Quiz 1 Preparation Answers

[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

[Sebastian]

Could someone please explain #7 part 5? Thanks

[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.

[Jocelyn_Dao_2N]

Hi I was also wondering if someone could explain #7 and #9 on quiz one?

[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.

[Chem_Mod]

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.

[Chem_Mod]

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.

[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.

[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?

[Chem_Mod]

***All equations and constants are given in all my quizzes and exams.***

[BrianaBarr2A]

For number 6, I'm a little stuck on which equation to use. Could someone help me with this?

[Chem_Mod]

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.

[Diego Cortez 2J]

Can someone show me the setup for #6? Please and thank you!

[Hue_Vo_1D]

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.

[Hue_Vo_1D]

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)

[Diego Cortez 2J]

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.

[Hue_Vo_1D]

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

[Hue_Vo_1D]

-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

[Diego Cortez 2J]

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

[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?

[Coleen_Leslie_2L]

What formulas will be included on the formula sheet for Quiz 1?

[Martha Xuncax 3G]

The same formulas from the back of your periodic table I believe.

[csebastiani_1B]

Why doesn't process 1 in #7 lead to an increase in entropy?

[MaddyMoore3A]

-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.

[Kira_Maszewski_1B]

What equation do we use for work in #3?

[Alina_Mitchell_2C]

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

[IssaelG_3E]

Can anyone explain how they calculated the enthalpy of vaporization for question 4?

[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?

[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!

[Natalie Hunt 1E]

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.

[Chem Student]

If entropy is a state function, why is DeltaSurr not equal to zero when DeltaSys is?

[Nayeon Kang 2M]

How can you solve #11? Is it right to use H=ncT here??

[melissapasao1B]

Can someone please explain how I should approach #11?

[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?

[Lily_Dermendjian_2B]

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

[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

[Tara_Shooshani_3N]

For #8, I don't understand why the reaction will not be spontaneous at any temperature.

Please help! Thanks!

[ahlam_ahmed_1d]

Can someone help me understand number 7?

[Emma_Green_2C]

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.

[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.

[Myra_Zhan_2N]

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.

[Noelle Min-1N]

Does anyone have any quizzes from previous year's coursereaders for extra practice?

[Vivian Nguyen 2A]

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.

[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

[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?

[Paul Adkisson 1D]

Quiz 1 Preparation, pg 113

[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

[Hogan Irwin 3A]

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.

[Hue_Vo_1D]

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.

[Kira_Maszewski_1B]

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?

[Kira_Maszewski_1B]

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?

[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.

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.

[TramHo2G]

For #12, what do I plug in for deltaS?
Is deltaH 2*(-824.2)?

[Kira_Maszewski_1B]

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?

[EmilyLeibovitch2E]

Is there a PDF that someone could post with exactly how to solve each problem on Quiz 1 Prep?

[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!

[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.

[RitaChang_2C]

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.

[MaddyMoore3A]

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?

[Lauren Trent 2A]

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.

[Myra_Zhan_2N]

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

[RitaChang_2C]

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.

[ariana_cruz_1C]

Can someone explain why in #11 does q go from -23.8 to 23.8?

[804748473]

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.

[Nikhita_Jaaswal_2H]

Can someone show me the setup for #6? Please and thank you!

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

[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?

[804748473]

If number 11 was on a quiz, would we be given the specific heat capacity of water?

[ariana_cruz_1C]

Why in #5 do we need to flip the reaction 2H2 + O2 = 2H2O and not the other ones ?

[ariana_cruz_1C]

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!

[kgiesch1N]

Can someone please explain why #8 is "no temperature"?

[Kiara1F]

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?

[Julia Hwang 3G]

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.

[904826427]

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

[Anna_Kim_2E]

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.

[John_Parks_3D]

Can someone please explain why 3 is an answer for #7? I understand why number 2 is an answer.

[Ara Yazaryan 1E]

Could someone explain to me how to get #10 please? I am confused because we do not know the final temperature.

[BetcyGaspar_3C]

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.

[Bronson_Barretto_2C]

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.

[Helena Vervaet 1N]

-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?

[Diana_Visco_1l]

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

[Chanvir_Singh_3E]

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.

[Xin Huang 3E]

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.

[Ara Yazaryan 1E]

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!

[Jaime_Chamberlain_3G]

Could someone please explain number 6?

[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

[Brian Huynh 3D]

Will we need a calculator for this test?

[Sharlene Reyes 2N]

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.

[Angela_Park_1F]

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..!

[deantuazon2G]

So for number 11, do we just assume that the deltaH given is per mole?

[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)?

[Angelica Nava-1E]

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?

[Chem_Mod]

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.

[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?

[BrianaBarr2A]

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

[BrianaBarr2A]

So for number 11, do we just assume that the deltaH given is per mole?

Yes just assume it's per mole

[BrianaBarr2A]

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

[404639305]

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.

[Fayt Sarreal 1G]

For #11, do the units not have to cancel out?

[Ashley Curtis 2O]

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?

[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.