CHEMISTRY COMMUNITY

Welcome back Chem 14B students!

For those of you who aren't familiar with me, my name is Karen Leung, one of your UAs this quarter. I will be hosting workshops every Sunday 7-9pm this quarter, and I use worksheets I have made to facilitate my workshops. They will be posted on this SAME post here every week (I'll try to have them posted by Friday night, but Saturday by the latest). I created these worksheets as extra practice for every week's material learned. (They are made to imitate Dr. Lavelle's question styles to the best of my ability.) If you plan on attending my workshops, please bring this worksheet (either printed or available on a computer/tablet). You do not need to complete it, we will go over the problems together during the session.

A breakdown of my workshop is follows:

-First review all course material learned in the previous week
-Do practice problems together from my worksheets
-Answer questions from lecture, homework, modules, etc. anything that anyone wants answered


The answers to the worksheets will be posted on this SAME post AFTER the session, but no work will be shown, so coming out to the session is necessary to understand the steps of getting to each solution. The worksheets for later weeks (Week 2,3,4..etc) will also be posted on THIS SAME post as well, so you will need to come back to this post every week for each week's worksheets.

And..... here is the first worksheet of the quarter!!

Key for worksheet 1.

Please note corrections:
1) #3 and #8, some reactants/products are in the solid phase.
2) 8c: I2 favors neither direction because it is in the solid phase, so it does not affect equilibrium, not because it is an inert gas. Inert gases are the last column of the periodic table (noble gases), and iodide is a halogen.

Here is the worksheet for this Sunday's session covering acid and base calculations:

Can someone please explain to me how number 4 part d and number 5 are solved for the the Chemical Equilibrium worksheet?

Key for Week 2. Good luck on quizzes!

Corrections:

4a is 4.40, not 4.30

Julia Lee wrote:Can someone please explain to me how number 4 part d and number 5 are solved for the the Chemical Equilibrium worksheet?

Hi Julia, can you please detail at which step you are stuck on? Or what concepts you think should be used?

Did anyone happen to get around 0.077g for #2 on the acid-base review worksheet? The key says 0.10g, so I am unsure whether Karen rounded or I did the question incorrectly.

I got .05 grams for number 2 on the acid base worksheet because I first found the hydronium concentration then I solved for X and from there multiplied the answer by the molar mass, and did not get .1. Can anyone help me on this?

Can someone explain how to do Worksheet 1 #4d? My friend did it with initial concentration of NH3 as 0 and she got the right answer. Why doesn't the initial concentration of NH3=5.82*10^-3 ?

For number 7 on worksheet 1, why does high temperature favor the formation of bromine atoms if we don't know if the reaction is endothermic or exothermic?

For #3 on Worksheet 2, how can you find the Kb value? I'm not quite what is the first step or what the base is.

Melissa Bu 1B wrote:Did anyone happen to get around 0.077g for #2 on the acid-base review worksheet? The key says 0.10g, so I am unsure whether Karen rounded or I did the question incorrectly.

I did the same thing at first. Make sure you use the initial concentration of HCOOH and not the equilibrium concentration of HCOOH, since the question asks for the grams needed to prepare the solution. You plugged in x into [x-5.5x10^(-4)] instead of [x].

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

On Worksheet 2, why is the concentration of CO3 the same as Ka2 for problem 4b?

annabel 2A wrote:Can someone explain how to do Worksheet 1 #4d? My friend did it with initial concentration of NH3 as 0 and she got the right answer. Why doesn't the initial concentration of NH3=5.82*10^-3 ?

I did this problem using the initial concentration of NH3 and ended up getting 11.6*10^-3 which is pretty close to the answer she got. Not sure if its right tho

Meigan Wu 2E wrote:On Worksheet 2, why is the concentration of CO3 the same as Ka2 for problem 4b?

^ I have the same question

Amy Dinh 1A wrote:For #3 on Worksheet 2, how can you find the Kb value? I'm not quite what is the first step or what the base is.

So if percent protonation was 6.7%, then 6.7 = 100 x (x/.1) should give you x. For the ice table I don't think you even need to know the base, so i just wrote " oct + h20 --> OH- + oct-" and set up the table like normal.
Kb= (x^2)/(.1-x) with the x you found originally should give a Kb of 4.8x10^-4

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

in the session she said the pH was 4.40 so you are correct.

lindsey_ammann_4E wrote:

Meigan Wu 2E wrote:On Worksheet 2, why is the concentration of CO3 the same as Ka2 for problem 4b?

^ I have the same question

Because some approximation was used. You can look at the example on page 488 on the 7th edition textbook.

Don't we need the volume to solve question 5 on worksheet 1?

Ray Guo 4C wrote:Don't we need the volume to solve question 5 on worksheet 1?

you can calculate the volume from the info given using PV=nRT!

What are the initial values and the Kc value that we have to use for 4d worksheet 1? are the initial values the values from 4b?

Can someone please indicate the main steps to solve worksheet 1 problem #5?

Meigan Wu 2E wrote:For number 7 on worksheet 1, why does high temperature favor the formation of bromine atoms if we don't know if the reaction is endothermic or exothermic?

Breaking a Br2 atom into 2 atoms is energetically not favorable, so heat is required to be inputted.

Meigan Wu 2E wrote:On Worksheet 2, why is the concentration of CO3 the same as Ka2 for problem 4b?

In the second ICE table, the concentration of CO3 is x, and because x is so small, it is essentially Ka2.

For 4b on worksheet 2, I keep getting 4.73x10^(-8) for the concentration of CO3. Can someone explain what I'm doing wrong? I got 3.99x10^(-5) for the concentration of H2CO3.

can you please explain the steps of this question on worksheet 2: How many grams of formic acid, HCOOH, do you need to prepare 1.00L of a pH 3.26
solution? Ka of formic acid = 1.8 x 10-4? I have no idea how to solve

For #5 on worksheet 1, don't we need the value of R to calculate the volume?

for problem 4 on the second worksheet, how do you find the concentration of CO3 2- you are given the concentration of H2CO3

Samantha Ito 2E wrote:Can someone please indicate the main steps to solve worksheet 1 problem #5?

First you would use PV=nRT to find the volume of the container, and then divide the .05 mol PCl5 by the volume to get the initial concentration of PCl5. Once you have that, you can do an ice table like u normally do and solve for x by using the equation for Kc, and can find the percentage deprotonation by dividing x/initial concentration of PCl5.

005199302 wrote:For #5 on worksheet 1, don't we need the value of R to calculate the volume?

R is a constant and since P is in atm, use the R constant that has atm in its units.

Can someone explain how to do number 4b on worksheet 1. I do not seem to understand the steps to take.
The Haber process is used to synthesize ammonia gas (NH ) from nitrogen gas ( ) 3 N2
and hydrogen gas ( H ). A system at equilibrium contains 1.85M H2 1.36M N2 and 2.91 x 10 ^−3 NH3 at a constant temperature. The volume is suddenly halved. What are the new concentrations for N2, H2, and NH3 ?

Cecilia Jardon 1I wrote:Can someone explain how to do number 4b on worksheet 1. I do not seem to understand the steps to take.
The Haber process is used to synthesize ammonia gas (NH ) from nitrogen gas ( ) 3 N2
and hydrogen gas ( H ). A system at equilibrium contains 1.85M H2 1.36M N2 and 2.91 x 10 ^−3 NH3 at a constant temperature. The volume is suddenly halved. What are the new concentrations for N2, H2, and NH3 ?

Concentration is moles / liter, so if volume is halved, the concentrations are doubled for H2, N2, and NH3. I believe the answer key only put the new concentration for NH3, but H2 should just be 3.70M and N2 should be 2.72M.

Meigan Wu 2E wrote:On Worksheet 2, why is the concentration of CO3 the same as Ka2 for problem 4b?

In the second ICE table, the concentration of CO3 is x, and because x is so small, it is essentially Ka2.[/quote]
I still don't understand this, I'm sorry could someone please explain more?
I'd greatly appreciate it.
What I did was take the x concentration found in part a of HCO3- and used it as my initial molarity for the second ICE table and set H+ to 0 and CO2- to 0 as well. I then did HCO3- equilibrium: 3.9 x 10^-5 - x and eq H+ to +x and eq CO3^2- to +x. Then I used Ka2 to find the x for this ICE table. But somehow x is just Ka2?
Thank you!

Can someone please explain why we need to square the Kc for worksheet 1 question #2a? I thought that the same molecules under the same conditions (same temp (973 K) would have the same Kc even if one has more "material" used. Why do I need to square it?
Thank you!

On worksheet 1 how exactly did they get the answer for number 3 because I am confused. I just thought it was Kc=[products]/[reactants] so Kc=x^5/1.0-x. Thank you

How did the NH3 become 11.3* 10^-7 On worksheet 1 4d

Worksheet for Week 3 (1/27) session. There was not much material learned this week due to the holiday, so the session should be quick.

If you can't make it, get notes/ full solutions from a classmate or friend that can attend. Full solutions are not posted because my work is messy and may cause unnecessary confusion. If you are stuck in solving a problem, please be detailed as to what you are stuck on so I can best help guide you in the right direction without blatantly solving it out for you.

Thank you so much, these really do help

For question 5, do we have to have whole numbers in the balanced equation to use bond enthalpies or can we keep fractions in there (for this and other questions like this).

Kenan Kherallah 2C wrote:For question 5, do we have to have whole numbers in the balanced equation to use bond enthalpies or can we keep fractions in there (for this and other questions like this).

You can use fraction, since bond enthalpy is a number per mol.

Sorry for the delay in last week's worksheet key! Got caught up in some of my own midterms.

Happy midterms season!

This week we will be going over the 1st law of thermodynamics. I will also cover 2 more equations that you will learn next week, so don't be alarmed if there are a problem or two you are struggling with in this worksheet. This is so that next week we will have more time doing review problems for the midterm.

Could someone explain the detailed reasoning for questions 2.a. and b. on worksheet 1?

Hello, to solve 2a and 2b in the first worksheet I would go about setting realizing the differences of the problems with the initial one with the given KC. If you square the KC you get the answer for 2a and if you inverse the KC you get the answer for 2b. I would go about setting the products/reactants with coefficients as the power and seeing the magnitude of change. Hope this helped!

Key for week 4!

Thank you

For question 6 on worksheet 4, how exactly are we supposed to determine which volume we use to find the pressure for the work component of the internal energy change?

To my understanding, the pressure was calculated using PV = nRT, with V = 998L, n = 31.9 mol, R = 0.08206 L * atm / mol * K, T = 311K. I understand why we're supposed to use the temperature (38 C) after the "sudden increase," but shouldn't the volume be 876L, prior to any expansion? I thought this made the most sense as the pressure changed after the temperature increase, but then remained constant throughout the process of expansion, but apparently not?

For the first law worksheet, how do you do #2?

For Worksheet 1 question 7, are the sig figs correct? Wouldn't it be 3 sig figs, not 4, because the lowest number given is 298 K? Or is it 4 sig figs because we don't use the temperature to solve this problem?

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=mC_s(Tf-Ti). Once you solve for mass you should get the right answer

Vincent Li 4L wrote:For question 6 on worksheet 4, how exactly are we supposed to determine which volume we use to find the pressure for the work component of the internal energy change?

To my understanding, the pressure was calculated using PV = nRT, with V = 998L, n = 31.9 mol, R = 0.08206 L * atm / mol * K, T = 311K. I understand why we're supposed to use the temperature (38 C) after the "sudden increase," but shouldn't the volume be 876L, prior to any expansion? I thought this made the most sense as the pressure changed after the temperature increase, but then remained constant throughout the process of expansion, but apparently not?

For work, all you had to do was internal energy-heat (because U=q+w)

Jeannine 1I wrote:For Worksheet 1 question 7, are the sig figs correct? Wouldn't it be 3 sig figs, not 4, because the lowest number given is 298 K? Or is it 4 sig figs because we don't use the temperature to solve this problem?

Yes, we use the smallest # of sig figs that we actually used.

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=mC_s(Tf-Ti). Once you solve for mass you should get the right answer

Which value should Cs be?

for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

I need help on worksheet #4 (First Law of Thermo). On the answer key, it said -631 kJ, but I got -631J. What is the correct unit? The given information is in J.

005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Were you able to solve 6a? I've tried doing it but keep getting 12.85 kJ, but I don't see where it went wrong.

Vincent Li 4L wrote:For question 6 on worksheet 4, how exactly are we supposed to determine which volume we use to find the pressure for the work component of the internal energy change?

To my understanding, the pressure was calculated using PV = nRT, with V = 998L, n = 31.9 mol, R = 0.08206 L * atm / mol * K, T = 311K. I understand why we're supposed to use the temperature (38 C) after the "sudden increase," but shouldn't the volume be 876L, prior to any expansion? I thought this made the most sense as the pressure changed after the temperature increase, but then remained constant throughout the process of expansion, but apparently not?

the pressure is given to you as 1.00 atm

Jeannine 1I wrote:

005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Were you able to solve 6a? I've tried doing it but keep getting 12.85 kJ, but I don't see where it went wrong.

I got the same answer and online I saw that its right so I think you're correct

005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Please see above answer to the same question

This is the worksheet for this upcoming Sunday before the midterm. It contains cumulative material from the beginning of the quarter. I will review the 2nd and 3rd law of thermodynamics, and we will move on quickly to the worksheet since it is longer than usual.

****This worksheet is not indicative of the structure, length, or format of the actual exam and may not contain absolutely everything you need to know for the midterm. These are just extra practice problems for you to refresh your memory of some (but not necessarily all) important concepts learned throughout the quarter.****

For the last question, Do we use 0.15 moles from the previous question since I think we have to use moles in the last question

Does anyone have the notes from her last session (feb. 3) that they could email me?

Here are the answers for the midterm practice worksheet.

Regarding #11, I had quite a lot of students ask why use Cv and not Cp. Here is the final verdict: You use Cv when calculating the temperature change because you are ignoring the change in volume for that part. So, in this question you are calculating the total change in entropy by calculating temperature change and volume change independent of one another. Therefore, when you change volume you can use a constant temperature and when you calculate temperature change you can use a constant volume. You can do this because entropy is a state function allowing you to calculate each independent of each other and add them together. Essentially we will use constant pressure only when its explicitly stated in the problem.

If there are any problems you don't understand how to get the answer for, ask a classmate or use this forum to discuss with each other!

Good luck!

Kenan Kherallah 2C wrote:For the last question, Do we use 0.15 moles from the previous question since I think we have to use moles in the last question

Yes #11 is a continuation of #10.

For question 3 on the midterm review worksheet, I was wondering why the answer is -312 kJ/mol. I'm sure I'm just making a calculator/stupid error, but I got +312 kJ/mol. Thanks for any help with this!

I got +312kJ initially too, but I think the question is more complicated in that you have to write out the combustion equations of the three equations given and then use Hess' Law. When you do that, you get -312kJ.

Melissa Bu 1B wrote:I got +312kJ initially too, but I think the question is more complicated in that you have to write out the combustion equations of the three equations given and then use Hess' Law. When you do that, you get -312kJ.

Yes this is correct.

For question 4, Im just wondering if the condition is temperature at zero or if the change (delta T) of temperature is zero so an isothermal condition?

Chem_Mod wrote:

005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Please see above answer to the same question

Why do we need to find a different pressure if pressure is constant?

I'm confused on #4, too. What equations are relevant?

For #6, how do we know that work is equal to 0?

for question 8, isn't the answer going to be Kb x 3^8?

Mya Majewski 1L wrote:for question 8, isn't the answer going to be Kb x 3^8?

the formula is S=KblnW, so you need to take the natural log of 3 to the 8th power

Can someone explain how to do #1. I found the heat gained by the water but without the final temperature of the metal how would I find the specific heat??

katherinemurk 2B wrote:Can someone explain how to do #1. I found the heat gained by the water but without the final temperature of the metal how would I find the specific heat??

Final temperature of the metal and water are equal

Hannah Pham 1D wrote:For #6, how do we know that work is equal to 0?

I was wondering the same thing. What I came up with was that is you use pv=nrt to find deltaV then you can see that since pressure, temperature, and R are constant the change in volume will depend on the change in moles. The equation has a 1:1 ratio for all gases, the moles of gas doesn't change either.

Can someone explain #4? Is this because q=0 when delta T=0, and w=0 when -PdeltaV=0?

For #6, if heat is negative and therefore losses heat why is the reaction endothermic?

Nevermind I don't know why I put a negative there. I know heat is positive now.

Can someone explain how to get the answer for question #6 part d?

katherinemurk 2B wrote:Can someone explain how to do #1. I found the heat gained by the water but without the final temperature of the metal how would I find the specific heat??

The final temperature of the metal is the same as the final temp of the water.

Can someone explain how to do #3?
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ

Can someone explain how to solve 10 and 11? For 11, I used the mol(cv)(ln(t2/t1)) and am not getting the correct answer. Can someone please outline the steps to solve?

Madeline Ho 1C wrote:Can someone explain how to do #3?
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ

Using Hess' Law will give the reaction
deltaHrxn = (1mol)(-1300kJ/mol) + (2mol)(-286kJ) + (1mol)(1560kJ) = -312kJ

Schuyler_Howell_4D wrote:

Hannah Pham 1D wrote:For #6, how do we know that work is equal to 0?

I was wondering the same thing. What I came up with was that is you use pv=nrt to find deltaV then you can see that since pressure, temperature, and R are constant the change in volume will depend on the change in moles. The equation has a 1:1 ratio for all gases, the moles of gas doesn't change either.

Is this a correct assumption?

Can someone explain 5c?

Should number 4 on the midterm review be when delta T = 0?

I'm still confused on #6 for worksheet 4. I get the same answer for work as the answer key if I used -nRT (n=31.9, R=.0821, T=38) but not if I use -PV (P=1, V=L). Why aren't they the same?

Kenan Kherallah 2C wrote:For question 4, Im just wondering if the condition is temperature at zero or if the change (delta T) of temperature is zero so an isothermal condition?

I was wondering the same thing.

How do you solve for the heat in #6a? Are we supposed to use standard enthalpy of formation or bond enthalpies? The standard enthalpy of formation of HI is not given and the bond enthalpy for C-O is not given.

*edit: nvm, I just realized change in internal energy is given.

Hannah Yates 1K wrote:Should number 4 on the midterm review be when delta T = 0?

Yes, I believe it should be deltaT=0 because temperature must be constant for q=-w.

can someone explain 6(d)? I tried using the additive equation for standard enthalpy of a reaction but I don't think I'm inputting the correct value for it or if that's even a possible way to get the answer

Hannah Yates 1K wrote:Can someone explain 5c?

You use work = -nRTln(V2/V1). Isolate ln(V2/V1) and then solve for V2/V1 by taking e of both sides

madisondesilva1c wrote:Can someone explain how to solve 10 and 11? For 11, I used the mol(cv)(ln(t2/t1)) and am not getting the correct answer. Can someone please outline the steps to solve?

For 10, use delta S = nRln(V2/V1)
For 11, you need to add up two different delta S's, one for the volume change and one for the temperature change. For the volume change, delta S = nRln(V2/V1). For the temp change, you'd use the equation that you used. Then just add the two separate changes together

Hope this helps!!

Jordan Lo 2A wrote:Can someone explain #4? Is this because q=0 when delta T=0, and w=0 when -PdeltaV=0?

When delta T = 0, the internal energy delta U = 0. Since delta U = q + w, if delta U = 0 then w = -q, aka the work done on the system is equal to the heat absorbed.

Angela Grant 1D wrote:can someone explain 6(d)? I tried using the additive equation for standard enthalpy of a reaction but I don't think I'm inputting the correct value for it or if that's even a possible way to get the answer

We know the total enthalpy needs to equal 42, so add the enthalpies of formation for all the other species (changing the signs for the reactants). The difference between the sum and 42 is the enthalpy of formation of HI.

Does anyone know how to solve #9?

Can someone explain how to do 6d?

On worksheet 3 Intro to Thermodynamics, problem 7, what equations are we supposed to use? My friend used the information given to make a cross multiplying equation but got an answer of 51.23g instead of 51.14g. How are we supposed to approach this problem?

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?

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.

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)