M][0.1M] = 2 100M-2 S -1 Exercise 1. In a solution at constant H + concentration, I - reacts with H2O2 to produce I2 and water. 2H + (aq) + 2I - (aq) + H2O2(aq) -----> I2(aq) + 2H2O(l) Experiment Conc. I - (M) Conc. H2O2 (M) rate(m/s) Write out the rate law and solve for the rate constant. rate = k[i - ][H2O2] 2 k = 100 1/M 2 s 2. The data below are for the reaction of NO with Cl2 to form NOCl. Experiment Conc. NO (M) Conc. Cl2 (M) rate(m/s) x x x 10-3 Write out the rate law and solve for the rate constant. rate = [NO][Cl2] k = 0.4 1/M s 91

21 Integrated rate law The rate law is used to determine the rate of a reaction, for a concentration of a reactant in the rate law. As we saw earlier rate is defined as a change in concentration divided by a change in time. We can use some principles of calculus to determine a concentration at some time, given the initial concentration and the rate law. We will not show how to derive the integrated rate laws but have summerized the results in the following table. Order Rate Law Integrated rate law 0 rate = K [ ]t - [ ]o = -kt 1 rate = k[ ] ln [ ] t [ ] 0 = - kt 2 rate = k[ ]2 1 [ ] t - 1 [ ] o = kt Example A first order reaction had a rate constant of 0.12 s -1. If the initial concentration was 0.25M how long will it take to reach 0.10M. Answer To solve this reaction we will use the first order integrated rate equation, which is rearanged to solve for time with the given conditionsl. t = ln [ ] o [ ] t k = ln s-1 = We can solve for any variable we choose as long as all the other parameters are known. Exercise 1. A first order reaction required 25 s to go from an initial concentration of 1M to 0.65M. What is the rate constant for this reaction. k = 1.7 x 10-2 s The decomposition of HI in the gas phase is a second order reaction. The rate constant for the reaction was found to be 1.2 x 10-3 M -1 s -1. If the initital concentration of HI is 0.36M how much will remain after 45 min. 0.17M 92

22 3. A first order reaction had a rate constant of 3.45 x 10-6 s -1. How long will it take for a 3.2 x 10-2 M solution to go to 1.0 x 10-4 M? 1.7 x 10 6 s Half life The time required for a reaction to reach half of its original concentration is called the half life of a reaction. If we started with a 1M concentration the time it takes to get to 0.5M would be called the half life. We can determine the half life for a reaction using the integrated rate law and the definition of the half life. The following example shows the derivation of the first order half life. Example To derive the first order half life we start with the integrated rate law and use the relationship 1/2[ ]o = [ ]t at t1/2. ln [ ] o [ ] t 1/2 = kt 1/2 t 1/2 = ln [ ] 0 1/2[ ] 0 k ln 2 k = k Notice that the first order half life is independent of concentration. Summary of half life equations Order Equation 0 t 1/2 = [ ] 0 2k 1 t 1/2 = k 2 t 1/2 = 1 k[ ] 0 93

23 Example A first order reaction had a k = 4.6 x 10-2 s -1 what is the half life for this reaction. Answer Exercise t 1/2 = k = x = 15 s s 1. If a 2nd order reaction had a half life of 265 s what is k for this reaction. k = / s [ ]o 2. The rate constant for the 1st order decomposition of N2O5 was found to be 0.35 min -1. How long will it take for a 0.80M concentration of N2O5 to decompose to 0.20M? 4.0 min 3 A 0 order reaction had a half life of 3.7 hours using the integrated rate law, how long it will take 0.30 atm to react to leave 0.2 atm of the reactant left. ( calculate k from the half life) 94

24 Linear plots The integrated rate laws can be used to determine the order of a reaction graphically by writting the rate law in a linear form. When the data is graphed the linearity of the graph will tell us the order. The following is the linear form for the 0 order reaction. [ ] 0 - [ ] t = kt Rearranging the equation into standard y = mx +b linear form gives the following equation. [ ] t = - kt + [ ] 0 y is the concentration at some time t and x is time t. The slope is the rate constant, and the intercept is the initial concentration. This means that if we graphed our data with the concentration vs time and the graph was liner it would be indicative of a 0 order reaction. [ ] If the data gave us this graph it would be a 0 order reaction, time The following is a summary of facts relating to rate laws, including what type of graph will give a linear plot. Order Rate law Integrated rate law t1/2 linear graph 0 rate = k [ ]o - [ ]t = kt t 1/2 = [ ] 0 2k [ ] vs time 1 rate = k[ ] ln [ ] o = kt t [ ] 1/2 = t k ln [ ] vs time 2 rate = k[ ] 2 1 [ ] t - 1 [ ] o = kt t 1/2 = 1 1 k[ ] 0 [ ] vs time 95

25 Exercise Using the data from the following table. graph the data and determine the order from the linear graph. Use the slope of the linear graph to detemine the rate constant and then write the complete rate law for the decomposition of N2O5. Data table from a concentration vs time experiment for the decomposition of N2O5. N2O5(aq) > 4NO2(aq) + O2(aq) concentration N2O5 (M) time(s) st order Activation energy Reactions involve the breaking and making of chemical bonds. The breaking of bonds require the input of energy which is provided by the collision of molecules. For the reaction to happen, a certain minimum amount of collision energy is required for the reaction to proceed. This minimum energy is called the activation energy (Ea). The following are some characteristics of activation energy. 1. Ea is a positive quantity. 2. Ea is independent of temperature and concentration. 3. Ea is dependent on the type of reaction. For any given reaction we find that activation energy will always be the same and will be greater for a slow reaction and smaller for a fast reaction. 96

26 Reaction Rate and Temperature Experimentally we find that reaction rate increases with temperature, and as a general and approximate rule for every 10 o C increase in temperature the rate will double. The following equation relates k and temperature. k = Ae -E a RT R = J/ Mole K T in K A = pre exponential factor For this equation to be useful we must know Ea and A, which must be determined by experiment. If we rearange the equation into a linear form this will allow us to use graphical methods along with experiment to determine A and Ea. The following is the the equation rearranged in linear form. ln k = -E a R 1 T + ln A If we graph the ln of k versus I/ T we can determine activation energy and the preexponential factor A. Ea will be the negative slope of the line multiplied by R, and the ln of A will be the y intercept. ln A ln k slope = -E a R 1/T 97

27 Exercise 1 The following data are for the gas-phase decomposition of acetaldehyde. 1 k M s T (K) Graph the data and determine the activation energy and the preexponential factor. 2. The following data were obtained for the reaction SiH4(g) -----> Si(s) + 2H2(g) k (s -1 ) T (K) Graph the data and determine the activation energy and the preexponential factor. We can also use the linear relationship to determine how the rate constant changes with temperature. We can with a little mathematics write the equation in the following form. Example ln k 2 k 1 = E a R 1 1 T 1 T 2 A certain reaction had an activation energy of 9.3 x 10 4 J. At 27 o C, k = 1.25 x / M s. Calculate k at 127 o C. Answer Using the equation above we can write the following. ln k x 104 = k =

28 k 2 k 1 = e 4.06 =1.15 x 10 4 k 2 =1.15x10 4 k 1 = (1.15 x10 4 )(1.25 x10-4 ) =1.44 x M s 1. A reaction had an activation energy of 6.2 x 10 3 J. k was measured to be 1.5 x 10-1 s -1 at 100C. The temperature was increased by 10 o C to 110 o C. What is the value of the rate constant at the higher temperature. ln k 2.15 = 6.2 x k2 = At 200 o C a reaction had a k= 1.2 x 10-1 s -1, and at 250 o C the k was found to be 3.8 s-1. What is the activation energy for this reaction. ln = E a Ea = 1.4 x