Transcript for:
AP Chemistry Kinetics Overview and Analysis

These are the answers to the AP Chemistry packet that covers topics 5.1 to 5.3. On this slide, you can see Unit 5 at a glance. Unit 5 is entitled Kinetics. In this video, we will take a look at topic 5.1, Reaction Rates, topic 5.2, Introduction to Rate Law, and topic 5.3, Concentration Changes Over Time. In the video description area, there is a link to the AP Chemistry course and exam description, which we call the CED for short, and there is also a link to the packet that accompanies this video. Topic 5.1, Reaction Rates. Here are some essential knowledge statements from the AP Chemistry course and exam description. The kinetics of a chemical reaction is defined as the rate at which an amount of reactants is converted to products per unit of time. The rates of change of reactant and product concentrations are determined by the stoichiometry in the balanced chemical equation. The rate of a reaction is influenced by reactant concentrations, temperature, surface area, catalysts, and other environmental factors. An experiment is performed in which a sample of N2O5 is added to a previously evacuated flask. A decomposition reaction occurs according to the chemical equation shown above. The table below represents data from the experiment as the reaction proceeds over time. Question 1. Plot the data from the table on the graph below. Draw a smooth line to connect the data points for each substance. Use a thick solid line. for N2O5, a thin solid line for NO2, and a dashed line for O2. So the first data point I'm going to use is at time equal 0 seconds, the concentration of N2O5 is 5.70 moles per liter. Then over time, that value decreases. So 4.30 moles per liter at 100 seconds, 3.10 moles per liter at 200 seconds. And then now we have 2.10, 1.30, 0.70, 0.30, and 0.10. So that line, that dark, thick, solid line is the change in concentration for N2O5. Now I'll move on to the data for the change in concentration for NO2. So 0 at time 0, and then 2.80 moles per liter, 5.20 moles per liter, 7. 0.20, 8.80, 10, 10.8, 11.2. So that is the change in concentration for NO2. And then finally the change in O2, starting at 0 and then 0.70 moles per liter at 100 seconds, 1.30 moles per liter at 200 seconds, 1.80, 2.20, 2.5, 2.7, 2.8. So that is the change in concentration for O2. Notice that the change in concentration over a certain period of time is proportional to the coefficients in the balanced chemical equation. So in this balanced chemical equation, 2 moles of N2O5 are consumed for every 4 moles of NO2 that are produced and for every 1 mole of O2 that is produced. So in this first 100 second period, the concentration of N2O5 decreases. from 5.70 moles per liter to 4.30 moles per liter. That's a change of 1.40. In that same period of time, the concentration of NO2 increases by 2.80, which is twice as much as the change in concentration for N2O5. That's consistent with the 4 moles of NO2 produced for every 2 moles of N2O5 consumed. So again, the coefficient ratios are consistent with the change in concentration. And then for O2, the change over that same period of time is 0.70 moles per liter, which is half as much as the change for N2O5. One mole of O2 is produced for every two moles of N2O5 that are consumed. All right, let's take a look at question two. Consider the reaction between NH3 and O2 that is represented by the equation shown above. At a certain point in time during the reaction, the rate of disappearance of NH3 was equal to 0.64 molarity per second. Part A, calculate the rate of disappearance of O2 at that same point in time. Include units in your answer. So we're going to use the coefficients in the balanced chemical equation to do this. We're using stoichiometry to do this calculation. So I have 0.64 moles per liter per second. and that is for NH3. Let's get rid of moles of NH3 by putting moles of NH3 on the bottom, and then we're converting into moles of O2 using the coefficients in the balanced chemical equation. So 5 moles of O2 are consumed for every 4 moles of NH3 that are consumed. And when I do this math, I will still have units of molarity per second, but now I'm switching from NH3 to O2. So it's 0.64 times 5 divided by 4, and on my calculator I get 0.80, and the units are moles per liter per second, but that would be for the disappearance of O2. So molarity per second for the rate of disappearance of O2. In part B, it's a very similar calculation, but now we're converting the given information, which is 0.64 moles of NH3 per liter per second into the rate of appearance of H2O, and that will also be molarity per second. Let's use the coefficients. Six moles of H2O are produced for every four moles of NH3 that are consumed. So 0.64 times 6 divided by 4, and when I do this math, I get 0.96. That would be moles of H2O per liter. per second. So molarity per second for the rate of appearance of H2O. So we're using stoichiometry and the coefficients in the balanced equation to convert the rate of change of one chemical into the rate of change of another chemical in that same reaction. Question three. The diagram below represents the data from an experiment in which the concentrations of both reactants and products are measured over time during the reaction. Part A. Label each curve in the diagram above as reactants or products. Justify your answer. So during the course of a chemical reaction reactants are being consumed so their concentration should decrease. Products are being formed so their concentration should increase. So I'm going to go ahead and label this curve as products. because the concentration is increasing over time. And I'll label this curve as reactants, because the concentration is decreasing over time. Now on to part B. Which of the following chemical equations is most likely to represent the reaction that occurred during this experiment? Justify your answer. So in our first reaction, it says XYZ goes to XZY. That would imply that one mole of reactant is turning into one mole of product. The second equation says 2x goes to x2. That tells us that 2 moles of reactant are changing into 1 mole of product. And the third reaction is the reverse. It says x2 goes to 2x. That tells us that 1 mole of reactant is being transformed into 2 moles of product. If we look closely at the change in concentration for the reactants and the products, we see that they are changing at different rates. So in the first 30 seconds, the reactant concentration changes from 2.8 to 2.0. During that same period of time, the first 30 seconds, the product concentration changes from 0 to 1.6. So the change in concentration for the product is twice as much as the change in concentration for the reactant. Well, if the change in concentration for the products is at a rate that is twice as much as the change in concentration for the reactants, then I'm going to go ahead and circle this reaction as my answer, that two moles of product are being produced for every one mole of reactant that is being consumed. According to collision theory, which will be discussed in topic 5.5, reactant particles must undergo successful collisions in order for products to be formed in a chemical reaction. Question four. A series of experiments is performed by combining solid magnesium with aqueous hydrochloric acid. The reaction occurs according to the equation shown above. The initial reaction rate is determined from the experimental data for each trial. In this particular question, variables are changing and we're going to take a look at concentration, temperature, and surface area. For each of the following, give an explanation that describes how the collisions between reactant particles are affected by the experimental conditions. Part A. Explain why the initial reaction rate in trial 2 is faster than the initial rate in trial 1. The change that was made from trial 1 to trial 2 is the concentration of hydrochloric acid was increased from 2.0 molar to 4.0 molar. Let's talk about the collisions between the particles when a higher concentration of reactant is used. A higher concentration of the reactant, HCl, will result in more frequent collisions between particles of Mg and HCl. The initial reaction rate in trial 2 should be faster than the initial reaction rate in trial 1. So it's greater number of collisions between reactant particles because of the higher... concentration of acid. Now on to Part B. This has to do with changing the temperature. Explain why the initial reaction rate in trial 3 is faster than the initial rate in trial 1. So the significant change here is going from a temperature of 20 degrees Celsius to a temperature of 40 degrees Celsius. So think about the behavior of the particles and how the collisions are affected. If the reaction is carried out at a higher temperature, The reactant particles should collide with each other more frequently and with more force. The initial reaction rate in trial 3 should be faster than the initial reaction rate in trial 1. So a greater number of collisions, so more frequent collisions, and with more force because of the higher kinetic energy of the particles. Now let's take a look at surface area in part C. Explain why the initial reaction rate in trial 4 is faster. than the initial rate in trial one. Now we go from having a metal strip of magnesium to having metal powder. So we're changing the surface area. Let's see how the collisions would be affected by an increase in surface area. A metal strip is a single piece of metal. Metal powder consists of lots of tiny pieces. The powder has more surface area. This results in more frequent collisions and more frequent collisions. between the particles of Mg and HCl, the initial reaction rate in trial 4 should be faster than the initial reaction rate in trial 1. So comparing 1 and 2, it was a higher concentration. Comparing trials 1 and 3, it was a higher temperature. And comparing trials 1 and 4, it was a greater surface area. All of those changes increase the frequency of the collisions between the reactant particles, leading to a faster reaction rate. Question 5. At what point in time during a chemical reaction should the reaction rate have the greatest value? Justify your answer based on the relationship between concentration and rate. So we already know that a higher concentration of reactants should lead to a faster reaction rate. So answering this question, number 5, you're thinking about the following. At what point in time would the concentration of the reactants... be at their highest value? And the answer is at the very beginning of the reaction, so at time equals zero. The reaction rate should have the greatest value at time equals zero. The initial rate is the fastest rate because at time equals zero the concentration of the reactants is at the highest value. Remember that a higher concentration of reactants results in more frequent collisions between reactant particles and a faster reaction rate. And one final comment before I move on to topic 5.2. In addition to the variables of concentration, temperature, and surface area, another factor that can affect the reaction rate is the presence of a catalyst. Catalysis will be discussed in topic 5.11. Topic 5.2, introduction to rate law. Here are some essential knowledge statements from the AP Chemistry course and exam description. Experimental methods can be used to monitor the amounts of reactants and or products of a reaction and to determine the rate of the reaction. The rate law expresses the rate of a reaction as proportional to the concentration of each reactant raised to a power. The power of each reactant in the rate law is the order of the reaction with respect to that reactant. The sum... of the powers of the reactant concentrations in the rate law is the overall order of the reaction. The proportionality constant in the rate law is called the rate constant. The value of this constant is temperature dependent and the units reflect the overall reaction order. Comparing initial rates of a reaction is a method to determine the order with respect to each reactant. One way to study the rate of a chemical reaction is to investigate how changes in the initial concentration of the reactants will affect the initial reaction rate. Question 6. An experiment is performed to study the kinetics of the reaction between nitrogen monoxide, NO, and hydrogen, H2, as represented by the equation above. The data from the experiment is summarized. in the table below. So we have information for four different trials in this experiment. When you see a chemical formula surrounded by brackets, this notation refers to the concentration of that substance in units of moles per liter. Part. Compare the data from trials 1 and 2. The initial concentration of hydrogen is doubled. while the initial concentration of nitrogen monoxide remains constant. As a result of this change, the initial rate of formation of nitrogen is also doubled. Therefore we can say that this reaction is first-order with respect to hydrogen. Now on to Part B. Compare the data for the trials 1 and 3. The initial concentration of nitrogen monoxide is doubled. while the initial concentration of hydrogen remains constant. As a result of this change, the initial rate of formation of nitrogen is actually quadrupled or multiplied by a factor of four. Therefore we can say that the reaction is second-order with respect to nitrogen monoxide. So the data from the experiment tells us that this reaction is first-order with respect to hydrogen and second-order with respect to nitrogen monoxide. Now that we know the order with respect to each reactant, we can write the rate law. So the rate law is rate equals little k or lowercase k, which is the rate constant, times the concentration of nitrogen monoxide raised to the second power because it's second order with respect to nitrogen monoxide, and then times the concentration of hydrogen raised to the first power. because it is first order with respect to hydrogen. Now before I move on to question 7 of this packet, I'd like to make a few comments about the general features of a rate law. In a rate law, the reaction rate is normally expressed in units of concentration over time. For example, molarity per second, molarity per minute, etc. The reaction order tells us how the reaction rate will change when the concentration of that reactant is changed. It is a number that is determined experimentally. So the exponent of 2 for nitrogen monoxide, this tells us that the reaction is second order with respect to nitrogen monoxide. And if the initial value of the concentration of nitrogen monoxide is doubled, then the reaction rate would increase by a factor of 2 raised to the second power, or 4. This exponent for hydrogen is understood to be 1, and this tells us that the reaction is first order with respect to hydrogen. If the initial value of the concentration of hydrogen is doubled, the reaction rate would increase by a factor of 2 raised to the first power, or 2. The overall order of the reaction is simply the sum of the exponents in the rate law. So we know it's second order with respect to nitrogen monoxide, and we know it's first order. with respect to hydrogen. So the sum of the exponents in this case would be 2 plus 1. This rate law is said to be third-order. overall. 2 plus 1 is 3. The rate constant, which is represented by lowercase k, is a proportionality constant that relates the reaction rate to the concentrations of the reactants. Its value is determined from the experimental data and the rate law. The units of the rate constant depend on the overall order of the reaction as you will see in question 9 of this packet. Here is the balanced chemical equation. and the coefficient for nitrogen monoxide happens to be 2. The coefficient for hydrogen also happens to be 2. Remember that the rate law must be determined experimentally. The order with respect to a particular reactant is not necessarily equal to the coefficient of that reactant in the overall balanced chemical equation. So the fact that it's first order with respect to hydrogen, but it happens to have a coefficient of 2 in the overall balanced chemical equation that reminds us that the rate law must be determined experimentally. So we knew it was first order with respect to hydrogen based on the data in the experimental table. All right, now on to question 7. Question 7 says the following. Use the rate law shown above and the information in the data table on the previous page in your packet to answer the following questions. Part A. Calculate the value of the rate constant, k, for this reaction. Include units in your answer. So we already have the rate law. The rate law is written as rate equals the rate constant, k, times the initial concentration of nitrogen monoxide squared times the initial concentration of hydrogen. So I'm going to start with the data for trial one. I'm going to plug in all of the values that I know. into the rate law and then solve for k. So based on the information from trial 1, the rate would be 1.2 times 10 to the negative 3 molarity per second. The rate constant is unknown at this point. The initial concentration of nitrogen monoxide is 0.10 moles per liter and that is also the initial concentration of hydrogen, 0.10 moles per liter. Remember we have to square the concentration of nitrogen monoxide because it is second order with respect to that chemical. Now when I do this math to solve for K, as in rearrange this equation to get K on a side by itself, here's what I get. 1.2 times 10 to the negative 3 molarity per seconds on the top and I get 0.001 molarity cubed. on the bottom. So when I do the math, I get 1.2 as my numerical answer. And when I consider the units of my answer, it's going to be molarity to the negative 2 power times seconds to the negative 1 power. Now another way you could write this is 1 over molarity squared, and then you have 1 over seconds, but I'm going to represent these exponents as negative exponents because that's a little bit cleaner easier for me to process. There's a variety of ways you could write these units, but another way would be 1 over molarity squared times seconds. All right, you might be wondering at this point, did I have to use the data from trial 1? Could I have used the data from trials 2 or 3? And the answer is yes. Let's take a look. If I had used the data for trial 2, I would have had a different rate. I would have had a different concentration for hydrogen. But if I do the math, I would still get a numerical value of 1.2, and my units would still be molarity to the negative 2 power, seconds to the negative 1 power. And if I had used the data from trial 3, I would have had a different rate, I would have had a different concentration of nitrogen monoxide. But if I had used the data values from trial 3, I still would have gotten 1.2. molarity to the negative 2 power, seconds to the negative 1. So I could have used the data from either trial 1, trial 2, or trial 3, and I would have gotten the exact same value for the rate constant and the units are inverse molarity squared and inverse seconds. All right, let's move on to part B of this question. In trial 4, we know the initial concentration of nitrogen monoxide and hydrogen, but we are not given the initial rate. of formation. So part B says calculate the initial rate of formation of nitrogen in trial 4. include units in your answer. So I'm going to go ahead and use the initial concentrations. And now we have a value for the rate constant K, which is 1.2 molarity to the negative 2 power seconds to the negative 1 power. So there's my concentration of nitrogen monoxide, 0.20 molarity squared. There's my concentration of hydrogen, 0.20 molarity. And when I do this math, I get 9.6 times 10 to the negative 3. My units... are still molarity per second or molarity times seconds to the negative one. Another way you could have figured out the answer to part B is to recognize that the only change that was made from trial 3 to trial 4 is they doubled the initial concentration of hydrogen. So when you double hydrogen, which is first order, you're going to double the initial rate. And so 4.8 times 2 is 9.6. So 9.6 times 10 to the negative 3 molarity per second is the answer to part B. Now, before I move on to question 8, let me go ahead and make some summary information about rate laws and the process of determining the order with respect to a certain reactant in a chemical reaction. The order with respect to a certain reactant is a numerical value that is usually a number equal to 0, 1, or 2. is determined from experimental data and indicates how the change in the initial concentration of that reactant will affect the initial reaction rate. The order with respect to a certain reactant may in fact be the same as the coefficient of that reactant in the overall balanced chemical equation. However, this is not always true. That is why the order with respect to a certain reactant must be determined from experimental data. The value of the rate constant k can be calculated from experimental data. The value of k is unique for a particular chemical reaction and is dependent on the temperature at which the reaction is carried out. If you change the initial temperature at which the reaction is performed, the value of the rate constant k will also change. If you only change the initial concentrations of the reactor, while keeping the initial temperature constant, the value of the rate constant will remain the same. Question 8. Experimental data indicates that the reaction represented by the equation above is first order with respect to N2O5. Write the rate law for this reaction. So even though there happens to be a coefficient of 2 next to the N2O5 in the balanced chemical equation, with respect to N2O5 is determined experimentally. And we are told that it is first order with respect to N2O5. So the rate law would be written as rate equals the rate constant K times the concentration of N2O5 raised to the first power. Again, first power because it's first order. And we got that from the experimental data that was given to us, not from the coefficient in the overall balanced chemical equation. Question 9. The overall reaction order is the sum of the orders with respect to each reactant in the rate law. The units of the rate constant K depend on the overall reaction order. For each rate law in the table below, assume that the units of the rate are molarity per second. Fill in the missing information in the table below. So in this table we have five different rate laws. When you see... Capital A and capital B, those are not referring to any specific elemental symbols. Those are generic labels for reactant A and reactant B. In our first rate law, it's very simple. It just says rate equals K, and that's it. So therefore, we can deduce that the overall reaction order would be zero. And that's because of the power of exponents in which anything raised to the zero power is equal to one. So now that we know that the overall reaction order is zero, we have a very simple rate law, let's figure out the units of the rate constant k. So we are told to assume that the units of the rate are molarity per second. So if I have units of molarity per second on the left side of the equation, then the units of the rate constant k would have to match that. So molarity per second would be my units of the rate constant for that particular rate law. Let's see what happens in our next example. We have rate equals k times the concentration of A raised to the first power. that means that the overall reaction order is 1, so first order. If on the left side of the equation we have molarity per second, and on the right side of the equation we have the rate constant k times something raised to the first power, where that's concentration, that's molarity, well then I would want the units of the rate constant k to be inverse seconds. That way, inverse seconds times the units of molarity would give me molarity per second. So the units of k can be whatever they need to be so that the units on the left side of the equation match the units on the right side of the equation. Okay, let's take a look at our next example. It is concentration of a raised to the second power. So the rate equals k times the concentration of a raised to the second power. That would be second order overall. Let's see what happens to our units of our rate constant. So on the left side of the equation, we still have the rate as molarity per second. But notice that now I have k times the concentration squared. So if I have molarity squared times k, what should the units of k be to make the units match on both sides of the equation? I would want the units of the rate constant k to be molarity. to the negative one power, or one over molarity, and then I still have to have the inverse seconds to make them match. So molarity to the negative one power times seconds to the negative one power. In our next example, we have first order with respect to A, first order with respect to B. Remember that the overall reaction order is the sum of the orders. So one plus one is two. I would still have molarity. times molarity on the right side of the equation. So I would have molarity squared. The units of K would have to be anything that they need to be to make the units match on both sides of the equation. So I would still have the same situation as if I had second order with respect to one chemical. I have second order overall. So I'm going to have molarity to the negative one power times seconds to the negative one power. And again, I do that so that the units match. of the rate constant will be whatever they need to be so that the units match on both sides of the equation. All right, now I have second order with respect to A. First order with respect to B, 2 plus 1 is 3. It's going to be third order overall. So what should my units of the rate constant be in this rate law? I've got molarity per second on the left, but now I have the units of the rate constant times molarity cubed. So the units of the rate constant are whatever they need to be to make the units match on both sides of the equation. So now it would be molarity to the negative 2 power times seconds to the negative 1 power. So as you can see in the answer to question 9, the units of the rate constant depend on the overall reaction order. And again, they can be whatever they need to be so that the units match on both sides of the equation. Hope that makes sense. All right, let's move on to question 10. Question 10 is just a summary of your understanding of reaction order. Answer the following questions related to the rate law for a generic reaction in which A changes to B, in which the identity of the reactant is represented by the letter A. Suppose that a reaction is zero order with respect to A. If the initial value of the concentration of A is doubled, the initial reaction rate will remain the same. Suppose that a reaction is first order with respect to A. If the initial value of the concentration of A is doubled, the initial reaction rate will be doubled. Suppose that a reaction is second order with respect to A. If the initial value of the concentration of A is doubled, the initial reaction rate will be quadrupled. All right, let's move on to question 11. A student studies the kinetics of the reaction between ClO2, which is chlorine dioxide, and OH-, which is hydroxide. As represented by the equation above, data from the experiment is shown in the following table. Part A. Determine the order of the reaction with respect to ClO2. justify your answer. So what I'm going to do is look for two different trials in this table in which the initial concentration of ClO2 has changed. but the initial concentration of hydroxide remains constant. Then I'm going to take a look at how the rate was affected by changing the concentration of ClO2 only. So I'm going to compare trial 2 and trial 1. So what I can see is that the initial concentration of ClO2 was changed from 0.02 to 0.06. That's an increase by a factor of 3. The initial concentration of the hydroxide remains constant in these trials. And now, how did the rate get affected by this change? So 0.0248 is a larger value than 0.00276. So by what factor did the rate increase? Was it a factor of 3? No. If you do the math, you realize that that's an increase by a factor of 9. So what does this tell us? that it must be second order with respect to ClO2. So compare the data for trial 1 with the data for trial 2. The initial concentration of ClO2 is tripled while the initial concentration of hydroxide remains constant. As a result of this change, the initial rate of appearance of ClO2-which is one of the products, is increased by a factor of 9. So therefore we can say that this reaction is second order with respect to ClO2 because 3 squared is 9. Now let's move on to Part B. Determine the order of the reaction with respect to hydroxide. Justify your answer. So now what I'm looking for is a comparison of two different trials in which the concentration of hydroxide has been changed, but the concentration of ClO2 remains constant. So I'm going to compare trial two and three. So as I compare these two trials, I see that the concentration of hydroxide changes from 0.03 to 0.09. That's an increase in the concentration of hydroxide by a factor of 3. The initial concentration of ClO2 remains constant. Let's see how the rate was affected. It was 0.00276, and then it changed to 0.0828. That's an increase by a factor of 3. So if we tripled the initial concentration of hydroxide and the rate was also tripled, then that would be first order with respect to hydroxide. So compare the data for trial 3 with the data for trial 2. The initial concentration of hydroxide is tripled while the initial concentration of ClO2 remains constant. As a result of this change, the initial rate of appearance of ClO2-, which is one of the products, is increased by a factor of 3. In other words, it's also tripled. Therefore, we can say that the reaction is first order with respect to hydroxide. So now that we know the order with respect to ClO2 and the order with respect to hydroxide, let's go ahead and write the rate law for the reaction. So rate equals rate constant K times concentration of ClO2 to... raised to the second power because we know its second order with respect to ClO2, and then times the concentration of hydroxide raised to the first power because we know its first order with respect to hydroxide. Part D. Calculate the value of the rate constant K for this reaction. Include units in your answer. So I can use the data from either trial 1, trial 2, or trial 3, and I should still get the same value for the rate constant. Let me go ahead and take a look at the data for trial 1. The rate is 0.0248 molarity per second, and that is equal to the rate constant times the initial concentration of ClO2, which happens to be 0.060 moles per liter, and that's squared because we know it's second order with respect to ClO2. and then times the initial concentration of hydroxide, which happens to be 0.030 moles per liter. So mathematically, when I solve for K, I get 0.0248 divided by 0.06 squared times 0.03. And that works out to be 1.08 times 10 to the negative 4 on the bottom. When I do that math, I get an answer of 230. And I'm just rounding this off to two significant figures. All right, what about my units of my rate constant? I have molarity per second on the top. I have molarity cubed on the bottom. So that's going to work out to be molarity to the negative 2 power times seconds to the negative 1 power. So there's my value for my rate constant, which is 230. And there's my units. Inverse molarity squared, inverse seconds. Or molarity to the negative 2 power times seconds to the negative 1 power. What would happen if I had used the data for trial 2? or trial 3. I still should have gotten the same value for the rate constant. Let's take a look. So in trial 2, it would have been 0.00276 molarity per second, and that equals the rate constant times 0.020 molar squared times 0.030 molar. If I do the math, I have a value of 230 And I still have the same units, which are molarity to the negative 2 power times seconds to the negative 1 power. So inverse molarity squared times inverse seconds. And then just to show you what would happen for trial 3, I have a different rate, so 0.00828. But when I do the math and I divide by the concentration of ClO2 squared and the concentration of hydroxide, Sure enough, when I do this math, I still get a value of 230, and my units are the same. Molarity to the negative 2 times seconds to the negative 1. So the answer to part D is going to be the same whether I had used the data for trial 1, trial 2, or trial 3. The numerical value is the same for the rate constant. Now on to part E of this question. Calculate the initial appearance. of ClO2-in trial 4. What I did here is I have initial concentration of ClO2, which is 0.050 moles per liter. I have the initial concentration of hydroxide, which is also 0.050 moles per liter. But now the question mark is, what is the rate? Since I already know the rate constant, which is 230, and there are my units, molarity to the negative 2 power times second to the negative 1. I multiply by the concentration of ClO2 squared, because it's second order with respect to that chemical, and then times the concentration of hydroxide. When I do this math, I get 0.0288 when I round this off to three significant figures. Alright, so there is my unit for rate, which is molarity per second. So that value, 0.0288 molarity per second, is my answer. to part E. Topic 5.3, concentration changes over time. Here are some essential knowledge statements from the AP Chemistry course and exam description. The order of a reaction can be inferred from a graph of concentration of reactant versus time. If a reaction is first order with respect to a reactant being monitored, a plot of the natural log Ln of the reactant concentration as a function of time will be linear. If a reaction is second-order with respect to a reactant being monitored, a plot of the reciprocal of the concentration of that reactant versus time will be linear. The slopes of the concentration versus time data for 0, first, and second-order reactions can be used to determine the rate constant for the reaction. So these three equations that are listed here are also included in the kinetics section of the AP Chemistry equations and constants sheet. Half-life is a critical parameter for first-order reactions because the half-life is constant. And related to the rate constant for the reaction by the following equation, half-life equals 0.693 divided by k. Radioactive decay processes provide an important illustration of first-order kinetics. Consider a generic reaction where reactant A is converted into product B, in which the identity of the reactant is represented by the letter A. Suppose that it has been determined from the analysis of experimental data that the reaction is first-order with respect to A. The rate law would be written as follows. rate equals k, the rate constant, times the concentration of A raised to the first power. We can use calculus to transform this rate law into the following expression. This equation is called the first-order integrated rate law equation. where k is the rate constant, t equals time. The concentration of A subscript zero is the initial concentration of A at time zero. The concentration of A subscript t is the concentration of A at some point in time equal to t. Question 12. Part A. Use the data for concentration of A in the table above to calculate the value of the natural log of the concentration of A. Record the calculated values for the natural log of the concentration of A in the data table. So when I pick up my calculator and I use the natural log button to figure out the natural log of 8, I get 2.08. When I take the natural log of 5.66, I get 1.73. When I take the natural log of 4, I get 1.39. And the natural log of 2.83 is 1.04. And I'm going to go ahead and fill in the rest of the data in this table by taking the natural log of the concentration. So 0.693, 0.344, 0. The natural log of 1 is 0. And then I now get negative values. So I get negative 0.342 and negative 0.693. So that is my answer to part A. In part B, plot the data for natural log of concentration of A versus time on the graph below. So at time equals zero, I have a value of 2.08. At 15 seconds, I have 1.73. At 30 seconds, I have 1.39. 45 seconds, 1.04. 60 seconds, 0.693. And I'm going to go ahead and plot the other four points. So there is my answer to part B. Natural log of concentration of A versus time. If a reaction is first order with respect to A, then the plot of the natural log of concentration versus time will be a straight line. And the absolute value of the slope of this line is equal to the value of the rate constant K. In part C, it says draw a straight line to the data points on the graph you created in part B and calculate the value. of the slope of this line and include units in your answer. Note that the concentration of A has units of molarity or moles per liter, but the quantity natural log of the concentration of A has no units. So when you take the natural log of a quantity the units disappear. They go away. So no units. Alright, well I've chosen two points on this straight line graph. I've chosen 0 seconds which is a value of 2.08, and then 60 seconds, which is the value of 0.693. I'm going to calculate the slope by doing the difference in y divided by the difference in x for this region. So my difference in y would be 0.693 minus 2.08, and I get negative 1.39 for my difference in y. And for my difference in x, it would be 60. minus 0, so 60 seconds. Again, I have no units on the y-axis, but I have units of seconds on the x-axis. So when I do this math of negative 1.39 divided by 60, I get negative 0.023, and my units are 1 over seconds, or inverse seconds. So my answer to part c is that the slope of that line is negative 0.023. inverse seconds. So if I wanted to determine the value of the rate constant K in this experiment, all I need to do is take the absolute value of the slope of this line. So if I plot natural log of concentration of A versus time with a first-order reaction, in this case the absolute value of the slope, which would be 0.023 inverse seconds, then that would equal the rate constant K. So remember this value, 0.023 inverse seconds. We're going to see this number show up again in part E of this question in the packet. All right, let's talk about half-life. The half-life of a reaction is defined as the time required for the concentration of a reactant to reach half of its original value. If a reaction is first order with respect to a particular reactant, then the half-life of that reactant should remain constant over time. during the course of the reaction. The converse of this statement is also true. If the half-life of a particular reactant remains constant over time during the course of the reaction, then the reaction should be first order with respect to that reactant. In part D, it says, based on the data for concentration of A versus time, shown on the previous page in your packet, determine the half-life of reactant A. Include units in your answer. So here is our graph of concentration of A versus time. Let's start with the first point at time zero. That's 8 moles per liter. Let's see how long it takes for the concentration of A to go from 8 to 4, because that would be half of the original value. Well, it looks like it takes 30 seconds for the concentration of A to change from 8 to 4. Let's see if the half-life remains constant over time. It takes another 30 seconds to go from a concentration of 4 to 2. It takes another 30 seconds to go from a concentration of 2 to 1. It takes another 30 seconds to go from a concentration of 1 to 0.5. So yes, the half-life is remaining constant over time, and the half-life is equal to 30 seconds. So that is our answer to Part D. On the next slide, I am showing you a portion of the AP Chemistry Equations and Constants sheet. This is the section entitled Kinetics. Now, there are four equations listed, but on the actual AP Chemistry Equations sheet that you use on the AP Chemistry exam, they are not labeled in this way. So, AP Chemistry students must know and understand which order goes with which equation because they're not labeled specifically in this way on the actual AP Chemistry exam. So the first equation listed is for zeroth order kinetics. The next equation listed is for first order kinetics. The third equation listed is for second order kinetics. And that last equation that's listed, which has half-life equals 0.693 divided by k, that equation is only used for first order kinetics. Again, AP Chemistry students have to know which order applies to which equation. All right, so number 12, part E. Use your answer to part D, which was a half-life of 30 seconds, and the half-life equation, half-life equals 0.693 divided by k, to calculate the value of the rate constant k. Include units in your answer. So all I'm going to do is plug in the value of 30 seconds for the half-life. I'm going to rearrange this equation so I can solve for k. So the value of k is equal to 0.693 divided by 30 seconds. When I do this math, I get 0.023, and my units are going to be inverse seconds. So recall that earlier in this packet, you had calculated the slope of the line. for the plot of natural log of concentration of A versus time. And that slope was negative 0.023 inverse seconds. So the absolute value of that slope is, in fact, the value of the rate constant for this experiment. So question 12 in this packet, with all of its parts, had to do with first-order kinetics. Now we're going to see in question 13 of this packet, it has to do with second-order kinetics. Let's take a look. Suppose that it has been determined from the analysis of experimental data that a reaction is second-order with respect to reactant A. The rate law would be written as follows. Rate equals the rate constant K times the concentration of A raised to the second power. We can use calculus to transform this rate law into the following expression. This equation is known as the second order integrated rate law equation, where K is equal to the rate constant, T equals time, and the concentration of A subscript 0 is the initial concentration of A at time equals 0, and the concentration of A subscript T is equal to the concentration of A at some point in time equal to T. So here's question 13. The first part says to use the data... for the concentration of A in the table above to calculate the value of the reciprocal of the concentration of A, or 1 over the concentration of A. Record the calculated values for 1 divided by the concentration of A in the data table. So 1 divided by 5 is 0.2. 1 divided by 1.67 is equal to approximately 0.6. 1 divided by 1 is 1. 1 divided by 0.71 is equal to 1.4. 1 divided by 0.56 is equal to 1.8. and I will continue that process of doing 1 divided by the concentration and fill in the rest of the data table. So I have values of 2.2, 2.6, 3.0, 3.4, 3.8, and then finally 1 divided by 0.24 is equal to 4.2. Now in the next part of the question, it says plot. the data for the reciprocal of the concentration of A versus time on the graph below. So we'll start with at 0 seconds, that would be 0.2, and then approximately 0.6 at 10 seconds, then we have a value of 1 at 20 seconds, 1.4, 1.8, and I'll just keep going, 2.2, 2.6, 3.0, 3.4, 3.8, and finally 4.2. at 100 seconds. Notice that in the graph we're looking at right now the units on the y-axis are inverse molarity 1 over molarity. When it was natural log there were no units at all but now we have units of inverse molarity, 1 over molarity and on the x-axis we still have units of seconds for time. In part C it talks about calculating the value of the slope of this line It says it's important to know that if a reaction is not first order but second order with respect to A, then the plot of the reciprocal of the concentration of A versus time will be a straight line. So the absolute value of the slope of this line is equal to the value of the rate constant K. Let's figure out the slope of this line. I've picked two points at 0 seconds and 60 seconds. So the difference in Y... divided by the difference in x will equal the slope of this line. So for the difference in y, it's going to be 2.6 minus 0.2, so I get a value of 2.4. Remember that the units on the y-axis are inverse molarity, and on the x-axis it's going to be 60 minus 0, so 60 seconds. 2.4 divided by 60 gives me the numerical value of the slope, which is 0.040. Now for my units of my slope, it's going to be inverse molarity times inverse seconds. So the value of the slope is 0.040, and the units are molarity to the negative 1 times seconds to the negative 1. So earlier in this packet, in question 12, I talked about first-order kinetics. In question 13, I just talked about second-order kinetics. Now in question 14, I'm going to talk about zeroth order kinetics. Let's take a look. Suppose that it has been determined from the analysis of experimental data that a reaction is zeroth order with respect to reactant A. The rate law would be written as follows. Rate equals the rate constant K, and that's it. Because if you take the concentration of A and you raise it to the zero power, anything raised to the zero power, power is just 1. So it's a very simple rate law. We can use calculus to transform this rate law into the following expression. This equation is known as the zeroth order integrated rate law equation, where k is the rate constant, t is time, concentration of A subscript 0 is the initial concentration of A at time 0, and concentration of A subscript t is the concentration of A at some point in time equal to t. So here we don't have to make any transformations like natural log or one over the concentration. It already is a straight line. So we have concentration on the y-axis, time on the x-axis. If a reaction is zeroth order with respect to A, then the plot of concentration of A versus time will be a straight line. And the absolute value of the slope of this line is equal to the value of the rate constant K. So... All we have to do here is to calculate the value of the slope of the line. We're not doing natural log. We're not doing reciprocal. So I'll pick two points at 0 seconds and 60 seconds. I'll calculate the slope by doing the change in y divided by the change in x. So my change in y would be 5.0 minus 9.8, and my units are molarity. My change in x would be 60 seconds minus 0 seconds. So negative 4.8 for the change in y, and I have 60 for the change in x. So my value for the slope is equal to negative 0.080, and my units are molarity per second. So my answer for the slope is negative 0.80 molarity per second. The absolute value of this slope would equal the rate constant K in this experiment. Question 15. Below is a summary of the information that has been presented so far. Fill in the missing information. So the order with respect to A is either 0, first, or second. And the first question is, what does a graph of concentration of A versus time look like? Is it a curve or a straight line? Well, the only graph of concentration versus time that's going to be a straight line is zero-order kinetics. For first-order and second-order kinetics, that's going to be a curve. So here's what those curves look like, or those straight lines. So... Straight line for concentration versus time for zero order. And then for concentration versus time for first order, you can see it's a curve. And for concentration versus time for second order, it's also a curve. Next question. Does the half-life of A remain constant over time? Well, the only situation in which the half-life of A remains constant over time is with first-order kinetics. So that would be no, yes, and no for that question. Next question. What type of graph should produce a straight line? If we take a look at the equation sheet, we can see that the kinetics equations are listed 0, first, and second order. So what type of graph should produce a straight line for zeroth order kinetics? Concentration versus time. What type of graph should produce a straight line for first order kinetics? natural log of concentration versus time. What type of graph should produce a straight line for second-order kinetics? The reciprocal or one over the concentration versus time. So those are the answers to that question. Next question, how do you determine the value of the rate constant k from the linear graph mentioned above? Well the answer is the same for all three. You take the absolute value of the slope of that straight line that linear graph. And then finally suppose that the reaction rate has units of molarity per second. What are the units of the rate constant? So if you're getting it from the slope of the line, that change in y over change in x, it's going to be either molarity per second. Remember natural log of concentration has no units, so just inverse seconds. And if I have one over the molarity, divided by the time, it would be inverse molarity times inverse seconds. So molarity to the negative 1 times seconds to the negative 1. All right, let's move on to question 16. Radioactive decay processes provide an important illustration of first-order kinetics. In nuclear decay, the composition of the atomic nucleus changes. In a chemical reaction, Chemical bonds are broken and or formed. Since the half-life remains constant during nuclear decay, it is classified as a first-order process. Question 16 says that nitrogen-13 is an isotope of nitrogen that undergoes a nuclear decay process in which it is converted into carbon-13. The half-life for this decay process is equal to 10 minutes. Part A, calculate the value of the rate constant K for the radioactive decay of nitrogen-13? Include units in your answer. So we know that radioactive decay is a first-order process because the half-life remains constant. So I'm going to use the first-order equation for half-life at the bottom here. I'm going to rearrange it to solve for K. So K is equal to 0.693 divided by the half-life, which we were told is 10 minutes. 0.693 divided by 10 minutes is 0.0693 and my units are inverse minutes. In part B of this question, it says, suppose that you have a pure sample of nitrogen-13 that has a mass of 64 milligrams. Calculate the mass in units of milligrams of nitrogen-13 that will be present in this sample after a period of 50 minutes. So what's nice about the math in part B is that 50 is a multiple of 10. So that means that we have exactly five half-life periods. If I start with 64 milligrams, one half-life period would go from 64 milligrams to 32 milligrams. Another half-life period would take it from 32 milligrams to 16 milligrams. Then I would have 8 milligrams, then 4 milligrams, and then finally 2 milligrams. That would be 5 half-life periods for a total of 50 minutes. So my answer to Part B is 2 milligrams. Now, could I use the equations on the equation sheet to get this answer? And the answer is yes. Let's take a look. So because radioactive decay is classified as a first-order process, I'm going to use the first-order equation involving natural log. The initial amount is 64 milligrams. And I'm going to use mass as opposed to molarity to do this calculation. So I have 64 milligrams at time zero. The amount at time t, which is... t equals 50 minutes is what I'm trying to find. The rate constant that I calculated in part A is 0.0693, and the units are inverse minutes. Now, when I do this math, I get that the natural log of the amount present at time t is equal to 0.694. So the opposite of natural log would be e to the x. I'm going to do e raised to the 0.694. on my calculator and I get 2 milligrams. So it was nice that I could get the same answer whether I just divided by 2 five times in a row or I used the natural log equation from the equation sheet. Alright, let's move on to part. Suppose that you have a pure sample of nitrogen 13 that has a mass of 125 milligrams. Calculate the mass in units of milligrams of nitrogen 13 that will be present in this sample after a period of oh, 37 minutes. 37 is not a multiple of 10. So I'm going to first estimate roughly what the amount of nitrogen-13 will be by simply dividing by 2 four times in a row. So why four times in a row? Well, that would be four half-life periods. That would be 40 minutes. Let's estimate. 125 is what I start with in milligrams. So the first 10 minutes, that would be dividing by 2, 62.5. Another 10 minutes, that would be 31.25, just dividing by 2. Another 10 minutes, I'm down to 15.625. And then finally, one more half-life period, I'm down to 7.8. Now, 7.8 is not my answer because that's 40 minutes. I'm trying to figure out how much will be present after 37 minutes. So I can estimate that the amount of nitrogen-13 that's present after 37 minutes is somewhere between 7.8 and 15.6 milligrams. Let's go ahead and use the natural log equation to get the exact answer. So here's my natural log equation. I know that the time is 37 minutes. I know the rate constant is 0.0693 inverse minutes. I start with 125 milligrams and I'm trying to figure out what I end with at time t. So when I do this math, I get natural log of the quantity present at time t is equal to 2.264. So that means that e raised to the 2.264 is my answer, which works out to be 9.6 milligrams. And that's consistent with the value that I had estimated. So that looks good. All right, well, that's the end of part C. Let's move on to question 17 in the packet. In question 17, we are given the following information. Blue food coloring, and there is a long chemical formula listed for that, reacts with household bleach, which is sodium hypochlorite, or NaOCl, to form colorless products, as represented by the equation above. A student used a spectrophotometer set at a wavelength of 635 nanometers to study the absorbance of the blue food coloring over time. during the course of the reaction. So on the y-axis in that first graph on the left, you see absorbance. Because blue food coloring has a particular absorbance at 635 nanometers, the higher the absorbance, the higher the concentration of the blue food coloring. So we're looking at the change in absorbance over time, which would be the change in the concentration of the blue food coloring over time. Now, this experiment is run with an initial concentration of bleach that is much greater than the initial concentration of the blue food coloring. This ensures that the concentration of the bleach remains essentially constant throughout the reaction. So we're focusing on the change in the concentration of the blue food coloring. We're not focusing on the change in concentration of the bleach. The student used the data from the experiment to generate three graphs as shown below. Number A. What is the order of the reaction with respect to the blue food coloring? Justify your answer. So again, we have absorbance on the y-axis in the first graph. We have natural log of the absorbance on the y-axis in the second graph. And we have the reciprocal or one over the absorbance on the y-axis in the third graph. On the x-axis, we have time in seconds. So remember that from topic 3.13, which is the Beer-Lambert law, also known as Beer's law, the absorbance of the food coloring is directly proportional to its concentration. So as the absorbance decreases, so does the concentration decrease. The only graph that we're looking at that is actually a straight line is the one that shows natural log of absorbance on the y-axis. And hopefully you remember that when natural log of concentration versus time is linear, then that indicates first order kinetics. So the answer to part A is that the reaction is first order with respect to the blue food coloring because the plot of natural log of absorbance versus time is linear. In part B, we are given the following information. It says the reaction is known to be first order with respect to bleach. In a second experiment, the student prepares solutions of food coloring and bleach with concentrations that differ from those used in the first experiment. When the solutions are combined, so when they combine the food coloring solution with the bleach solution, the student observes that the reaction mixture reaches an absorbance value of zero too rapidly. So the color is fading, but it's fading very fast. In order to correct this problem, The student proposes the following three possible modifications to the experiment. Which of these proposed modifications should correct the problem? Explain how that modification increases the time required for the reaction mixture to reach an absorbance value of zero. So remember that over time, as the reaction proceeds, that blue food coloring is getting lighter and lighter and lighter. So it's fading away as it reacts with the bleach. and eventually the absorbance goes down, down, down to a value of zero. So this reaction is happening too quickly. We're trying to slow down the reaction. So suppose you have a solution of blue food coloring. That's going to have a particular absorbance value. Then you add a solution of bleach, which starts a chemical reaction, and as that chemical reaction occurs, that blue food coloring gets lighter and lighter and lighter as it fades. over time. And again, the absorbance is decreasing over time. So we're trying to slow down the reaction so that that color does not fade so quickly. That's the problem that we're trying to correct. So our first example of a modification is to increase the temperature of the reaction. Well, we talked about this earlier in topic 5.1 that when you increase the temperature, you increase the collisions. and the force and the frequency of those collisions, and that would typically make a reaction occur even faster. So if we increase the temperature, then that process of fading away, that blue food color or reacting, it's going to happen even more quickly. So that is not going to be a good choice for us. We're trying to slow the reaction down, and a higher temperature would make it happen even faster. Another thing that we could do as a possible modification is to... increase the concentration of the bleach. But unfortunately, the bleach is what is causing the blue food coloring to fade away. So if we add an even higher concentration of bleach, then that's going to make the reaction happen more quickly. So again, that's not the problem that we're trying to solve. We're actually trying to slow it down and not speed it up. So increasing the concentration of the bleach would mean that that reaction will happen even more quickly and that blue food coloring would fade away at a faster rate. So I hope that the correct answer is the middle one, which is to increase the concentration of the blue food coloring. But it does say, explain, how does that increase the time? required for the reaction mixture to reach an absorbance value of 0. Well, suppose that we do increase the concentration of the blue food coloring. What is that going to do to the intensity of the blue color in that beaker? It's going to make it even darker. And if it makes it even darker, then the absorbance goes up to a higher value. So now you're starting at an even higher absorbance value. And it's going to take longer for that darker food coloring to fade away. So the correct answer is if we increase the concentration of the blue food coloring, then at that higher concentration the initial absorbance value should be higher at the very beginning of the reaction at time zero. So it should take a longer time for that darker blue color to fade away and reach an absorbance value of zero. So that is the correct answer to Part B. In part C of this question, we are given the following information. It says, in another experiment, a student wishes to study the reaction between red food coloring and bleach. Describe how the student should modify the original experimental procedure in order to determine the order of the reaction with respect to the red food coloring. So, let me remind you what that original experimental procedure was. We are told that the student used a spectrophotometer that was set at a wavelength of 635 nanometers to study the absorbance of the blue food coloring over time. Now, back in Unit 3, it was Topic 3.13, the Beer-Lambert Law, also known as Beer's Law, we learned about how a spectrophotometer works, where light is shining through the sample, and you can set your spectrophotometer at a particular wavelength based on that particular solution in terms of what color it is and what wavelength of light is going to represent the maximum absorbance. So hopefully you can see from this video, which is from the FET simulation on Beer's Law, that when we change the color of the solution that is being analyzed in the spectrophotometer, we actually have a different wavelength. setting on the spectrophotometer. So the correct answer to part C is based on that information. So describe how the student should modify the original experimental procedure to determine the order of the reaction with respect to the red food coloring. Well, the spectrophotometer should be changed to a different wavelength. You don't have to specify what that wavelength is, but just that would be the modification. Set that spectrophotometer to a different wavelength. when you go from blue food coloring to red food coloring. Now, let's take a look at the next question in the packet. This is question 18. A student studying the decomposition of this particular chemical represented by the equation shown above runs the reaction at 90 degrees Celsius. The student collects data on the concentration of this reactant as a function of time as shown by the data table. and the graph below. So in the data table we see time in hours and we see concentration of the reactant decreasing over time. We can also see in the graph that the concentration is on the y-axis, time in hours is on the x-axis, and we can see the change in the concentration of the reactant over time. Part A. What is the order of the reaction with respect to this reactant? Justify your answer. Now, there's actually two different strategies for determining the order of the reaction in this problem. So I'm going to show you one way, and it involves calculations for concentration, natural log of concentration, and the reciprocal, or one over the concentration. So I'm going to go ahead and fill in these two columns in the data table by picking up my calculator and taking... either the natural log of the concentration or doing 1 divided by the concentration. So the natural log of 0.1. works out to be negative 2.30. The natural log of 0.0707 is negative 2.65. The natural log of 0.0500 is negative 3. The natural log of 0.0354 is negative 334. And then I will go ahead and continue and fill in the remaining three values in this column. So negative 3.69, negative 4.03, and negative 4.38. Now I will fill in the other column by doing 1 divided by the concentration. So 1 divided by 0.1 is 10. 1 divided by 0.0707 is equal to 14.1. 1 divided by 0.05 is equal to 20. 1 divided by 0.0354 is equal to 28.2. And again, I'll continue this process and fill in the remaining values in this column, which would be 40, 56.5, and 80. So now what I have to do is consider which one of these three columns of data is going to help me determine the order of the reaction with respect to the reactants. So we're looking for a straight line. We're looking for a linear plot. Either it's concentration versus time, natural log of concentration versus time, or the reciprocal of concentration versus time. So which column of data would produce a straight line? Well, we already know from the given information that concentration versus time shows up as a curve. So that's not going to give us a straight line. Let's take a look at the reciprocal of the concentration. So every five hours, we have a new data point. Going from 10 to 14.1, that's a difference of 4.1. Going from 14.1 to 20, that's a difference of 5.9. Going from 20 to 28.2, that's a difference of 8.2. Going from 28.2 to 40, that's a difference of 11.8. Hopefully you can see that if you were to plot that data with reciprocal of concentration on the y-axis and time in hours on the x-axis, you would not get a straight line. So now that leaves us with natural log of concentration versus time. So every five hours, that value looks like it's getting lower and lower. It's decreasing, but I can tell that... In the first five hours, that value is decreasing by 0.35. In the next five hours, it's also decreasing by 0.35. And then I have a decrease of 0.34, but that's really close. So effectively, I can tell that this is going to be a linear relationship. Every five hours, the value of the natural log of the concentration is decreasing by approximately the exact same amount. And if you did plot this data, with natural log of concentration on the y-axis, time in hours on the x-axis, you would get a straight line. Remember that when natural log of concentration versus time is linear, this tells us we have first order kinetics. So the answer to part A is the following. The reaction is first order with respect to this reactant because the data for natural log of concentration versus time indicates a linear relationship. Now, as I had mentioned earlier, I said that there are actually two strategies for determining the order of the reaction. I'm now going to show you a much easier, much simpler, much faster way to get the answer to part A. So let's take a look. Earlier, when I was talking about the different orders of the reaction, how you can monitor the change over time, I said the following. If the half-life... of a particular reactant remains constant over time during the course of the reaction, then the reaction should be first order with respect to that reactant. So can we analyze half-life based on the given data in this problem? And the answer is yes, absolutely. So at time 0 hours, we start with the concentration of 0.1. So what's half of 0.1? Well, half of 0.1 is 0.05. So it takes 10 hours. for the concentration to decrease by half. It takes another 10 hours for the concentration to go from 0.05 to 0.025, which is half. of that amount. And then it takes another 10 hours to go from 0.025 to 0.0125. That's it! That's all you have to recognize to answer this question. So the reaction is first-order with respect to the reactant because the half-life remains constant over time. So what I would say to all the students who did generate natural log data and verify that that is a linear relationship... You got the right answer, but you took way too long. You wasted perhaps too much time, especially on the AP exam when you're a little bit stressed for time and trying to be very efficient with how quickly you solve a problem under a timed condition. This method of recognizing that the half-life is constant over time was a much simpler, much easier, much faster way to answer part A of this question. So now that we know that this reaction obeys first order kinetics, let's see what they're going to ask us in part B that's related to this. All right, part B. Determine the value of the rate constant K for this decomposition reaction at 90 degrees Celsius. Include units with your answer. So when you look on the equation sheet, we know that there are different equations in the kinetic section. We know it's first order. And because we know the half-life... I'm going to use this particular equation. You can only use that equation for first order kinetics, and that's what we have here. So I'm going to rearrange this equation to solve for K. So K is equal to 0.693 divided by the half-life. Again, we know the half-life from the data in the table. The half-life is 10 hours because it takes 10 hours for the concentration to decrease by half. So here's my equation. K is equal to 0.693 divided by the half-life. The half-life is 10 hours. So on my calculator, I get 0.693 divided by 10, 0.0693. Now, if you did write this as your answer, your final answer of 0.0693, and then you just stopped right there, you would not earn full credit for your answer because it says include units with your answer. The units... of the rate constant with first order kinetics is always going to be inverse time, but not inverse seconds, not inverse minutes. In this case, it's inverse hours. So that is the answer to part B, 0.0693, and a unit would be inverse hours or one over hours. Moving on to part C of this question. Calculate the expected value for the concentration of the reactant in this experiment after 35 hours have passed. Now we know it's first-order kinetics, so we can use the appropriate equation from the equation sheet for first-order. That's the natural log equation. We're trying to figure out the concentration of the reactant after 35 hours have passed. We know from part B that the rate constant is 0.0693 inverse hours. The time is 35 hours. We know the initial concentration of the reactant at time 0 is 0.1 moles per liter. So I'm plugging in all of that information into the natural log equation. What I don't know is the concentration at time t. So I do the math. And when I get to the point of saying that the natural log of the concentration at time t is equal to negative 4.729, all I have to do is the opposite of natural log, which is e to the x power. So e raised to the negative 4.729 on my calculator is the concentration at 35 hours. So 0.00884 is what I get. Here is another way to get that same answer or approximately the same answer by using your knowledge of half-life. So at 25 hours, the concentration of the reactant is 0.0177. And we're trying to figure out the concentration at 35 hours. If you remember that the half-life is 10 hours, and we already know that, then 25 plus 10... is 35. So after 10 hours you would have the concentration of the reactant at 35 hours, going from 25 to 35 is 10 hours. So what exactly is half of 0.0177? That would be my answer. So half of 0.0177 is 0.00885. So the answer to part C, depending on which way you solve it, is either 0.00884 moles per liter or 0.00885 moles per liter. I just wanted to show you two different ways to do that calculation in part C using the equations for first-order kinetics. Alright, let's move on to our next question in the packet. This is question 19. At high temperatures, the compound C4H6 reacts according to the equation above. The rate of the reaction was studied at 625 Kelvin in a rigid reaction vessel. Two different trials, each with a different starting concentration, were carried out. The data were plotted in three different ways, as shown below. So you can see that on each graph we have... two different trials, trial 1 and trial 2. We have concentration versus time, natural log of concentration versus time, and the reciprocal of concentration versus time. Part A. For trial 1, calculate the initial pressure in atmospheres in the vessel at 625 Kelvin. Assume that, initially, all the gas present in the vessel is C4H6. So on that graph of concentration versus time, for trial 1, which are the white circles in the diagram, the initial concentration at time 0 is 0.020 moles per liter. Now, we're trying to figure out pressure. If we can use the ideal gas law, PV equals NRT, we could hopefully solve for pressure if we can figure out the number of moles, the temperature, and the volume. Now as far as which value for the ideal gas constant we're going to use, since they want the pressure in atmospheres, then that would be R, the gas constant, of 0.08206 because that value has units of atmospheres. All right, so we know that the concentration at time zero was 0.020 moles per liter. n, the number of moles, can be 0.020. V, the volume in liters, can be 1 liter. So I'm going to go ahead and solve the ideal gas law for pressure. So p equals nRT over V, where n is the number of moles, and that's going to be 0.020. R is the gas constant that has units of atmospheres. Temperature is the Kelvin temperature, which is 625 Kelvin, and then the volume is one liter because we're setting that up, making the assumption that if it is a one liter vessel and the concentration is 0.020 moles per liter, I can figure out the moles and the liters by using that relationship. So I'm going to go ahead and do that math on my calculator, and I get 1.02575, but I think two significant figures is going to be sufficient here. So I'll round off my answer to part A, to two significant figures. and I'll just say 1.0 atmospheres. Now, some of you might be thinking, after this particular calculation in part A, wasn't that the ideal gas law? Isn't that unit three? I thought unit five was about kinetics and reaction rates. Don't worry. Sometimes it's nice to see a little bit of a review of topics that you learned earlier, but we are definitely talking about kinetics. Take a look at part B. What is the order? of the reaction with respect to C4H6 justify your answer. So we do have three graphs to examine. We have concentration versus time, natural log of concentration versus time, and the reciprocal of concentration versus time. What are we looking for? We're looking for a straight line or a linear relationship. So I can tell there's a lot of curvature in concentration versus time, so that's not linear. I can definitely see some curvature in trial one for natural log of concentration versus time. That does not look linear to me. And when I look closely at the reciprocal of concentration versus time, looks like nice straight lines. So that's linear. And therefore, when you plot reciprocal of concentration versus time, and you get a straight line, that would correspond to second order kinetics. So the answer to part B is the following. The reaction is second order with respect to C4H6 because the plot of the reciprocal of concentration versus time is linear. Let's take a look at Part C. Now that we know it's a second-order rate law, it says the initial rate of the reaction in trial 1 is equal to 0.0010 molarity per second. Calculate the value of the rate constant K for the reaction at 625 Kelvin. Include units in your answer. So be thinking about how you would write the second-order rate law. The initial concentration for trial 1 is 0.020 moles per liter, as I mentioned earlier in part A. The second order rate law would be rate equals K, the rate constant, times the concentration of C4H6, the reactant, raised to the second power because it's second order. Well, they gave us the rate. We know the initial concentration. Let's plug in those values into the second order rate law. and solve for k. So the rate that they gave us in part c is 0.0010 molarity per second. The rate constant is unknown. That's little k. And then we have 0.020 moles per liter. That's the initial concentration, and that's squared because it's second order. So I'll do this math, and when I do 0.0010 divided by 0.02 squared, What I get for my answer is 2.5. Now, if you write that as your final answer in part C, you would not earn full credit because they say include units in your answer. That's very typical when you're calculating a rate constant. The units are dependent on the order of the reaction. So that's very important to mention units. So I've got molarity per second on the top, and I've got molarity squared on the bottom. And so when I do that and I work out the units, it's going to be inverse molarity. So molarity to the negative one and then inverse seconds. So seconds to the negative one. Those are my units of the rate constant. That's also consistent with second order kinetics. So inverse concentration and inverse time are what you see for the units of the rate constant for a second order rate law. All right, let's move on to question 20 in the packet. In question 20, we are given the following information. The decomposition of H2O2, which is hydrogen peroxide, is represented by the equation above. A student monitored the decomposition of a sample of H2O2 at a constant temperature and recorded the concentration of H2O2 as a function of time. The results are given in the table below. So we have in the first column, time in hours. In the next column, we have the concentration of H2O2 in units of molarity, or moles per liter. And then we have two more columns that are blank. So in part A, calculate the missing values in the table above. Let's take the natural log of the concentration. So the natural log of 1 is 0. The natural log of 0.794 is equal to negative 0.231. The natural log of 0.6... is equal to negative 0.462. And the natural log of 0.5 is equal to negative 0.693. Now to finish off that final column, let's take the reciprocal of the concentration. So 1 divided by 1 is 1. 1 divided by 0.794 is equal to 1.26. 1 divided by 0.630 is equal to 1.59. and 1 divided by 0.5 is equal to 2. This process of figuring out the order of the reaction with respect to H2O2, this is similar to what I showed you in question 18 of this packet. What we're looking for to answer part B, determine the order of the reaction with respect to the reactant, is to look for a linear plot. So which column of data? Concentration. natural log of concentration or the reciprocal of concentration is going to produce a straight line when it's plotted versus time. So what I'm gonna do is start with the reciprocal of the concentration. Every two hours I get a new data point. Let's see if it's gonna increase by a regular interval. So from 1 to 1.26, that's a difference of 0.26. When I go from 1.26 to 1.59, that's a difference of... 0.33, so not exactly the same. And when I go from 1.59 to 2, that's a difference of 0.41. So this data in the far column, the reciprocal of concentration, will not produce a straight line when it's plotted versus time. Let's take a look at the natural log of concentration. It starts off at 0 and then after two hours it decreases to negative 0.231. Then it decreases again by a value of 0.231, and then it decreases again also by a value of 0.231. So therefore, looking for a straight line or a linear plot, I would say it's natural log of concentration. That's going to give me a linear plot when this data is plotted versus time, because every two hours, the value of the natural log of the concentration decreases. by the same amount. So I'm going to say in part B that I know it's first order with respect to H2O2 because the data for natural log of concentration versus time indicates a linear relationship. Now back in question 18 of this packet, you may recall that I showed you this long, complicated way of determining that it was first order. And then I said Here's a shortcut that is much simpler and easier and faster, and that in number 18, part A, we determined that it was first order because the half-life remains constant over time. So can I make that same claim in question 20? Let's take a look at our data. The answer is no. You cannot actually say that the half-life remains constant over time. And the reason why is because We don't have enough data in our table. Yes, we know that it takes six hours for the concentration of hydrogen peroxide to change from one molar to 0.5 molar. But that's it. We don't have successive data where we can say it decreases by 0.5, it decreases by 0.5. So therefore, if you say it's first order because the half-life remains constant over time, that's not a valid claim. based on this data table. We just simply don't have enough information to know if the half-life remains constant over time. Alright, so the correct answer to Part B is that it is first-order with respect to the H2O2, but that's because the data in the table shows us that natural log of concentration versus time is a linear relationship. Now on to Part C. We know that we have first-order kinetics. Part C says determine the value of the rate constant k for this decomposition reaction. Include units with your answer. So I've got four different equations on the equation sheet but I'm focusing on first order kinetics. So I'm going to use the natural log equation from the equation sheet to figure out the value of the rate constant k. I'm gonna focus on the data table and the first two hours. This is what I know. The time is two hours. The value of the rate constant is unknown to me. That's what I'm trying to find. The natural log of the initial concentration is 0 and the natural log of the concentration at two hours is negative 0.231. So I plug in all of that information into my first order integrated rate law and now I can solve for k. So 0.231 divided by gives me the numerical value which is 0.116. Of course the units of the rate constant are inverse time. For first-order kinetics this would be hours to the negative one or inverse hours. Another way you could figure out the rate constant is you take the absolute value of the slope when you plot natural log of concentration versus time. So let's take a look at that. We should get the same value. So in this situation, delta y over delta x, or the change in y over the change in x, it's changing by 0.231 every two hours. So the slope of that line would be 0.231 divided by 2, and my units would still be inverse hours. So I've answered part C. I have 0.116 inverse hours from my rate constant. Determine the half-life of H2O2 in this experiment. Include units in your answer. Well, the half-life... can be equal in a first order kinetic situation to 0.693 divided by the rate constant. In part C, we've already calculated the rate constant which is 0.116. Let's go ahead and use that half-life equation. 0.693 divided by 0.116 inverse hours. And if we do that math on our calculator, we get 6. Now, six hours is the half-life, which you could probably already figure out from the data in the table, because it does take six hours to go from a concentration of one molar to a concentration of 0.5 molar. All right, there's one more part in this question. This is number 20, part E. Calculate the expected value for the concentration of hydrogen peroxide this experiment after eight hours have passed. Now that means we have to go an extra two hours in the table, but again this is not about half-life. The half-life is six hours, so eight hours is not a multiple of six. Let's use the integrated rate law equation for first-order kinetics. Let's plug in everything we know. The time is eight hours. The rate constant is 0.116 inverse hours. We know the natural log of the concentration at time zero. is equal to 0. So natural log of 1 is 0. And then now we have the natural log of the concentration at time t, which we don't know that value yet. So let's do the math. If the natural log of the concentration at 8 hours is equal to negative 0.928, the opposite of that would be e to the x. So e raised to the negative 0.928 gives us our answer. And that works out to be 0.395 moles per liter. So that is the answer to Part E. It's also the last part of the last question in this packet. Therefore, this is the end of the video. I hope that you found these answers and explanations helpful.