Quiz 2 Winter 2017

[Chem_Mod]

Quiz 2 covers only Kinetics up to the end of page 73 in the course reader.

[Derek Lee 2K]

Does quiz 2 just cover up to the end of kinetics? Is it cumulative?

[Maya Kandah 3M]

From the quiz 2 preparation in the course reader, it seems that it will just cover kinetics.

[Melanie Martinez 1K]

How many questions will be on quiz 2?

[Chem_Mod]

We are still working on quiz 2 (and working our tails off as the midterm was last night, and answering questions on Chemistry Community, etc., got home long after midnight).

[Chem_Mod]

Quiz 2 has 6 questions.

[Sarai_Ferrer_3C]

Hey guys, so does that mean that the concepts generally covered on the quiz include ad follows: 1st order, second order, third order, and 1/2 life? Thanks in advance for the clarification.

[Winnie Liu 1M]

Yup! That's what is outlined in the coursereader :D

[704709603]

Would it be possible to get the answers to the practice quiz please?

[Kaitlin_Ryan_ 3K]

Will answers to the practice Quiz 2 in the back of the course reader be posted to Chemistry Community as the answers to practice Quiz 1 were?

[Nikhita_Jaaswal_2H]

Will Quiz 2 also cover materials from before the midterm (in between Quiz 1 and midterm)?

[Katie McCombs 3G]

Will Quiz 2 also cover materials from before the midterm (in between Quiz 1 and midterm)?

No quiz 2 will only cover kinetics

[Chem_Mod]

Quiz 2 Answers

1. 2 N₂O₅(g) → 4 NO₂(g) + O₂(g)
5.78 (mol NO2)⋅L^–1⋅s^–1

2. False

3. 2 N₂O(g) → 2N₂(g) + O₂(g)
120 sec

4. first-order
2.00 s^–1

5. has a rate that does not change much with temperature.

6. 0.30 M

7. 9.88×10⁻³ min⁻¹ or 1.65 ×10⁻⁴ s⁻¹

8. 6.06 min

9. D

10. false

11. 100 min⁻¹

[Brian Huynh 3D]

How is question one done from quiz 2?

[704709603]

Can some one explain question 1 and 5 please? Also, can we do calculations with any unit of time?
Thank you!

[Diwana Lucero 3K]

How is question one done from quiz 2?

When you balance the reaction, you get 2 mol N2O5, so you multiply the rate of decomposition of N2O5, 2.89, by 2 to get the rate of formation of NO2. So, this would be 2(2.89)= 5.78 (mol NO2)⋅L–1⋅s–1

[nicoleclarke_lec1M]

Do we need to know how to do #9 and #11 from Quiz 2 Prep?

[Annie Tagvoryan 3A]

I know the quiz covers material up to page 73 in the course reader, but does that correspond to up to 15.11 in our textbook?

Thanks.

[Ariana de Souza 4C]

for number 8, is it okay if we express it in seconds, since the question gave the rate constant in seconds?
Additionally, do we need to know how to do numbers 5, 9, and 11 ? They all have to do with enzymes and the arrhenius equation, and i thought we didn't have to know that

[Leo_Navejas_3B]

For rxn mechanisms will we have to infer the intermediate for a rxn or will it be given to us?

[Ariana de Souza 4C]

For rxn mechanisms will we have to infer the intermediate for a rxn or will it be given to us?

Pretty sure the mechanism will be given to us. The intermediate will be the compound that is in the products of the first reaction and in the reactants of the second, and so will be canceled out -- it should be clear what it is

[Diana_Anum1G]

I know that quiz 2 covers up to page 73 in the course reader, but where does it cover up to in the homework problems? Is it up to 15.55 because 61 starts talking about activation energy?

[Madison_Moore_1I]

Could someone explain to me how to do #6?

[Ariana de Souza 4C]

Could someone explain to me how to do #6?

So this one you need to split into two problems. The first step will be to determine what k is.
1) you have your initial concentration (0.95 M) and your final concentration (0.44 M) and the time it takes (11 minutes)
you use your equation 1/(At) - 1/(A0) = k*t, and from there, find k.

2) now you know k. You now need your final concentration. You have your initial concentration (still 0.95 M) and your k and your time (21 minutes). So you plug everything into that same second order integrated equation, and you find At.
Hope that helps! :)

[Chem_Mod]

Do we need to know how to do #9 and #11 from Quiz 2 Prep?

9 yes
11 no

[Chem_Mod]

for number 8, is it okay if we express it in seconds, since the question gave the rate constant in seconds?
Additionally, do we need to know how to do numbers 5, 9, and 11 ? They all have to do with enzymes and the arrhenius equation, and i thought we didn't have to know that

8 okay to give answer in seconds

5 no
9 yes (see comments on zero order reactions)
11 no

[Mannat Sukhija 1E]

Does anyone know on the dates of the first and last lecture that will be covered on Quiz 2?

[Abigail_Low_1A]

I know that quiz 2 covers up to page 73 in the course reader, but where does it cover up to in the homework problems? Is it up to 15.55 because 61 starts talking about activation energy?

Yes, I think so. That's where I stopped.

[Morgan_McMasters_3A]

For #1, I used the unique rates with stoichiometric coefficients and set it up as -(1/a)(deltaA/deltat)=(1/c)(deltaC/deltat) and got -(1/2)(2.98)=(1/4)(x) would this be incorrect? I got -5.78mol NO2/L/s and I noticed the answer is +5.78.

[Calderon_Humberto_3E]

What will the quiz cover? Is it the content of the midterm till now or is it cummalative since day one?

[Melanie Martinez 1K]

What will the quiz cover? Is it the content of the midterm till now or is it cummalative since day one?

The quiz only covers Kinetics, which up to the end of page 73 in the course reader.

[Chem_Mod]

For #1, I used the unique rates with stoichiometric coefficients and set it up as -(1/a)(deltaA/deltat)=(1/c)(deltaC/deltat) and got -(1/2)(2.98)=(1/4)(x) would this be incorrect? I got -5.78mol NO2/L/s and I noticed the answer is +5.78.

Rates are always positive values.

[204177206]

Will Quiz 2 be 40 minutes as well, and of similar format as Quiz 1?

[Madeline Fox 2B]

Will we have to derive equations for quiz 2?

[Jaret Nishikawa]

Will it ask us to use specifically to use pre-equilibrium approach for reaction mechanisms or can we use either pre-equilibrium or steady state? Thanks

[104781135]

Hi everyone and sorry, this might be a dumb question, but will we need to know page 65 in the course reader? I don't want to be surprised if radioactivity shows up. Thank you so much!

[Jerry_Dar_3C]

Will we be quizzed on any reactions past second order? Or do we only need to know up to second?

[nicoleclarke_lec1M]

Do we need to know how to do #9 and #11 from Quiz 2 Prep?

9 yes
11 no

could you please explain #9? It has to do with enzymes and I don't really know anything about them.

[904621096]

What is the unique rate of reaction?

[Yasmine Choroomi 2C]

Hi! I know you said we need to know number 9, but it has to do with enzymes and I thought we didn't need to know that?

Also are there any examples of using the equilibrium approach problems in the book?

Thanks :)

[Kira_Maszewski_1B]

It is important to distinguish between the reaction and the rate constant. The units of rate constant are always M * s^-1. True or False.

I think the answer is True, however I'm not sure about this question. Could someone please explain?

[Arshia_Sabir_3E]

Can someone please help me with #3? I used the integrated rate law for first order and I got t=205 seconds

[Christian Hardoy 3F]

It is important to distinguish between the reaction and the rate constant. The units of rate constant are always M * s^-1. True or False.

I think the answer is True, however I'm not sure about this question. Could someone please explain?

The rate constant can be M * s^-1 for zero order, s^-1 for first order, and M^-1 * s^-1 for 2nd order. The units of the rate are constant; it will be M * s^-1 for zero order, first order, and second order. Because the units of the rate are constant, k must change units when the reaction is a different order because there is another term in a 2nd order rate as compared to a 1st order rate. 1st order: Rate = k*[A], 2nd order: Rate=k*[A]^2. If the 2nd order rate is in M/s and the 1st order rate is in M/s, then k must be different because the 2nd order rate has an additional [A], which is in M.

[Sarah_Kremer_1A]

Can someone please help me with #3? I used the integrated rate law for first order and I got t=205 seconds

For #3, it is a first order reaction, so you use the corresponding equation
ln[A] = -kt + ln[A0]
so, ln(.33) = -(.0068)t + ln(.75)
ln(.33/.75) = -.0068t
t = ln(.33/.75)/-.0068
t = 120 sec

[Morgan_McMasters_3A]

What parts of chapter 15 will be covered in the quiz? Only until 15.10 or..?

[ChristineCastanon_1B]

Can someone please help me with #3? I used the integrated rate law for first order and I got t=205 seconds

For #3, it is a first order reaction, so you use the corresponding equation
ln[A] = -kt + ln[A0]
so, ln(.33) = -(.0068)t + ln(.75)
ln(.33/.75) = -.0068t
t = ln(.33/.75)/-.0068
t = 120 sec

Thank you! I had the same question

[eric_beraut_2B]

Does anyone know on the dates of the first and last lecture that will be covered on Quiz 2?

Quiz two covers the lectures from Monday 2/6/17 up to the lecture on Friday 2/17/17

[BrianaBarr2A]

Will Quiz 2 be 40 minutes as well, and of similar format as Quiz 1?

Yes this is correct, Dr. Lavelle formats all his quizzes this way.

[Andrea Medina 1A]

Can someone explain the graphing part of question 4 on the practice quiz?

[danny_nguyen_3C]

If the answer to #2 if false, what other possible units can be used to signify rate constant?

[Ariana de Souza 4C]

Can someone explain the graphing part of question 4 on the practice quiz?

It tells you that you have a graph where you're plotting the natural log of NOBr against time, and you get a straight line. on page 63 in the course reader, you can see that that's basically how you know that the reaction is first order. You can also check this with the equation
ln (A) = -kt + ln(A)
from here you can see the formula of a line, where -k is the slope and ln(A) is the y intercept. Therefore, ln(A) must be what you plot against t.

The other part asks that if you have a slope of -2, what k must be. Since the slope of the graph would be -k, and -k = -2 /s you can see that k=2/s

[Ariana de Souza 4C]

If the answer to #2 if false, what other possible units can be used to signify rate constant?

It depends on what order the reaction is. you can just check this with the rate equations.

ZERO ORDER:
Rate=k
the units of the rate are M/s
so k must be M/s too for zero order reactions

FIRST ORDER:
Rate = k[A]
M/s = ? (M)
so k must be /s

SECOND ORDER:
Rate = k[A]^2
M/S = ? * (M^2)
so k must be 1/s*M

[ChristineCastanon_1B]

Could anybody briefly explain how you did number 11?

[Ann Zhang_1M]

how many questions are included in quiz2?

[Madeline Fox 2B]

Can someone explain why question 10 is false? I thought half-lives are always the same.

[Chem_Mod]

how many questions are included in quiz2?

See above answer.

[Chem_Mod]

Will Quiz 2 be 40 minutes as well, and of similar format as Quiz 1?

All quizzes are 40 minutes. See Syllabus.

[Chem_Mod]

Will we have to derive equations for quiz 2?

Derivations are not asked on quizzes.

[Chem_Mod]

Will it ask us to use specifically to use pre-equilibrium approach for reaction mechanisms or can we use either pre-equilibrium or steady state? Thanks

You can use either approach to solve a reaction mechanism problem.
However, as I discussed in class you are only expected to know the pre-equilibrium approach.

[Chem_Mod]

Hi everyone and sorry, this might be a dumb question, but will we need to know page 65 in the course reader? I don't want to be surprised if radioactivity shows up. Thank you so much!

Page 65 will not be asked. Since you are Life Science majors I include it for students to see the biological relevance.

[Chem_Mod]

Will we be quizzed on any reactions past second order? Or do we only need to know up to second?

That would be super mean and 'bad' chemistry if I were to ask students topics we have not covered.
In class we covered: zero, first, and second order reactions

[Flamingos 3L]

Up to what chapter in the textbook are advised to study up to for this quiz? I see that the Arrhenius equation shows up during 15.11 so are we advised to just study the chapters before that?

[Chem_Mod]

Up to what chapter in the textbook are advised to study up to for this quiz? I see that the Arrhenius equation shows up during 15.11 so are we advised to just study the chapters before that?

All of kinetics except the Arrhenius equation and catalysis.

[Ashley Van Belle 2B]

What problems out of the textbook given in the course reader would be sufficient in covering kinetics? Because some of the problems begin to cover activation energies, which we are not being tested on in this quiz. Would it be up until 5.55?

[Jonathan Orozco 1A]

Can anyone help me with number 6 from Quiz two preparation? I don't understand how the answer is 0.30 M
I used the second order integrated Rate law, and my answer for K was 0.11092 i was wondering if i am going in the right direction and if so can you help me finish peals and thank you very much.

[Chem_Mod]

Monday night of the holiday weekend and I don't have the textbook on me.
Just omit any questions related to the Arrhenius equation and catalysis.

[Imani Johnson 1H]

Do we have to know how to use the pre-equilibrium approach since that is where we left off or no?

Thanks.

[Shelby Slaughter 3D]

Will we need to know the details of radioactive decay?

[Henry_Shin_3B]

Will we need to know the details of radioactive decay?

No, I do not believe so. I think the radioactive decay thing was just an example to show what real-life scenarios the concept can be applied.

[Chem_Mod]

Do we have to know how to use the pre-equilibrium approach since that is where we left off or no?

Thanks.

Yes you do. See previous Q&A on this.

[Natalie Hunt 1E]

Can anyone help me with number 6 from Quiz two preparation? I don't understand how the answer is 0.30 M
I used the second order integrated Rate law, and my answer for K was 0.11092 i was wondering if i am going in the right direction and if so can you help me finish peals and thank you very much.

Yes, you are going in the right direction! To finish the problem, you use the second order integrated rate law again with [Ao] = 0.95 M, k=0.11092 (which is what you just found), and t = 21 minutes. You then solve for [At] to get 0.30 M.

[Natalie Hunt 1E]

Will we need to know the details of radioactive decay?

No, I do not believe so. I think the radioactive decay thing was just an example to show what real-life scenarios the concept can be applied.

No, we do not need to know that information. As Dr. Lavelle said in an earlier post, "Page 65 will not be asked. Since you are Life Science majors I include it for students to see the biological relevance."

[Chem_Mod]

Thank you Natalie. Introduce yourself at the end of class.

[ben maryamian]

which homework problems will the quiz cover?

[Simone Seliger 1C]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I am also confused by this question. I know that the equation is second-order, but do not understand why the decomposition would take different amounts of time.

[rebecca_handler_1B]

Can someone explain why question 10 is false? I thought half-lives are always the same.

You know this one is a second order reaction because rate=[A]^2, and if you set up two equations using the 2nd order integrated rate law (one for 1.0 M and 0.5 M and the other for 0.5 M and 0.25 M), you can solve for t knowing k is constant. Once you solve, you can see that t will change values.

[Terry1A]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I am also confused by this question. I know that the equation is second-order, but do not understand why the decomposition would take different amounts of time.

For number 10 you use the integrated rate law for the second order reaction which is 1/[R]=kt+1/[R]0. You can then manipulate the equation to solve for time, so t=(1/[R]-1/[R]0)/k. The k constant will be the same whether the concentration is decreasing from 1 M to 0.5 M or 0.5 M to 0.25 M. If you plug in the concentrations, you'll get t=1/k for 1 M --> 0.5 M, and t=2/k for 0.5 M --> 0.25 M. This means that the time it takes for the concentrations to decrease are both different, therefore the statement is false.

[danny_nguyen_3C]

Do we need to know how to do #9 and #11 from Quiz 2 Prep?

9 yes
11 no

could you please explain #9? It has to do with enzymes and I don't really know anything about them.

It is not possible to explain everything about enzymes, but I will try to explain why the answer is D. First of all, enzymes are catalysts, meaning they increase the reaction rates of chemical reactions. A and B speak about the concentration of substrate. What is important to know is that the rate of reaction of the enzyme-catalyzed reaction depends on the amount of substrate (reactants) you are given. This means that as you are given more substrate, enzymes become "active" in order to breakdown more of the substrate. On the other hand, if you are given less substrate, the enzymes aren't required to catalyze as many reactions than if there were large amount of substrates. I cannot explain C just because you need to know that all enzymes are proteins, but not necessarily the other way around (just like how a square is a rectangle, but a rectangle is not a square). Therefore, the answer is D.

[Aditya Mamtora 2C]

Could anybody briefly explain how you did number 11?

We do not need to know how do do #11 for Quiz #2

[Angela To 2B]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I think half-lives are the same for only 1st order reactions

[Veronica_Giap_1C]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I think half-lives are the same for only 1st order reactions

To expand on this, if you look at the half reactions formulas, you will see how both the 0th order rxn and the 2nd order rxn are dependent on the initial reactant concentration. So when the quiz problem gives you two ranges of concentrations [1.0,0.50] and [0.50,0.25], it is in a way giving you a "new" initial concentration for the 2nd equation and because the initial concentrations are different, your half life values will be too: t_1/2=1/k[R]₀

[Katelyn_Nakamura_1B]

Can anyone help me with number 6 from Quiz two preparation? I don't understand how the answer is 0.30 M
I used the second order integrated Rate law, and my answer for K was 0.11092 i was wondering if i am going in the right direction and if so can you help me finish peals and thank you very much.

So this problem you need to split into two parts. You use the same equation but with diff #s

First part of the question you need to find k. With what you're given you know that it is a 2nd order and you have time involved (that is not in regards to half life) so you know to use the integrated rate law. You plug in all the numbers known to get k which you found was correct.

With k known you can use the same equation and plug in the new numbers given ([A]initial is known because that does not change). The final answer is rounded to .30M using the sig fig rules.

I hope this helped :)

[Jennifer_Lam_1A]

Can someone please explain #5 on the practice quiz? I thought the answer would be A because you have to take into account temperature when finding the activation energy.

[704709603]

For the pre-equilibrium approach: What if there are more than two steps in the mechanism? Also, how did you go from rate = k1 [CHBr3] ((k2[Br2])/k' 2) to rate = k [CHBr3][Br2]?

[Jordan Pace 2B]

For #11:
I'm not 100% sure this is how to approach this problem, but this is what a friend and I did to solve it!

ln k2/k1= Ea/8.314 [1/T1 - 1/T2]

k2=35 min^-1
k1=25 min^-1
T1=298
T2=350

Solve for Ea

0.336= 0.000498Ea/8.314

Ea=5617.3

Now find rate constant at 770K

ln x/25= 5617.3/8.314 [1/298 - 1/770]

ln x/25 = 1.3898

x=25e^1.38, so x=99.37 min^-1 or 100 min^-1

Hope this helps! If this isn't the correct way to solve this problem, please feel free to correct it!

[ariana_cruz_1C]

Hi, can someone show me how they did #7, I know you use the first order integrated rate law, but when I work it out I am not getting the answer that is posted.

[Chem_Mod]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I think half-lives are the same for only 1st order reactions

To expand on this, if you look at the half reactions formulas, you will see how both the 0th order rxn and the 2nd order rxn are dependent on the initial reactant concentration. So when the quiz problem gives you two ranges of concentrations [1.0,0.50] and [0.50,0.25], it is in a way giving you a "new" initial concentration for the 2nd equation and because the initial concentrations are different, your half life values will be too: t_1/2=1/k[R]₀

Correct. Half-life formulas with [A]o depend on the initial concentration (as discussed in class). See pages 64, 66, and 67 in the lecture notes.

[ariana_cruz_1C]

Hi, can someone show me how they did #7, I know you use the first order integrated rate law, but when I work it out I am not getting the answer that is posted.

nevermind!

[Diana Catapusan 1N]

Hi, can someone show me how they did #7, I know you use the first order integrated rate law, but when I work it out I am not getting the answer that is posted.

First, you need to rearrange the first order integrated rate law and isolate k. (k=ln(A/A₀)/t) After, plug in each value. k=ln(0.622M/0.773M)/22.0 min, which should give you 9.88x10^-3min^-1. Hope this helps, I'm sorry I couldn't figure out how to upload a picture!

[Tiffany Wee 2D]

This might be dumb question, but could someone quickly explain the difference between the reaction rate and the rate constant?

[AnnaTong1E]

Is there any place that all of the course reader quiz 2 answers (written out, long, and explanatory) are compiled in one place? I know there are the answers on this site.

[Melanie_Wong_1K]

How is question one done from quiz 2?

When you balance the reaction, you get 2 mol N2O5, so you multiply the rate of decomposition of N2O5, 2.89, by 2 to get the rate of formation of NO2. So, this would be 2(2.89)= 5.78 (mol NO2)⋅L–1⋅s–1

Does this mean the coefficient in front of the reactants doesn't matter for this question?

[Mizuno_Mikaela_1D]

Can someone explain why question 10 is false? I thought half-lives are always the same.

I am also confused by this question. I know that the equation is second-order, but do not understand why the decomposition would take different amounts of time.

Since it is second order, the equation for t_1/2 is t_1/2 = 1/([A]₀ *k) which means it depends on [A]₀ . Since in #10, the half lives you are comparing have different initial concentrations, the half life is different. Initial concentration also affects goes a zero order reaction but not a first order.

[SarahTian3k]

Could someone please explain number 7? I keep getting 7.46x10-3 min-1

[Hirmand_Sarafian_2N]

Can somebody explain to me why 10 on the quiz 2 prep is false?

[604735966]

Can someone please explain number 7! I know it's a straight forward problem, but I must be missing a step because I can't seem to get the correct answer.

[604735966]

As for number 7, when do we know to use the rearranged integrated rate law and when to use the differential rate law?

[Amy_Hoang_1E]

Can someone please explain number 7! I know it's a straight forward problem, but I must be missing a step because I can't seem to get the correct answer.

So for number 7, you can use the 1st order integrated rate law, which is ln[A] = -kt + ln [A]initial and then isolate k, since you're solving for k
ln [A] is 0.622M
ln [A] initial is 0.773 M
t is 22 mins
and you're solving for k

all plugged in, it should be ln 0.622= -k (22) + ln 0.773
so, - (ln 0.622 - ln 0.773)/ 22 = k
then it should be straight algebra after that. The answer I got is 0.009877, which is the same as 9.88 x 10^-3

[christy_zambrano_2E]

Can someone please help me with #3? I used the integrated rate law for first order and I got t=205 seconds

For #3, it is a first order reaction, so you use the corresponding equation
ln[A] = -kt + ln[A0]
so, ln(.33) = -(.0068)t + ln(.75)
ln(.33/.75) = -.0068t
t = ln(.33/.75)/-.0068
t = 120 sec

Thank you! I had the same question

the answer I got was 120.73... are we not suppose to round to three sig figs? So wouldn't it be 121 sec?

[chris2E]

Is there a reason why number six's answer has three significant figures despite the given concentration 0.35M having two significant figures? Do we not bother with sig figs when ln is involved?