CHEMISTRY COMMUNITY

The title is called "Hot dog" so that you can search "Hotdog" and easily find this post!

This is your UA, Lyndon Bui, with information on the much anticipated HOTDOG Review Session! As always I am creating a packet of practice problems that stem from past exam questions and course outlines. You are covering some major topics for the midterm this quarter so there is no possible way to cover everything. Definitely use course outlines to make sure you are ready! I am doing my best to cover ALL TOPICS for the midterm.

Please complete these problems before my review session. My review session this quarter (W19) will be Monday, Feb 11, 7-10pm, in Franz 1178.
I will go over each problem in detail during my review session. Due to private concerns/engagements, please do not expect answers to be posted. Any errors in the test will be posted below.

Link to Download Problems: Problems no longer available. Quarter has ended. See Course website and syllabus for most up-to-date material.

***This is not indicative of the structure, length, or format of the actual midterm. Treat these as extra practice problems. ***


[b][u]This contains few questions about acid/base equilibria. I will go over concepts and problem solving for various acid/base type problems in detail at the review session but you should find more acid/base problems for practice.

[i]Errors: #11 should say M for the units of the solution concentration

Happy Studying and Good Luck!
-Lyndon Bui, UA

Printed copies will NOT be provided at the review session.

For 4C, are we supposed to find/know the enthalpy of formation for the reactant O2, or should that be given?

Thanks!

The enthalpy of formation of O2 is zero since it is in its standard state.

Thank you for doing this for us! Greatly appreciated :)

Venya Vaddi 1L wrote:The enthalpy of formation of O2 is zero since it is in its standard state.

Thank you!!!

How do you calculate internal energy for #6 without knowing the temperature?

Thank you thank you for all this :D

Yah for number six, im confused as well. Internal energy is dependent on temperature and the work function for a reversible reaction has temperature for one of the unknown variables. Is there a way around this?

Temperature is supposed to be given for #6 I believe.
EDIT: Okay the review system I went to did end up giving us the temperature but you can calculate it n your own using PV=nRT where pressure and volume is the pressure and volume of the system after the isochoric step so V=10l and P=10atm

You can solve for temp using PV=nRT. I keep getting a wack answer for this question, however. I'm assuming the overall delta U and delta S will be zero because they are state functions, but I am not getting the right q and w quantities. Can someone who feels confident put down their values so I can compare?

There is great discussion here! There is no mistake for #6, the temperature does not need to be given.

Thanks for posting this again, Lyndon! See you tomorrow at the review session!

See you all 7pm Franz 1178!

Thank you for posting, see you soon.

Hello, Unfortunately I won't be able to attend the review, will answers be posted? If not could someone please upload a pic of the answers after the review?? I would greatly appreciate it!! Thank you!

Thank you, I had such a hard time finding it haha!

I will not be able to make the review session tonight. Can someone post the answers given?

when will you post the answers to hotdog?

For #6, how did we get w=9.119 x 10^3 J?

Mya Majewski 1L wrote:For #6, how did we get w=9.119 x 10^3 J?

For the first part it says you preform an isobaric compression to 10L. Isobaric is constant pressure so work is equal to -PdV. So pressure is 1.00 atm and the change of V is 10-100 so -90. This comes to -90 L.atm you then convert this to Joules

Megan_Ervin_1F wrote:when will you post the answers to hotdog?

He mentioned that he won't be posting answers this time

If he said he isn't gonna post the answers, could someone post them so I can check my answers?

what was the full answer for w on #6? I got that part of it is 2.33KJ from the w=-nrtln(v2/v1)

I also got that the other part of w=9.12 from w=-pdv. so i got a total of 11.45kj or 1.145x10^5j for w. is this correct?

with a corresponding q=-11.45kj?

Can someone please explain why 3E is false?

MMoreno3K wrote:Can someone please explain why 3E is false?

Because heat is required during melting or boiling (phase change transition), temperature of a sample can remain constant even though heat is being supplied.
For example, if you have boiling water, it will remain at 100 C even though heat is being added, until it has enough heat/energy to go through vaporization.

for #6, how do we know that change in entropy is zero?

005199302 wrote:for #6, how do we know that change in entropy is zero?

Since the system starts and ends at the same state, the entropy doesn't change.

Help with question 2c)? I understand that since we're cooling it kc will also drop because it's dependent on temperature, but what is the detailed explanation?

can someone explain how to do #5

Are there answers to these problems?

Philipp_V_Dis1K wrote:can someone explain how to do #5

For this problem, we have to find the change in entropy, delta S. To do this we want to change the Helium and Krypton to mols.
We get 2.25mol He and 1.49mol Kr. Then we can plug them into the equation DeltaS=nRln(V2/V1).

For the Helium, it is placed in a compartment that is 1/3 the size of Kr, so lets assume Volume that He is placed in is 1L and Volume that Kr is placed in is 3L. So when they are combined the final Volume is 4L.

Now we can finish the DeltaS of He and DeltaS of Kr:

He: DeltaS1= (2.25mol)(8.3145 J*mol-1*K-1)(ln 4L/1L)= Use ur calc im lazy :P
Kr: DeltaS2= (1.49mol)(8.3145 J*mol-1*K-1)(ln 4L/3L)= Use ur calc im lazy :P

However, not only did the volume change, the temperature changed too, so we need to use the equation DeltaS=Cp,m(ln T2/T1) where Cp,m= 3/2(8.3145 J*mol-1*K-1). This can be found on the equation sheet. We need to add the mols of He and Kr together as they are not independent reactions.

DeltaS3= (2.25+1.49mol)(3/2)(8.3145 J*mol-1*K-1)(ln 348K/323K)= Use ur calc im lazy :P

Now we add DeltaS1+DeltaS2+DeltaS3= Total DeltaS of the whole system which should get u 32.97J*K-1 or 33.0J*K-1 (sig figs).

Hope this helps. Correct me if I'm wrong!!

lukezhang2C wrote:Are there answers to these problems?

He said he wouldn't be posting answers for this practice test.

Could someone explain for 4a how he got 5.0x10^3?? When I worked it out on my own I got -498 and all my elements cancelled out.

pamcoronel1H wrote:Could someone explain for 4a how he got 5.0x10^3?? When I worked it out on my own I got -498 and all my elements cancelled out.

Your answer is correct. He gave the answer DeltaHrxn= -5.0*10^5 Joules which is the same thing as -498kJ. Just make sure you use your sig figs as there are only 2 sig figs in the DeltaH of Vodka+Sweat->2 Honey. :)

can someone explain the reasoning behind the three steps to solve for enthalpy of #12B

Please feel free to come in at 1pm to CS50 on Wednesday (tomorrow) and Lyndon will be one of the UA's overseeing the drop in session

For #10, should both heat capacities (of ice and of water be the same), or should we use the heat capacity of ice on the left side of the equation. I remember at the review he used 4.184 for both, but shouldn't we take into account C of ice as well?

paytonm1H wrote:can someone explain the reasoning behind the three steps to solve for enthalpy of #12B

So we are trynna calculate the DeltaHrxn(total) of the human body at 37C and we know the DeltaHrxn=-2756kJ at 200C

First, we need to heat up the reactants C6H12O6 and 6O2 to 200 which will be your DeltaH1
Then we use the given DeltaHrxn at 200C which will be your DeltaH2. Now we have the products instead of the reactants.
Then we need to cool down the products, 6 CO2 and 6 H20 to 37 which will be your DeltaH3

DeltaHrxn at 37C=DeltaH1+DeltaH2+DeltaH3.

005199302 wrote:For #10, should both heat capacities (of ice and of water be the same), or should we use the heat capacity of ice on the left side of the equation. I remember at the review he used 4.184 for both, but shouldn't we take into account C of ice as well?

Because the Ice is at 0C, it will melt into water first, then start heating up using the constant 4.184J*mol-1*K-1.
So in this problem, we are actually not at any point using heat capacity of ice because we don't need to heat it up to the temperature it vaporizes, because it is already at 0C. All we need to use is the DeltaHfusion= 6.01kJ/mol

Now... if the Ice was at -50C, we would need to use the specific heat capacity of ice in order to heat it up to 0 degrees C and then use the DeltaH of fusion.

Hope this helps

He whose name cannot be spoken wrote:Help with question 2c)? I understand that since we're cooling it kc will also drop because it's dependent on temperature, but what is the detailed explanation?

Since the reaction is breaking bonds, we know that it will be endothermic. When we learned how temperature affects a system, we learned that heating an endothermic reaction will favor the forward reaction. Therefore, if we are cooling an endothermic system, it will favor the reverse reaction.

Can anyone share the answers to questions 1-3 on Lyndon's review? I was an hour late to the review session bc I had a midterm

Can anyone explain 3b?

For 3b: the first step is to calculate the amount of heat need to raise the temperature of all the ice cream to 0 degrees C. Then, you subtract this value from the total heat that was given. Next, you set the value for heat from the subtraction equal to the half the mass of the ice cream multiplied by the enthalpy of fusion for the ice cream. Then just solve for the enthalpy of fusion.

Does anyone have the solutions?

dgerges 4H wrote:I also got that the other part of w=9.12 from w=-pdv. so i got a total of 11.45kj or 1.145x10^5j for w. is this correct?

I also got that for work

Thanks again Lyndon for the practice exam! Helped a lot! C:

Can anyone explain 1D?
Thanks!

Thank you