Unit 5 is often a stand alone unit and offers a unique perspective on reactions. Episode 57 discusses the questions 2010B - Question 6 and 2019 - Question 6.
Unit 5 is often a stand alone unit and offers a unique perspective on reactions. Episode 57 discusses the questions 2010B - Question 6 and 2019 - Question 6. These are released FRQs from previous exams and copyright of the College Board (0:44).
Question 6 of the 2010B exam focuses on the synthesis of hydrogen chloride from its elements (1:33). We are asked to determine the order of reaction for both reactants and justify the answer.
Question 6 of the 2019 exam looks at a decomposition reaction and the data collected is graphed in three graphs. We can use the graphs to identify the reaction as second order with respect to nitrogen dioxide and write the rate law. In part c we have two possible mechanisms, and have to determine if they are both consistent with the rate law. The first mechanism has a slow first step, the second mechanism has a slow second step and therefore requires substitution in the justification (7:43).
Question: What is the order of ALL half-life reactions on the AP Chemistry exam? A. zeroth order B. first order C. second order
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Unit 5 Free-response questions.
Let’s Zoom Out
The FRQ questions on the AP exam often combine content from two or three different units. In today’s episode, we are focusing on questions that require knowledge and skills from Unit 5: Kinetics. Unit 5 is quite the opposite of Unit 4: it often is a stand alone, but can also be found in combination with Thermochemistry and Equilibrium. In the curriculum, Unit 5 can pretty much be taught at any point in the year. However, it is our author Sarah’s favorite unit because it offers a unique perspective on reactions.
The questions we are using today are online accessible. Our suggestion: Answer the questions yourself and then listen to this episode to hear the explanations, as well as do’s and dont’s for answering questions of Unit 5. The questions used are: 2019 - Question 6 and 2010 Form B - Question 6. These are released FRQs from previous exams and copyright of the College Board.
Lets Zoom in:
We’ll start with 2010 - Form B - Question 6. This question refers to the reaction between gaseous hydrogen and gaseous chlorine to form gaseous hydrogen chloride. The question stem provides us with data showing three trials with initial concentration of reactants as well as the initial rate of formation of hydrogen chloride. In questions a) and b) we are asked to use the data to determine the rate of reaction with respect to each reactant. For this problem, we are using the method of initial rates and comparing how the change in concentration affects the rate of formation. For hydrogen we can compare trials 1 and 2, where chlorine is constant, and the concentration of hydrogen doubles. We can see that this leads to a doubling in the rate of formation. This means, the reaction is first order with respect to hydrogen. In the same manner, we can derive and justify the order of reaction with respect to chlorine. We compare trials 2 and 3, in which the concentration of hydrogen is held constant and the concentration of chlorine in 3 is half of the concentration of chlorine in 2. That change reflects in the initial rate of formation of product, which shows that trial 3 has half of the rate of formation than trial 2. This 1:1 relationship lets us determine that the order of the reaction with respect to chlorine is also first order.
Important with these types of questions is your justification. Refer to the data provided! Therefore, our overall rate law, which we are asked to write in c) is: rate = k [H2][Cl2]. Keep in mind: we don’t write the exponent “1” in Chemistry; both of them are first order. Also: Don’t forget: the rate constant is a LOWER case k. The equilibrium constant is an uppercase K. Part d) is also a speciality of Unit 5: Rarely, we are asked about or even earning points for units. So why do we use them anyway? Because: UNITS ARE FRIENDS! They help us set up our calculations and double-check our work. But in Unit 5, we can be asked about the units for the rate constant. You do not have to memorize it, you can easily derive it using your established rate law and solve for k. Focusing on the units we have: k equals rate over concentration of hydrogen times concentration of chlorine. That means: mol per liter second over mol per liter times mol per liter. Setting it up that way shows that mol per liter in the numerator cancels out with one set of mol per liter in the denominator. Therefore, our unit for k will be: liter per mol second.
In part e) we are asked to predict the initial rate of reaction with the given concentrations for hydrogen and chlorine. Comparing it with our data table, we can see that the concentration for Chlorine is the same as the concentration in trials 1 and 2, but the concentration given for hydrogen is three times higher than the concentration in trial 1. Since we know that double the concentration of hydrogen will double the initial rate, we can infer that tripling the concentration of hydrogen will triple the initial rate of formation given in trial 1. Therefore, our prediction would be 3 times 1.82 x 10-12 mol per liter seconds, which is 5.46 x 10-12 mol per liter seconds. You do not have to justify your answer, the point is earned for the correct numerical value.
The second part of this question uses gas-phase decomposition instead of synthesis: The decomposition of nitrous oxide. The reaction is represented as a two-step reaction. In the first part, (f), we are asked to write the balanced equation for the overall reaction. We can see that overall, 2 moles of nitrous oxide enter the reaction and yield 2 moles of diatomic nitrogen and 1 mole of diatomic oxygen. As a follow up, question (g) asks if the monatomic oxygen is a catalyst or an intermediate. Don’t forget: catalysts enter the reaction and are reformed so that you have them in the beginning and the end. Intermediates are formed during the reaction and used up. In this reaction, we can see that the monatomic oxygen is formed in step 1 and used in step 2. Therefore, it is an intermediate. Last, but not least, we are given a rate law and are asked to determine which step is rate-determining. We know that the slow step is rate-determining and therefore has to be consistent with the rate law. In this example, this is the case for the first step, in which we have nitrous oxide as a single reactant. This aligns with the rate law given as rate equals k times the concentration of nitrous oxide. 2010 - Form B - Question 6, parts a through g. The End.
The second question we are looking at is 2019 - Question 6. Here we have nitrogen dioxide that is decomposing to form nitrogen monoxide and diatomic oxygen. The concentration of nitrogen dioxide is monitored as it decomposes and the data is graphed. We are asked to explain how the graphs indicate that the reaction is second order. In those cases, we are looking for a LINEAR graph, which we have for the second graph. Be careful, READ the y-axis. Don’t think they are in order from zero to second order. The second graph has 1 over concentration of nitrogen dioxide, which is the y-axis for a second order graph. Therefore, in part (b) the rate law is: rate equal k times concentration of nitrogen dioxide squared. In part (c) we have two possible mechanisms. Here you now have labels showing you the slow and the fast step. Remember: the slow step is rate-determining and therefore should align with the rate law you proposed in (b). We can see that this is consistent, because we have two nitrogen dioxide molecules as reactants and therefore a rate law for the elementary step of “rate equal k times concentration of nitrogen dioxide times concentration of nitrogen dioxide” is consistent with b.
In the second part of (c) we have another mechanism and are again asked to evaluate if this is consistent with our rate law. Here, the second step is the slow step and it shows dinitrogen tetroxide. This seems confusing at first glance, but looking at the entire mechanism, you can see that the dinitrogen tetroxide is an intermediate, formed in the first step from 2 nitrogen dioxides. Here we need some knowledge from Unit 7, equilibrium. Intermediates cannot appear in the rate law, but the first step is in equilibrium. Therefore, we can write the equilibrium constant K to be concentration of dinitrogen tetroxide over concentration of nitrogen dioxide squared. Solving for the concentration of dinitrogen tetroxide, we get: equilibrium constant K times concentration of nitrogen dioxide. Now we can substitute this term in the rate law of the second step and get: rate is equal to rate constant times Equilibrium constant times the concentration of nitrogen dioxide squared, which is consistent with part (b). This substitution process can be a bit confusing, so make sure you practice it. We recommend using colored pencils to guide your work.
To recap…
Unit 5 offers us a unique perspective on chemical reactions by focusing on the kinetics of a reaction. A couple of key aspects when answering questions from unit 5: Practice justifying your answer using data as well as theoretical background - like when justifying how the mechanism corresponds with the rate law. In those cases, mention that the slow step is rate-determining. When writing rate laws, don’t forget the form: rate equals k times… and be careful: the rate constant is a lower case k. Units are friends, especially with these concepts. Make sure you always include them! Carefully read the y-axis to determine the order of a reaction with respect to a specific reactant. Don’t assume they are in the order from zero to second. Practice substitutions to justify mechanisms with a slow second step.
Coming up next on the APsolute RecAP Chemistry Edition: Energy of Phase Changes
Today’s Question of the day is about half-life.
What is the order of ALL half-life reactions on the AP Chemistry exam?