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Posted: Tue Feb 12, 2019 10:22 pm
Celine Hoh 2L wrote:
Karyn Nguyen 1K wrote:For worksheet 4, is the answer for 6a wrong? I did (3/2)(31.9 mol)(8.314 J/Kmol)(311.15 K) = 124 kJ. When I didn't convert the temperature from C to K I got 15.1 kJ which is the answer given in the key.

We don’t have to convert it to kelvin as it is change in temperature (38-0)

Posted: Tue Feb 12, 2019 10:24 pm
Celine Hoh 2L wrote:
Karyn Nguyen 1K wrote:For worksheet 4, is the answer for 6a wrong? I did (3/2)(31.9 mol)(8.314 J/Kmol)(311.15 K) = 124 kJ. When I didn't convert the temperature from C to K I got 15.1 kJ which is the answer given in the key.

We don’t have to convert it to kelvin as it is change in temperature (38-0)

Don't we need to convert it to K to cancel out the units and end up with J?

Posted: Wed Feb 13, 2019 12:13 am
Celine Hoh 2L wrote:Quick question for Worksheet 1 ques8:

When I2 is added, neither products nor reactants is favored as I2 is a solid.
What happens when I2 is removed, does the reaction not shift too?

Yeah, since it is a solid it doesn't affect it either way

Posted: Wed Feb 13, 2019 1:48 pm
Karyn Nguyen 1K wrote:
Celine Hoh 2L wrote:
Karyn Nguyen 1K wrote:For worksheet 4, is the answer for 6a wrong? I did (3/2)(31.9 mol)(8.314 J/Kmol)(311.15 K) = 124 kJ. When I didn't convert the temperature from C to K I got 15.1 kJ which is the answer given in the key.

We don’t have to convert it to kelvin as it is change in temperature (38-0)

Don't we need to convert it to K to cancel out the units and end up with J?

We don't need to convert to K because the temperature difference in terms of celsius and in terms of kelvin are the same value.

Posted: Wed Feb 13, 2019 1:51 pm
Is there answer where it shows the work for these worksheets.

Posted: Wed Feb 13, 2019 3:36 pm
Anjali_Kumar1F wrote:Is there answer where it shows the work for these worksheets.

No, she only posts the answer key online, but during her workshop, she goes over how to do the problems!

Posted: Thu Feb 14, 2019 3:40 pm
Why is for delta H combustion, the equation is the the delta h combustion of reactants minus delta h combustion products whereas for delta h of formation it is the opposite?

Posted: Thu Feb 14, 2019 11:21 pm
Hi students, there will be no workshop on Sunday because of the long weekend. Sessions will resume 2/24.

Enjoy the weekend after your midterm!

-K

Posted: Sat Feb 16, 2019 9:35 am
So is there no worksheet for this week?

Posted: Sat Feb 16, 2019 5:45 pm
Julia Lee wrote:So is there no worksheet for this week?

No worksheet this week! The next session will cover all topics on Test 2.

Posted: Thu Feb 21, 2019 1:15 am
This is so helpful. Are the attachments basically study guides?

Posted: Fri Feb 22, 2019 6:05 pm
This week we will go over Gibbs free energy and most of electrochemistry (everything your test will cover). This is two sessions worth of material so please bear with me, see you there!

Posted: Sun Feb 24, 2019 11:00 pm
Thanks for the worksheet it helped a lot!

Posted: Mon Feb 25, 2019 12:55 am
Here are the keys to the Gibbs free and electrochem worksheets. As promised, for electrochem, I tried including as much detail as I can to follow through since there was not enough time to go over it.

NOTE:
For Gibbs free energy worksheet #3a, I made a mistake today in session saying that the conditions are at equilibrium. That is INCORRECT - the conditions are at standard conditions but not equilibrium. This follows through such that delta G at equilibrium is 0 and shouldn't be shifting to any side of the reaction. In 3a, it just so happens that P is 1, because 1/1^2 is 1, but P does not equal K (which is based on concentrations). So instead of writing RTlnK, I should've wrote RTlnP. Sorry for confusion.

If there are any other further confusions/mistakes, please let me know! I am still in the process of editing and perfecting these worksheets.

Posted: Mon Feb 25, 2019 5:18 pm
Can someone please explain #4 on the electrochem worksheet? I don't understand why the answer is Ag+ not Fe2+

Posted: Mon Feb 25, 2019 5:56 pm
For the worksheet 7 (electrochem) question 2, I don't get why the E value would cause C to be false. Can someone explain this to me please?

Posted: Mon Feb 25, 2019 8:34 pm
When we balance the equations on worksheet 7, how do we know which side of the reaction to add H2O to?

Posted: Mon Feb 25, 2019 10:12 pm
Julia Lee wrote:Can someone please explain #4 on the electrochem worksheet? I don't understand why the answer is Ag+ not Fe2+

Since you want copper to spontaneously reduce, then you need to check the oxidation values of elements/options given. In order to have a spontaneous reaction with copper reducing, the cell potential must be positive, so the oxidation potential of the substance must be positive, or the reduction potential of the substance must be negative. Fe2+ and Al3+ both work since they fulfill this criterion. I think the #4 answer just has an error.

Posted: Tue Feb 26, 2019 12:45 am
On #1 of the Gibbs to find the standard enthalpy and entropy of formation you have to do the sum of products minus the sum of the reactants using the numbers given right? For some reason I am not getting the same answer.

Posted: Tue Feb 26, 2019 1:00 am
In the Gibbs WS #4 how do you find T? or can you just assume its at 25 C?

Posted: Tue Feb 26, 2019 12:35 pm
Vincent Li 4L wrote:
Julia Lee wrote:Can someone please explain #4 on the electrochem worksheet? I don't understand why the answer is Ag+ not Fe2+

Since you want copper to spontaneously reduce, then you need to check the oxidation values of elements/options given. In order to have a spontaneous reaction with copper reducing, the cell potential must be positive, so the oxidation potential of the substance must be positive, or the reduction potential of the substance must be negative. Fe2+ and Al3+ both work since they fulfill this criterion. I think the #4 answer just has an error.

Actually, when something says plain "copper" or "iron" use the actual solid and not the ion form (Cu2+, Fe3+, etc). With that said, because the question is asking for what can copper reduce, copper is acting as a reducing AGENT, which means it is oxidized, so we will use -0.52V for its oxidation potential. For reduction potentials, we leave as is on the chart. For Ag+, it is 0.80V. Total potential = reduction potential + oxidation potential, so 0.80 + (-0.52) leaves you with a positive potential value, which is what we want in order for a reaction to be spontaneous because delta G = -nFE. The values for the other ions, all have an absolute value less than 0.52 which would cause the total E to be negative, which is NON-spontaneous.

Posted: Tue Feb 26, 2019 12:37 pm
Karyn Nguyen 1K wrote:For the worksheet 7 (electrochem) question 2, I don't get why the E value would cause C to be false. Can someone explain this to me please?

In order for a spontaneous reaction to occur, total E must be a positive value and when you add up the values for answer choice c, it is negative.

Posted: Tue Feb 26, 2019 12:38 pm
Karyn Nguyen 1K wrote:When we balance the equations on worksheet 7, how do we know which side of the reaction to add H2O to?

Usually add H2O to the opposite side you added H+ or OH- in order to balance the number of O and H atoms

Posted: Tue Feb 26, 2019 3:30 pm
on electrochem worksheet #8d, does it matter if on the cell diagram NO is listed first or NO3 is listed first? Or can the order of these two be reversed and the diagram still denotes the same reaction

Posted: Tue Feb 26, 2019 5:47 pm
Madison Hurst wrote:In the Gibbs WS #4 how do you find T? or can you just assume its at 25 C?

yeah you just assume 25 C, so 298 K

Posted: Tue Feb 26, 2019 5:50 pm
Madison Hurst wrote:On #1 of the Gibbs to find the standard enthalpy and entropy of formation you have to do the sum of products minus the sum of the reactants using the numbers given right? For some reason I am not getting the same answer.

the values are given per mole, are you multiplying them by the respective moles for each compound?
so delta H for example would be [-657.0 + 2(-110.5)] - [-910.0 + 2(0) + 2(0)] = 32 kJ/mol

Posted: Tue Feb 26, 2019 5:52 pm
for #4 on gibbs free energy, does it matter that the standard delta G is given in kJ instead of kJ/mol? i thought we'd have to divide by 2 to get the delta G per mole of NH3, but that didn't give me the right answer so i'm a little confused because plugging in the given value does give me the right answer even though it's not technically per mole

Posted: Tue Feb 26, 2019 7:19 pm
For #4 on the Gibbs worksheet, I keep getting -42.2 kj/mol and I can't figure out what I'm doing wrong

Posted: Tue Feb 26, 2019 7:34 pm
For #6 on the Gibbs free energy worksheet, how do we know that delta G is zero? Also, if delta G is zero and delta S of system is zero, why isn't delta S of surroundings also 0?

Posted: Tue Feb 26, 2019 9:24 pm
Can someone please explain #4 on the Gibbs Free Energy worksheet?

Posted: Wed Feb 27, 2019 12:40 am
could someone please explain #8 part a on the electrochemistry worksheet? How do we know that S2O8 2- is reduced and NO is oxidized?

Posted: Wed Feb 27, 2019 9:45 am
Did Karen post her most recent handout answer key that we weren't able to go over in session?

Posted: Wed Feb 27, 2019 2:07 pm
Could someone explain #3 on Worksheet 6? How would we find K without using partial pressure? Thanks!

Posted: Wed Feb 27, 2019 2:18 pm
Samantha Ito 2E wrote:Can someone please explain #4 on the Gibbs Free Energy worksheet?

I used deltaG = deltaGo + RTlnQ. I found Q with the pressure values that were given and assumed temperature to be 25oC, or 298K.

Posted: Wed Feb 27, 2019 5:52 pm
Can someone pls explain how to calculate the equilibrium constant for Gibbs Free Energy wkst #1d? Thank you!

Posted: Wed Feb 27, 2019 6:32 pm
Can someone pls explain how to do #3 from the Gibbs Free Energy worksheet.Thanks!

Posted: Thu Feb 28, 2019 2:21 pm
For Worksheet 7 #8d, why is OH-(aq) not included in the cell diagram?

Posted: Thu Feb 28, 2019 5:43 pm
Josephine Lu 4L wrote:could someone please explain #8 part a on the electrochemistry worksheet? How do we know that S2O8 2- is reduced and NO is oxidized?

I also have this question

Posted: Fri Mar 01, 2019 3:21 pm
ariana_apopei1K wrote:
Josephine Lu 4L wrote:could someone please explain #8 part a on the electrochemistry worksheet? How do we know that S2O8 2- is reduced and NO is oxidized?

I also have this question

You look at the chart of reduction potentials. The one with the higher reduction potential will be reduced. For the other half reaction, it will become oxidized and all you need to do at that point is change the sign of the reduction potential for the half reaction that will become oxidized.

Posted: Fri Mar 01, 2019 3:47 pm
Here is this week's worksheet focusing on reaction rates specifically differential rate law.

Posted: Mon Mar 04, 2019 10:48 am
Key for differential rate laws.

Posted: Sat Mar 09, 2019 6:49 am
Does anyone know if Karen posted worksheet 9 yet?

Posted: Sat Mar 09, 2019 2:48 pm
Julia Lee wrote:Does anyone know if Karen posted worksheet 9 yet?

no it would have been posted here, week 8 is the most recent

Posted: Sun Mar 10, 2019 11:48 am
Sorry for the delay in this week's worksheet! Didn't realize my last post didn't go through. This week we are reviewing integrated rate laws and if time permits, I will introduce reaction mechanisms. On Wednesday I will post a last worksheet for reaction mechanisms since I won't be seeing you next week due to your final on Sunday.

See you all tonight!

Posted: Wed Mar 13, 2019 12:56 pm
Here is the key for last Sunday's worksheet and another worksheet focusing on reaction mechanisms. Full solutions are also given.

Good luck on your final!!!!!! It has been a great quarter!

Best,
K

Posted: Wed Mar 13, 2019 9:24 pm
Thank you :)

Posted: Wed Mar 13, 2019 11:04 pm
For the integrated rate law worksheet #5a, I found the answer using
1/((7.0*10^9 M*s)*(120 sec) + 1/0.086M)
= 1/ (8.4*10^11 M*s^2 + 11.63 M^-1)
and I got the right answer but I'm confused by the units. How do the units work out to be M? And how do you add 8.4*10^11 M*s^2 with 11.63 M^-1?

### writing rate laws: "rate"or "d[A]/dt?"

Posted: Wed Mar 13, 2019 11:46 pm
when we write the rate laws, do we write it as "rate=..." or ""d[A]/dt=..."? Specifically in question #3 of the Reaction Mechanisms worksheet, could we replace "d[F]/dt" with "rate"?

Posted: Thu Mar 14, 2019 12:30 am
For worksheet 8 #4d I am getting 144 M/s instead of 140 M/s. Am I doing something wrong?

Posted: Thu Mar 14, 2019 12:56 am
How do you do #8 on worksheet 9? I tried to do .99[A]=.5^n[A] and solving for n by ln(.99)/ln(.5)=n, but this didn't work.

Posted: Thu Mar 14, 2019 9:08 pm
On Worksheet 9, how would you solve for #7?

Posted: Thu Mar 14, 2019 11:46 pm
Semi Yoon wrote:On Worksheet 9, how would you solve for #7?

For a first order reaction, you can use the equation (1/2)^n=whatever fraction of A is left because the half life is constant throughout the reaction. For this problem, use (1/2)^n=1/8 which means n is 3 (n is the number of half lives). We know the length of each half life is 355 seconds so the total time taken is 3x355=1065 s

### Re: writing rate laws: "rate"or "d[A]/dt?"

Posted: Thu Mar 14, 2019 11:47 pm
Josephine Lu 4L wrote:when we write the rate laws, do we write it as "rate=..." or ""d[A]/dt=..."? Specifically in question #3 of the Reaction Mechanisms worksheet, could we replace "d[F]/dt" with "rate"?

Yes they are synonymous. Writing it as d[A]/dt is just better because it specifies which species we are referring to.

Posted: Thu Mar 14, 2019 11:54 pm
Why is it that in #3 in the reaction mechanisms worksheet we're able to use the pre-equilibrium assumption to substitute and replace [C]? Wouldn't the approximation fail since the first and second steps are both fast, so there's no bottleneck to stop [C] form being consumed instead of forming A and B again?

Posted: Fri Mar 15, 2019 1:36 am
Celine Hoh 2L wrote:Quick question for Worksheet 1 ques8:

When I2 is added, neither products nor reactants is favored as I2 is a solid.
What happens when I2 is removed, does the reaction not shift too?

Yes! Because it is a solid it is not included in the equilibrium equation so any change in concentration would not affect the reaction.

Posted: Fri Mar 15, 2019 2:11 am
for number 5 of karen's 9th worksheet (kinetics), it says that the rate constant is 7.0x10^0 M.s. Does M stand for molarity or for minutes?

Posted: Fri Mar 15, 2019 2:18 am
for number 6 of karen's 8th worksheet (kinetics pt 1),

why does k not depend on the order of the reaction? wouldn't the reaction order depend on the value of K?
If there is a higher order then wouldn't the k value be higher?

Posted: Fri Mar 15, 2019 2:26 am
how do you solve 5a on worksheet number 9?
I know the second order equation is 1/[A]=kt+1/[A]0. If I were to use this equation, wouldn't I just plug the values given into the equation?
The answer that i got was 7.14x10^-11, which is different from the answer. pls help thxxxx

Posted: Fri Mar 15, 2019 2:38 am
how do you solve for worksheet 9 number 6?

I used the half-life equation for a first-order reaction and solved for k when the half-life was 6 days.
my k value was 0.1155.
In order to find the percent of Hg(II) left in the body, would I have to find the k value for the 30-day half-life and divide the two k's?

Posted: Fri Mar 15, 2019 2:57 am
can someone please explain how to get the equilibrium constant for 1d in Gibbs free energy/worksheet 6? I keep getting 1.014.

Posted: Fri Mar 15, 2019 3:02 am
Katie_Duong_1D wrote:can someone please explain how to get the equilibrium constant for 1d in Gibbs free energy/worksheet 6? I keep getting 1.014.

I forgot to convert delta G from kj/mol to j/mol to match units. Now I get the correct answer 1.663 x 10^6.

Posted: Fri Mar 15, 2019 1:08 pm
Emily Tam 1k wrote:for number 5 of karen's 9th worksheet (kinetics), it says that the rate constant is 7.0x10^0 M.s. Does M stand for molarity or for minutes?

M stands for molarity, or mol/L

Posted: Fri Mar 15, 2019 1:11 pm
Emily Tam 1k wrote:how do you solve for worksheet 9 number 6?

I used the half-life equation for a first-order reaction and solved for k when the half-life was 6 days.
my k value was 0.1155.
In order to find the percent of Hg(II) left in the body, would I have to find the k value for the 30-day half-life and divide the two k's?

For first order reactions you can use the equation (1/2)^n=whatever fraction of A is remaining, with n being the number of half lives, since the half life is constant throughout the reaction. In this case we know 5 half lives would have passed (30/6=5), so we do 1/2^5 which yields 1/32 (3.1%) which means 3.1% of A is left after 30 days.

Posted: Fri Mar 15, 2019 7:08 pm
sonalivij wrote:For worksheet 8 #4d I am getting 144 M/s instead of 140 M/s. Am I doing something wrong?

She rounded to 140 because you need to have two sig figs.

Posted: Fri Mar 15, 2019 9:54 pm
how would you solve for worksheet 10 number 2 part c and d?

Posted: Fri Mar 15, 2019 10:47 pm
Amy Dinh 1A wrote:On #4a Worksheet 2, I keep getting pH=4.40, while the answer key says 4.30, when both using the quadratic formula and the shortcut way. I don't understand how to get the answer to

The pH is 4.40

Posted: Sat Mar 16, 2019 12:14 pm
005199302 wrote:
ariana_apopei1K wrote:
005199302 wrote:For the first law worksheet, how do you do #2?

First you have to do stoichiometry to find how many moles of Pb is made from 49.7 g PbO. You multiply the number of moles of product times the enthalpy value given, which gives you the q to plug into your regular q=mCs(Tf-Ti). Once you solve for mass you should get the right answer

Which value should Cs be?

it should be the specific heat capacity for water but you must convert it to kilojoules

Posted: Sat Mar 16, 2019 12:39 pm
Karyn Nguyen 1K wrote:
Celine Hoh 2L wrote:
Karyn Nguyen 1K wrote:For worksheet 4, is the answer for 6a wrong? I did (3/2)(31.9 mol)(8.314 J/Kmol)(311.15 K) = 124 kJ. When I didn't convert the temperature from C to K I got 15.1 kJ which is the answer given in the key.

We don’t have to convert it to kelvin as it is change in temperature (38-0)

Why do we use that equation for the change in internal energy and not q+w

Posted: Sat Mar 16, 2019 12:44 pm
How do you solve #8 on worksheet 9? How do you find the half-life given that 99% of the reactant decomposes in 137 min?

Posted: Sat Mar 16, 2019 12:55 pm
Is there any worksheets for kinetics?

Posted: Sat Mar 16, 2019 1:05 pm
Ashley P 4I wrote:Is there any worksheets for kinetics?

Nevemind I just found them, thank you for your help! Do you by chance have a final practice test? Or does anyone else have one?

Posted: Sat Mar 16, 2019 6:24 pm
For the electrochem worksheet 8d, why don't we include the OH- in the cell diagram?

Posted: Sat Mar 16, 2019 8:52 pm
Yes, it is further up in this thread and there is a key as well. There's also two parts!

Posted: Sat Mar 16, 2019 10:53 pm
Chem_Mod wrote:
ariana_apopei1K wrote:
Josephine Lu 4L wrote:could someone please explain #8 part a on the electrochemistry worksheet? How do we know that S2O8 2- is reduced and NO is oxidized?

I also have this question

You look at the chart of reduction potentials. The one with the higher reduction potential will be reduced. For the other half reaction, it will become oxidized and all you need to do at that point is change the sign of the reduction potential for the half reaction that will become oxidized.

Thank you !!!

Posted: Sat Mar 16, 2019 11:07 pm
in worksheet 5 problem 6, how do we know that work is 0?

Posted: Sat Mar 16, 2019 11:08 pm
thank you so much karen! these worksheets were so helpful!

Posted: Sun Mar 17, 2019 12:39 am
For kinetics ws pt 1 number 5 why isn't C included in the rate law? Is it an intermediate? if so, how can you tell?