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Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 2:51 pm
by Chem_Mod
1. ΔU = 310 J

2. ΔU = 0 kJ

3. w = –5.48 kJ
q = +5.48 kJ
ΔU = 0

4. +29.5 kJ⋅mol-1

5. N2H4(l) + H2(g) → 2NH3(g) –151 kJ⋅mol-1

6. 51.2 g

7. 2 and 3

8. no temperature

9. ΔG and ΔSsys

10. +0.683 J⋅K-1

11. 75.9 g

12. 2Fe2O3(s) + 3C(s, graphite) → 4Fe(s) + 3CO2(g)

301 kJ/mol

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:06 pm
by Sebastian
Could someone please explain #7 part 5? Thanks

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:39 pm
by Hue_Vo_1D
In an isolated system, neither matter nor energy is exchanged with the surroundings->no change in internal energy->deltaU=0

Got it! Thank you.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:47 pm
by Jocelyn_Dao_2N
Hi I was also wondering if someone could explain #7 and #9 on quiz one?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:49 pm
by Tyler Miao 1A
5kJ of energy is the internal energy itself, not the change in internal energy. The question wants to know how much the internal energy CHANGE is. Just because the starting internal energy is 5 doesn't mean anything in an isolated system, because there is no change, so the change of internal energy is 0.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:49 pm
by Chem_Mod
Hue_Vo_1D wrote:For #2,

If NEITHER matter nor energy is exchanged with the surroundings in an isolated system, then why does the internal energy goes down to 0? It must have something to do with the long period of time, 100 years? But what happen to the 5kJ of heat as time goes by?

What's the main concept behind this question?

Thanks in advance.


Please read the question. Their is no change in the internal energy, that is why ΔU = 0.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:50 pm
by Chem_Mod
Sebastian wrote:Could someone please explain #7 part 5? Thanks


5) A glass of water loses 100 J of energy reversibly at 30°C.

This is a decrease in entropy as energy left the system.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 3:57 pm
by Chem_Mod
Q9. Degeneracy at a minimum can be 1, never zero.

Changes in state functions of the system are zero when the system returns to the same initial conditions.

Q&A already posted here https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=133&t=18400&sid=0f7019b4dcfbdf821d0730f03b39cecb answers additional questions.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:01 pm
by Xin Huang 3E
I am also wondering about #9, specifically about ΔS(surr), why is it not 0? Isn't it a state function, too? I know that ΔS(sys) + ΔS(surr) = ΔS(univ) > 0, but that is only for a spontaneous process? Why is #9 a spontaneous process?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:01 pm
by Chem_Mod
***All equations and constants are given in all my quizzes and exams.***

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:03 pm
by BrianaBarr2A
For number 6, I'm a little stuck on which equation to use. Could someone help me with this?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:05 pm
by Chem_Mod
Xin Huang 3E wrote:I am also wondering about #9, specifically about ΔS(surr), why is it not 0? Isn't it a state function, too?


Please read carefully: Changes in state functions of the system are zero when the system returns to the same initial conditions.

Delta S (total) > 0 for this process. Therefore delta S (surroundings) > 0.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:17 pm
by Diego Cortez 2J
Can someone show me the setup for #6? Please and thank you!

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:53 pm
by Hue_Vo_1D
Diego Cortez 2J wrote:Can someone show me the setup for #6? Please and thank you!


-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 4:55 pm
by Hue_Vo_1D
BrianaBarr2A wrote:For number 6, I'm a little stuck on which equation to use. Could someone help me with this?


Use the equation for the combustion of propane.
C3H8 + 5O2 => 3CO2 + 4H2O (balanced)

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 5:08 pm
by Diego Cortez 2J
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:Can someone show me the setup for #6? Please and thank you!


-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

I balanced the equation, but I'm still confused on how to get the moles.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 5:14 pm
by Hue_Vo_1D
Diego Cortez 2J wrote:
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:Can someone show me the setup for #6? Please and thank you!


-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

I balanced the equation, but I'm still confused on how to get the moles.


2220 kJ/1 mol=2580 kJ/X mol

Solve for X

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 5:15 pm
by Hue_Vo_1D
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:
Hue_Vo_1D wrote:
-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

I balanced the equation, but I'm still confused on how to get the moles.


Use stoichiometry.
2220 kJ/1 mol=2580 kJ/X mol

Solve for X

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 5:22 pm
by Diego Cortez 2J
Hue_Vo_1D wrote:
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:I balanced the equation, but I'm still confused on how to get the moles.


Use stoichiometry.
2220 kJ/1 mol=2580 kJ/X mol

Solve for X

Thanks

Re: Quiz 1 Preparation Answers (#11)

Posted: Sun Jan 29, 2017 5:41 pm
by Hogan Irwin 3A
How many grams of water can be heated from 25.0 degrees Celsius to 100 degrees Celsius by the heat released from converting 49.7g of PbO to Pb?

The converting reaction is: PbO(s) + C(s) --> Pb(s) + CO(g) deltaH = -106.9kJ

How do I go about setting up this problem in order to solve it?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 6:10 pm
by Coleen_Leslie_2L
What formulas will be included on the formula sheet for Quiz 1?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 6:44 pm
by Martha Xuncax 3G
The same formulas from the back of your periodic table I believe.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 7:39 pm
by csebastiani_1B
Why doesn't process 1 in #7 lead to an increase in entropy?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 8:43 pm
by MaddyMoore3A
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:
Hue_Vo_1D wrote:
-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

I balanced the equation, but I'm still confused on how to get the moles.


2220 kJ/1 mol=2580 kJ/X mol

Solve for X


Hi, I'm also stuck on this problem. I've been trying to solve this but I keep getting a larger number than the given answer. I've done it five times now and get the same answer every time. Is this for sure the way to solve the problem? I don't understand where that ratio is coming from.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 9:10 pm
by Kira_Maszewski_1B
What equation do we use for work in #3?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 9:16 pm
by Alina_Mitchell_2C
Kira_Maszewski_1B wrote:What equation do we use for work in #3?

you use the equation w = -nRT ln (V2/V1) because the reaction is reversible and isothermal

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 9:49 pm
by IssaelG_3E
Can anyone explain how they calculated the enthalpy of vaporization for question 4?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 9:50 pm
by Claire_Zhou_1A
I read previous replies and still confused about #9. Could someone explain to me? I know that △S(system) is 0 because it's a state function, but why is △S(surroundings)>0? Also, w and q are not state functions, but the processes in the question seem to be the opposite, shouldn't the q for each step just cancel out and give w=0?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 10:08 pm
by Natalie Hunt 1E
Can anyone explain how they calculated q in #3? I was able to calculate w, and I know how to calculate deltaU if I can determine q. Do we assume that deltaU = 0, which would make q=-w? Thank you!

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 10:19 pm
by Natalie Hunt 1E
IssaelG_3E wrote:Can anyone explain how they calculated the enthalpy of vaporization for question 4?


deltaH sublimation = deltaH vaporization + deltaH fusion so by manipulating the equation to make deltaH vaporization = deltaH sublimation - deltaH fusion, you can then calculate the enthalpy of vaporization given the values of enthalpy of sublimation
and enthalpy of fusion.

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 10:28 pm
by Chem Student
If entropy is a state function, why is DeltaSurr not equal to zero when DeltaSys is?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 10:34 pm
by Nayeon Kang 2M
How can you solve #11? Is it right to use H=ncT here??

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 10:55 pm
by melissapasao1B
Can someone please explain how I should approach #11?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 11:16 pm
by Colleen_VeraCruz_3A
Could someone please explain how you get the answer for #11(How many grams of H20 can be heated from 25 degrees to 100 degrees by the heat released from converting 49.7g of PbO to Pb)? What equation(s) are we supposed to use?

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 11:25 pm
by Lily_Dermendjian_2B
melissapasao1B wrote:Can someone please explain how I should approach #11?


You divide 49.7 by the molar mass of PbO to get the number of moles, then multiply that by the deltaH to get the heat released by the reaction of that amount of PbO. You then use q=mCdeltaT and solve for m.

49.7g/(223.2g/mol)=0.223mol; 0.223mol*-106.9kJ=-23.8kJ
m=q/CdeltaT=(23.8*10^3 J)/((4.184 J/Cg)(75 C))=75.9 g H2O

#11 approach

Posted: Sun Jan 29, 2017 11:29 pm
by Veronica_Giap_1C
#11 approach

First you want to find the amount of heat released by the PbO reaction
-you do that by converting PbO grams -> mols
-set up ration of (-106.9kJ/1mol)=(q_p kJ/n mols of PbO)

With q_p that you just found, that is the amount of heat you have to heat m grams of H2O from 25C -> 100C
-find mass using q_p=mC[delta]T

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 11:43 pm
by Tara_Shooshani_3N
For #8, I don't understand why the reaction will not be spontaneous at any temperature.

Please help! Thanks!

Re: Quiz 1 Preparation Answers

Posted: Sun Jan 29, 2017 11:46 pm
by ahlam_ahmed_1d
Can someone help me understand number 7?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 12:43 am
by Emma_Green_2C
Tara_Shooshani_3N wrote:For #8, I don't understand why the reaction will not be spontaneous at any temperature.

Please help! Thanks!


If you plug in the given values into the equation deltaG= deltaH-T*deltaS and set delta G to zero, you will get a negative number for the temperature (-985.5K). You cannot have a temperature less than 0K, so a temperature less than -985.5K (which you would need for delta G to be negative and spontaneous) is impossible.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 12:53 am
by Simone Seliger 1C
Can someone verify that the equation needed to solve #12 is the one we learned for calculating standard free energy change? I am getting caught up in the wording: "Calculate the standard Gibbs free energy". Should I be using a different equation (even though we're given standard change in enthalpy of formation and standard entropy)? I am not getting the correct answer.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 10:28 am
by Myra_Zhan_2N
Simone Seliger 1C wrote:Can someone verify that the equation needed to solve #12 is the one we learned for calculating standard free energy change? I am getting caught up in the wording: "Calculate the standard Gibbs free energy". Should I be using a different equation (even though we're given standard change in enthalpy of formation and standard entropy)? I am not getting the correct answer.



You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 10:54 am
by Noelle Min-1N
Does anyone have any quizzes from previous year's coursereaders for extra practice?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 11:05 am
by Vivian Nguyen 2A
Natalie Hunt 1E wrote:Can anyone explain how they calculated q in #3? I was able to calculate w, and I know how to calculate deltaU if I can determine q. Do we assume that deltaU = 0, which would make q=-w? Thank you!


We can assume that deltaU = 0 because we are looking at an ideal gas that expands isothermally and reversibly.
Thus, like you stated, 0 = q + w and q = -w.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 12:09 pm
by Paul Adkisson 1D
Did anyone else notice that the initial conditions to Problem #3 are not consistent with PV = nRT?
(3atm)(6L) ≠ (2mol)(0.0826L-atm/mol-K)(300K)
18L-atm ≠ 49.56L-atm

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 12:13 pm
by Anna_Kim_2E
Where do you find the prep quiz?
I only see the midterms from past years.

Do I have to buy a separate booklet or is it in the course reader? Then what page is it in?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 12:17 pm
by Paul Adkisson 1D
Quiz 1 Preparation, pg 113

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 1:28 pm
by Jazmin_Morales_3J
Can someone please explain why on #5 the answer has 2NH3 on the product side? When I did the question and canceled everything, I divided 2NH3 by 2 so that it would be the same as in the target equation N2H4 + H2 = NH3. This gave me an enthalpy of -105.1kJ/mol, instead of -151 kJ/mol. Thanks

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 1:34 pm
by Hogan Irwin 3A
Jazmin_Morales_3J wrote:Can someone please explain why on #5 the answer has 2NH3 on the product side? When I did the question and canceled everything, I divided 2NH3 by 2 so that it would be the same as in the target equation N2H4 + H2 = NH3. This gave me an enthalpy of -105.1kJ/mol, instead of -151 kJ/mol. Thanks


I believe that is a typo. The 2 should not be in front of the NH3. Try the problem without the 2 and see what you get.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 2:05 pm
by Hue_Vo_1D
Jazmin_Morales_3J wrote:Can someone please explain why on #5 the answer has 2NH3 on the product side? When I did the question and canceled everything, I divided 2NH3 by 2 so that it would be the same as in the target equation N2H4 + H2 = NH3. This gave me an enthalpy of -105.1kJ/mol, instead of -151 kJ/mol. Thanks


I don't think it's a typo. It's just that you are supposed to balance the equation.

N2H4 + O2 => 2NH3

This will get you the right answer.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 2:15 pm
by Kira_Maszewski_1B
Chem_Mod wrote:
Xin Huang 3E wrote:I am also wondering about #9, specifically about ΔS(surr), why is it not 0? Isn't it a state function, too?


Please read carefully: Changes in state functions of the system are zero when the system returns to the same initial conditions.

Delta S (total) > 0 for this process. Therefore delta S (surroundings) > 0.


What are q and w equal to in this case? What equations would I to solve or these values?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 3:20 pm
by Kira_Maszewski_1B
Lily_Dermendjian_2B wrote:
melissapasao1B wrote:Can someone please explain how I should approach #11?


You divide 49.7 by the molar mass of PbO to get the number of moles, then multiply that by the deltaH to get the heat released by the reaction of that amount of PbO. You then use q=mCdeltaT and solve for m.

49.7g/(223.2g/mol)=0.223mol; 0.223mol*-106.9kJ=-23.8kJ
m=q/CdeltaT=(23.8*10^3 J)/((4.184 J/Cg)(75 C))=75.9 g H2O


Why does the q go from -23.8 to 23.8?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 3:42 pm
by Simone Seliger 1C
Myra_Zhan_2N wrote:
Simone Seliger 1C wrote:Can someone verify that the equation needed to solve #12 is the one we learned for calculating standard free energy change? I am getting caught up in the wording: "Calculate the standard Gibbs free energy". Should I be using a different equation (even though we're given standard change in enthalpy of formation and standard entropy)? I am not getting the correct answer.



You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.




That is the equation I am using. And I remembered to convert J to kJ. Still got the wrong answer.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 3:45 pm
by TramHo2G
For #12, what do I plug in for deltaS?
Is deltaH 2*(-824.2)?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 3:50 pm
by Kira_Maszewski_1B
Myra_Zhan_2N wrote:
Simone Seliger 1C wrote:Can someone verify that the equation needed to solve #12 is the one we learned for calculating standard free energy change? I am getting caught up in the wording: "Calculate the standard Gibbs free energy". Should I be using a different equation (even though we're given standard change in enthalpy of formation and standard entropy)? I am not getting the correct answer.



You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.


For some reason I am getting -301.49 kJ/mol... Do you have any idea why?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 6:40 pm
by EmilyLeibovitch2E
Is there a PDF that someone could post with exactly how to solve each problem on Quiz 1 Prep?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 8:47 pm
by Katrina_Domingo_3G
Hello,

Can someone please explain Question #9 on the quiz prep?

A sample of 1 mol of gas initially at 1 atm and 298 K is heated at constant pressure to 350 K, then the gas is compressed isothermally to its initial volume and finally it is cooled to 298 K at constant volume. Which of the following values is 0?
DeltaSsurr and DeltaG

DeltaSsys and DeltaSsurr

q and w

W

DeltaG and DeltaSsys

Can someone please explain why the other choices are not 0 and why Delta G and Delta Ssys have a value of 0?

Thank you so much!

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 8:58 pm
by Raul Hernandez
For number 3, what equation do we use since there seems to be a change in pressure? I can't seem to get the answer posted here by any method I've tried.
Thanks.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 9:10 pm
by RitaChang_2C
104450116 wrote:For number 3, what equation do we use since there seems to be a change in pressure? I can't seem to get the answer posted here by any method I've tried.
Thanks.


Use w=-nRT ln V2/V1
So then you'd get -(2.00 mol)(8.314 J/K mol)(300 K)(ln 18.00/6.00) to get -5480 J or -5.48 kJ.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 9:28 pm
by MaddyMoore3A
Simone Seliger 1C wrote:
Myra_Zhan_2N wrote:
Simone Seliger 1C wrote:Can someone verify that the equation needed to solve #12 is the one we learned for calculating standard free energy change? I am getting caught up in the wording: "Calculate the standard Gibbs free energy". Should I be using a different equation (even though we're given standard change in enthalpy of formation and standard entropy)? I am not getting the correct answer.



You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.




That is the equation I am using. And I remembered to convert J to kJ. Still got the wrong answer.

I also do not understand this question. The answer I am getting is not even close to 310, are we sure the equation is right?

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 9:44 pm
by Lauren Trent 2A
Hogan Irwin 3A wrote:
Jazmin_Morales_3J wrote:Can someone please explain why on #5 the answer has 2NH3 on the product side? When I did the question and canceled everything, I divided 2NH3 by 2 so that it would be the same as in the target equation N2H4 + H2 = NH3. This gave me an enthalpy of -105.1kJ/mol, instead of -151 kJ/mol. Thanks


I believe that is a typo. The 2 should not be in front of the NH3. Try the problem without the 2 and see what you get.


The equation given in the question is not balanced; once you balance the equation there will be a product of 2NH3.

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 11:06 pm
by Myra_Zhan_2N
Maddy Moore 1H wrote:
Simone Seliger 1C wrote:
Myra_Zhan_2N wrote:

You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.




That is the equation I am using. And I remembered to convert J to kJ. Still got the wrong answer.

I also do not understand this question. The answer I am getting is not even close to 310, are we sure the equation is right?


Following the equation: ΔG^o=ΔH^o−TΔS^o, I got ΔG^o=467.87 kJ/mol−(298K) (0.55832 kJ/K*mol) = 301 kJ

Re: Quiz 1 Preparation Answers

Posted: Mon Jan 30, 2017 11:10 pm
by RitaChang_2C
Maddy Moore 1H wrote:
Simone Seliger 1C wrote:
Myra_Zhan_2N wrote:

You would use ΔG^o=ΔH^o−TΔS^o after you find ΔH^o (rxn) and ΔS^o (rxn) values from the tables provided. Make sure your units match. You will need to convert ΔS^o to KJ for the final answer.




That is the equation I am using. And I remembered to convert J to kJ. Still got the wrong answer.

I also do not understand this question. The answer I am getting is not even close to 310, are we sure the equation is right?



Make sure to multiply each delta G by the number of moles in the balanced chemical equation. So multiply delta G for Fe by four, CO2 by three, and so on.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 6:01 am
by ariana_cruz_1C
Can someone explain why in #11 does q go from -23.8 to 23.8?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 8:21 am
by 804748473
csebastiani_1B wrote:Why doesn't process 1 in #7 lead to an increase in entropy?

This is because the pressure is doubled, meaning the system does not expand and thus takes up less space and the overall disorder in the system is not increased.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:07 am
by Nikhita_Jaaswal_2H
Diego Cortez 2J wrote:Can someone show me the setup for #6? Please and thank you!

(2580) /(2220)= value
Multiply this value by the molar mass of propane

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:27 am
by Alyssa_Hsu_2K
Hi, I was wondering why the sig figs for number 10 is 3 instead of 2? Doesn't the 20 degrees celsius mentioned in the problem only have 2 sig figs?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:42 am
by 804748473
If number 11 was on a quiz, would we be given the specific heat capacity of water?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:14 am
by ariana_cruz_1C
Why in #5 do we need to flip the reaction 2H2 + O2 = 2H2O and not the other ones ?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:14 am
by ariana_cruz_1C
804748473 wrote:If number 11 was on a quiz, would we be given the specific heat capacity of water?





It's on our formula sheet so I assume so!

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:36 am
by kgiesch1N
Can someone please explain why #8 is "no temperature"?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:39 am
by Kiara1F
RitaChang_2C wrote:
104450116 wrote:For number 3, what equation do we use since there seems to be a change in pressure? I can't seem to get the answer posted here by any method I've tried.
Thanks.


Use w=-nRT ln V2/V1
So then you'd get -(2.00 mol)(8.314 J/K mol)(300 K)(ln 18.00/6.00) to get -5480 J or -5.48 kJ.


how do you then find the heat to find the internal energy?
With q=nCdeltaT ..?
What would be the value of C?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 12:51 pm
by Julia Hwang 3G
Kiara1F wrote:
RitaChang_2C wrote:
104450116 wrote:For number 3, what equation do we use since there seems to be a change in pressure? I can't seem to get the answer posted here by any method I've tried.
Thanks.


Use w=-nRT ln V2/V1
So then you'd get -(2.00 mol)(8.314 J/K mol)(300 K)(ln 18.00/6.00) to get -5480 J or -5.48 kJ.


how do you then find the heat to find the internal energy?
With q=nCdeltaT ..?
What would be the value of C?


Rather than actually doing a new calculation, you analyze your results with the equation delta U = q + w.
Since work was equal to -5.48 kJ, you can use w = -q to know that q = +5.48 kJ. q and w then cancel each other out and delta U (change in energy) is 0.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:02 pm
by 904826427
ariana_cruz_1C wrote:Can someone explain why in #11 does q go from -23.8 to 23.8?


You cannot have negative mass so it's in thermodynamics that when it is transferred the energy is now positive

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:06 pm
by Anna_Kim_2E
ariana_cruz_1C wrote:Can someone explain why in #11 does q go from -23.8 to 23.8?


Because -23.8kJ is negative because it was released from converting PbO to Pb.
Then, that amount of heat is transferred into the system, which is water. When heat is put into an system, it is an endothermic process and the sign should be positive.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:08 pm
by John_Parks_3D
Can someone please explain why 3 is an answer for #7? I understand why number 2 is an answer.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:36 pm
by Ara Yazaryan 1E
Could someone explain to me how to get #10 please? I am confused because we do not know the final temperature.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:45 pm
by BetcyGaspar_3C
Ara Yazaryan 1E wrote:Could someone explain to me how to get #10 please? I am confused because we do not know the final temperature.


I do believe we use the equation deltaS=qrev/T,by plugging in 200J/(273+20) you will get the answer.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 1:53 pm
by Bronson_Barretto_2C
kgiesch1N wrote:Can someone please explain why #8 is "no temperature"?


The reasoning is based on the chart (pg. 37 in the course reader) Dr. Lavelle presented in lecture that according to the +/- values of ∆H and ∆S, we can assume the +/- of ∆G, negative being spontaneous and positive being non-spontaneous. For #8, the ∆H value is positive and the ∆S value is negative; based on the chart, these values give a ∆G that will NEVER be spontaneous. As such, the answer to the question is then none or n/a.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 4:28 pm
by Helena Vervaet 1N
Hue_Vo_1D wrote:
Diego Cortez 2J wrote:
Hue_Vo_1D wrote:
-Write out the equation for the combustion of propane and make sure to balance the eqn.
-From the given standard enthalpy of combustion (kJ/mol), figure out how many moles of propane is burned in order to release 2580. kJ of heat.(use stoichiometry).
-Convert moles to mass using MM.

Just let me know if you're still confused with any step, I can be more specific.

I balanced the equation, but I'm still confused on how to get the moles.


2220 kJ/1 mol=2580 kJ/X mol

Solve for X


Your explanation helped a bunch - Thanks! I was just wondering if anything would be different about this process if there had been a stoichiometric coefficient with the propane? Right now it's balanced with a 1, but if there had been a 2 instead would there be another calculation involved?

Re: Quiz 1 Preparation Answers (#11)

Posted: Tue Jan 31, 2017 4:46 pm
by Diana_Visco_1l
Hogan Irwin 3A wrote:How many grams of water can be heated from 25.0 degrees Celsius to 100 degrees Celsius by the heat released from converting 49.7g of PbO to Pb?

The converting reaction is: PbO(s) + C(s) --> Pb(s) + CO(g) deltaH = -106.9kJ

How do I go about setting up this problem in order to solve it?


q = (n)ΔH

(49.7 g PbO/22 g PbO ) (-106.9) = -23.8 KJ
-23.8 KJ x (1000 J / 1 KJ) = -23800 J

q= mcΔT

-23800 J = m (4.184) (100°-25°)

m= 75.9 g H20

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 8:00 pm
by Chanvir_Singh_3E
ariana_cruz_1C wrote:Why in #5 do we need to flip the reaction 2H2 + O2 = 2H2O and not the other ones ?


We need to flip the reaction because we need 2H2O on the reactant side so we can cancel it with the 2H2O on the product side in the given reaction N2H4 + O2 = N2 + 2H2O.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 9:01 pm
by Xin Huang 3E
Chem Student wrote:If entropy is a state function, why is DeltaSurr not equal to zero when DeltaSys is?

Because Entropy of surrounding is not a state function as entropy of the system does. And since deltaTotal is always > 0 (since this is the universal trend haha the world is always tending to confusion) while deltaTotal = deltaSurr + deltaSys and deltaSys = 0; => therefore, deltaSurr > 0.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:17 pm
by Ara Yazaryan 1E
BetcyGaspar_3C wrote:I do believe we use the equation deltaS=qrev/T,by plugging in 200J/(273+20) you will get the answer.


Thank you so much!

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:22 pm
by Jaime_Chamberlain_3G
Could someone please explain number 6?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:27 pm
by Minu Reddy
Hi Everyone,
I have a question on #10.

So when I did the problem, I divided 200/20, but I was wondering if someone can please explain the conversion factor to get the answer on the key?

Thank you

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 10:56 pm
by Brian Huynh 3D
Will we need a calculator for this test?

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:27 pm
by Sharlene Reyes 2N
Brian Huynh 3G wrote:Will we need a calculator for this test?


Yes, unless you're fine with the pain of doing all the calculations by hand. No graphing calculators though, so I scientific calculator or a more simple calculator.

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:39 pm
by Angela_Park_1F
Alyssa_Hsu_2K wrote:Hi, I was wondering why the sig figs for number 10 is 3 instead of 2? Doesn't the 20 degrees celsius mentioned in the problem only have 2 sig figs?


I had the same question..!

Re: Quiz 1 Preparation Answers

Posted: Tue Jan 31, 2017 11:43 pm
by deantuazon2G
So for number 11, do we just assume that the deltaH given is per mole?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 12:29 am
by Jake_Susi_2J
I know people have already posted how to find work (w) for #3 but what do we do to find heat (q)? I know the problem consists of an ideal gas, change in volume and pressure, etc.. but is it just a modified version of q=nC(delta_T)?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 2:16 am
by Angelica Nava-1E
BetcyGaspar_3C wrote:
Ara Yazaryan 1E wrote:Could someone explain to me how to get #10 please? I am confused because we do not know the final temperature.


I do believe we use the equation deltaS=qrev/T,by plugging in 200J/(273+20) you will get the answer.



Where did you get the 273 from? I know q=200J and the temp. is 20 degrees C but where did the 273 come from?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 9:39 am
by Chem_Mod
Angelica Nava-1E wrote:
BetcyGaspar_3C wrote:
Ara Yazaryan 1E wrote:Could someone explain to me how to get #10 please? I am confused because we do not know the final temperature.


I do believe we use the equation deltaS=qrev/T,by plugging in 200J/(273+20) you will get the answer.



Where did you get the 273 from? I know q=200J and the temp. is 20 degrees C but where did the 273 come from?


To convert degrees C to K.

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 11:01 am
by chris2E
Hi, I had some questions about sig figs.

For #1, shouldn't the answer be 310.J instead of 310J since both 254J and 564J have three significant figures?

For #10, shouldn't the answer be 0.7J/K instead of 0.683J/K since both 200J and 20°C have one significant figure?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 11:20 am
by BrianaBarr2A
Jake_Susi_2J wrote:I know people have already posted how to find work (w) for #3 but what do we do to find heat (q)? I know the problem consists of an ideal gas, change in volume and pressure, etc.. but is it just a modified version of q=nC(delta_T)?


From what I was told, for this q=-w so it's just the negative of whatever you got when solving for w

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 11:21 am
by BrianaBarr2A
deantuazon2G wrote:So for number 11, do we just assume that the deltaH given is per mole?


Yes just assume it's per mole

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 11:23 am
by BrianaBarr2A
Minu Reddy wrote:Hi Everyone,
I have a question on #10.

So when I did the problem, I divided 200/20, but I was wondering if someone can please explain the conversion factor to get the answer on the key?

Thank you


I'm not sure if the conversion factor is on the key for this one or not but you first have to convert degrees C to K so add 273.25 to 20

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 2:40 pm
by 404639305
John_Parks_3D wrote:Can someone please explain why 3 is an answer for #7? I understand why number 2 is an answer.


I think increasing temperature increases entropy because heat excites the molecules and causes more random movements.

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 5:38 pm
by Fayt Sarreal 1G
For #11, do the units not have to cancel out?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 6:05 pm
by Ashley Curtis 2O
Natalie Hunt 1E wrote:
IssaelG_3E wrote:Can anyone explain how they calculated the enthalpy of vaporization for question 4?


deltaH sublimation = deltaH vaporization + deltaH fusion so by manipulating the equation to make deltaH vaporization = deltaH sublimation - deltaH fusion, you can then calculate the enthalpy of vaporization given the values of enthalpy of sublimation
and enthalpy of fusion.


Why does this work? Why would you not have to factor in all the temperature changes to get to a temperature of vaporization or so?

Re: Quiz 1 Preparation Answers

Posted: Wed Feb 01, 2017 7:23 pm
by Valeria Quintana 1G
#8 on Quiz 1 Preparation says, "For reaction: 2C+2H2=C2H4, DeltaH= +52.3 kJ/mol and DeltaS= -53.07 J/K*mol at 298 K. At what temperature will this reaction be spontaneous?"

Initially, I thought the formula needed to solve for the temperature was DeltaG= DeltaH - T*DeltaS but then I saw that the answer is "no temperature." Could someone explain why the answer is "no temperature"? Please and thank you.