### Re: Quiz 2 Winter 2017

Posted:

**Tue Feb 21, 2017 9:29 pm**Are the formulas like lnA= lnA(0)-kt going to be given or do we have to remember them

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=160&t=19565

Page **2** of **2**

Posted: **Tue Feb 21, 2017 9:29 pm**

Are the formulas like lnA= lnA(0)-kt going to be given or do we have to remember them

Posted: **Tue Feb 21, 2017 10:00 pm**

chris2E wrote: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?

Number six's answer is 0.30M, which has two sig figs.

Posted: **Tue Feb 21, 2017 10:10 pm**

Can someone please explain #2?

Posted: **Tue Feb 21, 2017 10:17 pm**

For Quiz 2 Prep, how do you answer #1, where it asks to find the rate of formation of NO2 when the rate of decomposition of N2O5 is 2.89 mol/L*s?

Posted: **Tue Feb 21, 2017 10:17 pm**

604735966 wrote:As for number 7, when do we know to use the rearranged integrated rate law and when to use the differential rate law?

You use the integrated rate law when time is involved.

Posted: **Tue Feb 21, 2017 10:30 pm**

Yes, 40 minutes long with one point per minute given.

Posted: **Tue Feb 21, 2017 10:48 pm**

ariana_cruz_1C wrote: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.

ln[A] = -kt+ln[A](initial)

ln(0.622M) = -k(22min) + ln(0.773M)

ln(0.622M/0.773M) = -k(22min)

-0.00988min^(-1) = -k

k= 9.88 *10^(-3) min^(-1)

Posted: **Tue Feb 21, 2017 11:07 pm**

For number 5 wouldn't a lower activation energy make the rate of reaction more sensitive to an increase in temperature? because not as much thermal energy is needed to get the reaction going? Can someone clarify? thanks

Posted: **Tue Feb 21, 2017 11:11 pm**

chris2E wrote: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?

When dealing with natural logs, only the digits after the decimal point are counted as significant figures. Thus, 6.06 has two sig figs because of the two digits after the decimal point. The rules for sig figs can be found in the back of the course reader, as stated by Dr. Lavelle, on pages 122-123.

Posted: **Tue Feb 21, 2017 11:45 pm**

Tiffany Wee 2D wrote:This might be dumb question, but could someone quickly explain the difference between the reaction rate and the rate constant?

Reaction rate is defined as the change in concentration of one of the reactants or products divided by the time interval over which the change takes place. The rate constant, k, is the constant of proprortionality ina rate law; it is characteristic of the reaction (so different reactions have different rate constants).

The chart on this website helps clarify as well if the book definitions don't help as much!

http://www.thebigger.com/chemistry/chem ... -constant/

Posted: **Wed Feb 22, 2017 1:04 am**

So are we being quiz on knowing Activation energy? If so, does the equation ln[k2/k1]=Ea/R ((1/t2)-(1/T1)) work for all rate laws?

Posted: **Wed Feb 22, 2017 5:39 am**

Thanks for the explanation on #10

Posted: **Wed Feb 22, 2017 7:56 am**

christy_zambrano_2E wrote:ChristineCastanon_1B wrote:Sarah_Kremer_1A wrote:

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?

For this problem, it would only be 2 sig figs (because of .75M, .33M, and k=6.8 x10^-3). So even though it would normally round to 121 seconds, we only need 2 sig figs, making it 120 seconds. Hope that helps!

Posted: **Wed Feb 22, 2017 9:56 am**

Calderon_Humberto_3E wrote:What will the quiz cover? Is it the content of the midterm till now or is it cummalative since day one?

The quiz just covers kinetics. So from 15.1 to 15.11 or from the beginning of kinetics in the Course Reader to page 73.

Posted: **Wed Feb 22, 2017 10:47 am**

are we going to need to memorize the reaction order equations or will they be given to us on the formula sheet?

Posted: **Wed Feb 22, 2017 10:59 am**

This is a really basic question, but is it fine to switch the sign of a negative rate at the end? So, if my answer is negative, could I just make it positive for full credit, or do I have to set up the equation to get a positive answer from the very beginning?

Posted: **Wed Feb 22, 2017 11:16 am**

Priscilla_Covarrubias_HL wrote:For Quiz 2 Prep, how do you answer #1, where it asks to find the rate of formation of NO2 when the rate of decomposition of N2O5 is 2.89 mol/L*s?

I believe this should be correct:

So the reaction you are given is: N2O5(g) → NO2(g) + O2(g) but it is not balanced. After balancing the reaction, you should get 2 N2O5(g) → 4 NO2(g) + O2(g)

The decomposition rate of one mol of N2O5 is 2.89 L^-1 s^-1. But after balancing the reaction, you have 2 mol of the reactant (N2O5), so you multiply the decomposition rate by 2 as well. This should give you +5.78 (mol N2O5) L^-1 s^-1.

Posted: **Wed Feb 22, 2017 11:19 am**

andrea mejia wrote:are we going to need to memorize the reaction order equations or will they be given to us on the formula sheet?

I think that we'll be provided with the same formulas as those on the formula sheet.

Posted: **Wed Feb 22, 2017 12:59 pm**

chris2E wrote: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?

I think the rule with logarithms and, by extension, natural logs, is to count out the number of sig figs in decimals.

e.g. if we're solving log

(I think.)

Posted: **Wed Feb 22, 2017 1:03 pm**

Is there a question about the differential rate law on the quiz?

Posted: **Wed Feb 22, 2017 3:00 pm**

604735966 wrote:As for number 7, when do we know to use the rearranged integrated rate law and when to use the differential rate law?

I think the differential rate law talks about the overall rate while the Integrated Rate Law talks about rate as a function of time (usually a particular moment in time). So,when time is involved I think we generally use the integrated rate law.

Posted: **Wed Feb 22, 2017 4:37 pm**

AnnaTong1E wrote: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.

I believe Chemistry Community is the only place you can find the practice quiz 2 answers and detailed explanations!

Posted: **Wed Feb 22, 2017 8:22 pm**

#10 is false because it is a second order reaction, correct? To figure it out, do you have to solve the second order reaction equation for time, or do you just know?

Posted: **Wed Feb 22, 2017 8:30 pm**

For #8, how do you get your units to become 6.06 minutes?

I know you have to use the first-order integrated rate law to find the time.

I did ln(0.35 M/1.50 M)/-4.00*10^-3 s^-1

Is this correct?

I know you have to use the first-order integrated rate law to find the time.

I did ln(0.35 M/1.50 M)/-4.00*10^-3 s^-1

Is this correct?

Posted: **Wed Feb 22, 2017 8:40 pm**

604735966 wrote:As for number 7, when do we know to use the rearranged integrated rate law and when to use the differential rate law?

When an initial concentration is given along with its concentration at a given time, the integrated rate law is easiest to use in order to find the constant "k" of the reaction.

Posted: **Wed Feb 22, 2017 10:05 pm**

I'm not sure why I keep getting this wrong but for number 6 I keep getting 1.66 for the value of k. Could somebody please help?

Posted: **Wed Feb 22, 2017 10:48 pm**

I was wondering the same thing ^ (sig figs in #6)

Posted: **Wed Feb 22, 2017 11:06 pm**

Kira_Maszewski_1B wrote: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?

I think the rate units are always M*s^-1, but the units of rate constant changes according to the order of the reaction so that the units always cancel out and equal to the rate's unit (M*s^-1)

Posted: **Wed Feb 22, 2017 11:18 pm**

Could somebody explain #5 to me? Why is it that the temperature doesn't change the rate much?

Posted: **Wed Feb 22, 2017 11:52 pm**

SarahTian3k wrote:Could someone please explain number 7? I keep getting 7.46x10-3 min-1

I think you accidentally plugged in 0.733 M instead of 0.773 M.

Posted: **Wed Feb 22, 2017 11:58 pm**

DanielaWong_2G wrote:I'm not sure why I keep getting this wrong but for number 6 I keep getting 1.66 for the value of k. Could somebody please help?

For #6, you have to use the 2nd order integrated rate law and just plug everything in:

1/[0.44M] = k(11 min) + 1/[.95M].

Then subtract 1/[.95M] on both sides to get 1.22= 11k and isolate the k to get 0.11 for the rate constant.

Posted: **Thu Feb 23, 2017 12:15 am**

Hirmand_Sarafian_2N wrote:Can somebody explain to me why 10 on the quiz 2 prep is false?

This reaction is second order because you are given k=[A]^2.

If you look at the equation we derived in class for the half-life of a second order reacion (which is also on the formula sheet), you will find t(1/2) = 1/(k*[A] initial). This equation tells us that the half-life of a second order reaction changes as [A] initial changes. Therefore, the half-life for [A] decreasing from 1.00M to 0.50M is shorter than the half-life for [A] decreasing from 0.50M to 0.25M ( [A initial] is in the denominator for the half life expression so a higher [A initial] means a shorter half-life.

Posted: **Thu Feb 23, 2017 2:37 am**

SarahTian3k wrote:Could someone please explain number 7? I keep getting 7.46x10-3 min-1

Are you positive you're using the correct order equation? For first-order reactions it's ln[A] = -kt + ln[A

Posted: **Thu Feb 23, 2017 11:51 am**

Byron_Corado_3D wrote:Could somebody explain #5 to me? Why is it that the temperature doesn't change the rate much?

Dr. Lavelle said that we don't need to know how to do #5. But to answer your question, refer to page 75 on the course reader and look at the bottom graph. Reactions with low activation energies have flatter slopes than reactions with high activation energies on a plot of 1/T vs. ln k. Therefore, the rates of reactions with low activation energies don't change as much when the temperature changes.

Posted: **Thu Feb 23, 2017 11:53 am**

TramHo2G wrote:For #8, how do you get your units to become 6.06 minutes?

I know you have to use the first-order integrated rate law to find the time.

I did ln(0.35 M/1.50 M)/-4.00*10^-3 s^-1

Is this correct?

Yes your equation is correct. Look at the rate constant units, it's expressed in seconds. So the answer you get, which should be around 364, is in terms of seconds. You just divide 364/60 to get 6.06 minutes.

Posted: **Thu Feb 23, 2017 5:41 pm**

ariana_cruz_1C wrote: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.

Immediately you know that this problem is dealing w first order (given). The goal is to find the rate constant (k) and you are given both the concentration of A, the initial concentration of A, and the time (22.0 mins). You use the equation of the integrated rate law for first order and with the given information listed (also in prev. sentence) you plug in each into the corresponding part of the equation.

If you're not getting the right answer perhaps you are using the wrong integrated rate law, or maybe you are plugging in the concentrations in the wrong place? The initial is the starting concentration [0.773 M] and it goes into the [A]o spot. Maybe units but still I don't see how units can be an issue?

Posted: **Thu Feb 23, 2017 5:43 pm**

For number 4 , the rate constant was 2.00s-1. Is the rate constant always positive ?

Posted: **Thu Feb 23, 2017 7:35 pm**

Tiffany Wee 2D wrote:This might be dumb question, but could someone quickly explain the difference between the reaction rate and the rate constant?

The reaction rate is the rate at which the reaction proceeds, dependent on the rate constant (k) and the concentrations of the reactants. The rate constant is a an independent variable that affects the reaction rate.

Posted: **Thu Feb 23, 2017 8:04 pm**

Do we have to know how to draw any graphs for quiz 2?

Posted: **Thu Feb 23, 2017 8:22 pm**

Number 6's answer is 0.30M, that's 2 sig figs. And there is no ln involved

Posted: **Thu Feb 23, 2017 8:42 pm**

On question 6 of the practice quiz, do you need to convert the minutes given to seconds in order to solve the problem?

Posted: **Thu Feb 23, 2017 8:58 pm**

Which equation would you use for number 11 on the practice quiz? I know it is a first order reaction, but I'm not sure which formula to use to solve what the question is asking for.

Posted: **Thu Feb 23, 2017 11:31 pm**

Madeline Fox 2B wrote:Will we have to derive equations for quiz 2?

yes, possibly

Posted: **Fri Feb 24, 2017 12:01 am**

On the quiz if mmol's are given do you have to change it to mols or can you keep it this way? Also on number 8 on the practice quiz do you have to change the answer to minutes or if on the actual quiz if I were to keep it in seconds would it still be correct?

Posted: **Fri Feb 24, 2017 1:04 pm**

Problem 8

the answer you gave has 3 sig figs but when I look at the values given there is a 0.35 M... to my understanding, any 0 before a decimal does not count as a sig fig. Therefore, the answer should be 6.6 min not 6.06 min

please advise

the answer you gave has 3 sig figs but when I look at the values given there is a 0.35 M... to my understanding, any 0 before a decimal does not count as a sig fig. Therefore, the answer should be 6.6 min not 6.06 min

please advise

Posted: **Fri Feb 24, 2017 1:10 pm**

danny_nguyen_3C wrote:nlc_lec1 wrote:Chem_Mod wrote:

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.

I would like to add that C is also not true. Not all enzymes are proteins, some are RNA molecules. I think where this question is poorly worded. But all in all, D is MORE WRONG than C

Posted: **Sat Feb 25, 2017 8:26 am**

Hi,

How do you do number 11?

Thanks!

How do you do number 11?

Thanks!

Posted: **Sat Feb 25, 2017 10:20 pm**

SIG FIG questions.

See page 123 in the course reader.

When taking log of a number with 2 sig fig the answer is reported with 2 decimal places. Otherwise taking the antilog of the answer with one decimal place does yet result in the original (correct) 2 sig fig number.

Therefore taking log of number with 2 sig fig results in answer with 2 decimal places (and 3 sig fig).

See page 123 in the course reader.

See page 123 in the course reader.

When taking log of a number with 2 sig fig the answer is reported with 2 decimal places. Otherwise taking the antilog of the answer with one decimal place does yet result in the original (correct) 2 sig fig number.

Therefore taking log of number with 2 sig fig results in answer with 2 decimal places (and 3 sig fig).

See page 123 in the course reader.

Posted: **Sat Feb 25, 2017 10:22 pm**

For those with reaction mechanism questions, see page 652 in the textbook.

Posted: **Sun Feb 26, 2017 10:05 pm**

This may have been asked before, but for the quiz, are the two points for correct units and two points for significant figures per question or for the whole quiz?

Posted: **Sun Feb 26, 2017 11:42 pm**

0.30 is only two sig figs.

Posted: **Sun Feb 26, 2017 11:56 pm**

z_hernandez_loza_3J wrote:This may have been asked before, but for the quiz, are the two points for correct units and two points for significant figures per question or for the whole quiz?

For the whole quiz!

Posted: **Fri Mar 03, 2017 11:21 am**

Can someone who received full credit on the question where we had to label the graph please post a picture of their answer? I am having trouble understanding why I only received 3/5 points.

Posted: **Fri Mar 03, 2017 2:03 pm**

Will the quiz answers be posted?

Posted: **Fri Mar 03, 2017 4:47 pm**

My TA told me today that there are some weird rules that apply to sig figs with logs and natural logs and that we do not need to know them so dont worry about sig figs

Posted: **Sat Mar 04, 2017 3:16 pm**

What was the average for Quiz 3?

Posted: **Sun Mar 05, 2017 12:10 pm**

For Quiz 2, does ln(A) represent the final concentration (what's left over) or does it represent the change in concentration, and then you would have to subtract from the original amount to get the final concentration?

Posted: **Mon Mar 06, 2017 8:25 pm**

Will the answers of quiz 2 (the actal quiz) be posted?

Posted: **Sun Mar 12, 2017 9:57 pm**

For number 11 on quiz 2 preparation, how do you use the rate constants and temperatures to find the rate constant at another temperature?

Posted: **Fri Mar 17, 2017 7:53 pm**

Are the answers posted on Chem Community?

Posted: **Sun Mar 19, 2017 11:53 am**

If given an initial concentration and asked to solve for it in a second order reaction do we need to take the e of the ln or are we answering leaving it in 1/A?