### Re: Quiz 1 Preparation Answers

Posted:

**Wed Feb 01, 2017 7:26 pm**Can someone please explain #5? I can't seem to get the right answer.

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=160&t=18554

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Posted: **Wed Feb 01, 2017 7:26 pm**

Can someone please explain #5? I can't seem to get the right answer.

Posted: **Wed Feb 01, 2017 9:49 pm**

for #1 I don't understand why the answer is 310J not -310J

Posted: **Wed Feb 01, 2017 10:11 pm**

In the email we received about information on significant figures, I came across this:

"When your calculation involves more than one operation, you will need to follow the standard order of operations. You may have learned the order of operations as PEMDAS (parentheses, then exponents, then multiplication/division, and finally addition/subtraction). You should only round your final answer (see the caution below in the next section), but you will need to keep track of the sig figs carried through each step of the calculation if you want to get to the correct answer."

Will we need to do this on quizzes and exams?

"When your calculation involves more than one operation, you will need to follow the standard order of operations. You may have learned the order of operations as PEMDAS (parentheses, then exponents, then multiplication/division, and finally addition/subtraction). You should only round your final answer (see the caution below in the next section), but you will need to keep track of the sig figs carried through each step of the calculation if you want to get to the correct answer."

Will we need to do this on quizzes and exams?

Posted: **Wed Feb 01, 2017 10:12 pm**

On question #7, #5, a glass of water loses 100J of energy reversibly at 30 degrees C. How would this affect entropy?

Posted: **Wed Feb 01, 2017 10:39 pm**

John_Parks_3D wrote:Can someone please explain why 3 is an answer for #7? I understand why number 2 is an answer.

I believe 3 is an answer because by following PV = nRT, if temperature increases and pressure and number of moles stays constant, the volume must also increase to match the temperature and therefore the entropy increases because of more possible states in the system.

Posted: **Wed Feb 01, 2017 10:59 pm**

For #12, should the units of free energy be in kJ, not kJ/mol, because you had to multiply by the number of moles of each reactant and product in the calculation?

Posted: **Wed Feb 01, 2017 11:27 pm**

Can someone please explain the answer to #8? Thank you!

Posted: **Wed Feb 01, 2017 11:31 pm**

Valeria Quintana 1G wrote:#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.

There is no temperature because when the enthalpy is positive and the entropy is negative the reaction is non spontaneous. This can be found on page 37 of your course reader.

Posted: **Thu Feb 02, 2017 12:11 am**

kgiesch1N wrote:Can someone please explain why #8 is "no temperature"?

since delta G = delta H - T* delta S, and delta S is negative, the value of delta G will always be positive, no matter what value is inserted. In order for a process to be spontaneous, delta G has to be negative, and in this case, no matter what T is, it'll always be positive.

Posted: **Thu Feb 02, 2017 12:21 am**

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 you balanced the equation!

Posted: **Thu Feb 02, 2017 12:40 am**

Valeria Quintana 1G wrote:#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.

That's the right equation that you use. In this case, you have to set DeltaG=0 because that's the critical point where anything less will be negative, and therefore spontaneous.

So now you have 0= DeltaH - T*DeltaS. Isolate T and solve for it.

T=DeltaH/DeltaS. You find that you get a negative temperature for your answer which is impossible. This reaction will not be spontaneous for any temperature.

Posted: **Thu Feb 02, 2017 2:43 am**

Angela_Park_1F wrote: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..!

You have to change the temp to kelvin, which has three sig figs

Posted: **Thu Feb 02, 2017 4:02 am**

When delta H is positive and delta S is negative, there is no way for there to be spontaneity at any temperature. This is is because entropy of the system is decreasing, and the enthalpy of heat is increasing. naturally, this would never occur in the real world. If it was vice versa, a positive Delta S and negative Delta H would result in spontaneity at all temperatures.

Posted: **Thu Feb 02, 2017 7:34 am**

Valeria Quintana 1G wrote:#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.

This is in the lecture notes and in the course reader.

See the table discussed in detail in class: positive Delta H and negative Delta S therefore Delta G can only be positive because

Go through table I discussed in class for the signs of Delta G.

Posted: **Thu Feb 02, 2017 9:10 am**

For #3 on the practice quiz, why do we ignore the change in pressure? To solve it, we use the reversible equation for work which regards a change in Volume, but not a change in pressure.

Posted: **Thu Feb 02, 2017 1:40 pm**

for #12, do we multiply the coefficients of the balanced equation by both the delta h and s values?

Posted: **Thu Feb 02, 2017 3:49 pm**

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.

In order for the reaction to be spontaneous, DeltaG has to be negative. In this case, since you are subtracting a negative DeltaS from a positive DeltaH, DeltaG would yield a positive answer. Since it's positive, there is no way it can be spontaneous, so the answer is "no temperature".

Posted: **Thu Feb 02, 2017 6:53 pm**

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.

The answer is no temperature because when you solve for G=0, you get around -900K. The Kelvin scale does not go below 0K, therefore it is impossible.

Posted: **Thu Feb 02, 2017 9:54 pm**

For #6, why does the answer have 3 sig figs? We didn't use the 298.K, so shouldn't the sig figs be 4 (for 2580kJ and -2220.kJ/mol)?

Posted: **Thu Feb 02, 2017 9:54 pm**

For #6, why does the answer have 3 sig figs? We didn't use the 298.K, so shouldn't the sig figs be 4 (for 2580kJ and -2220.kJ/mol)?

Posted: **Fri Feb 03, 2017 12:00 am**

This entire thread has saved my life :)

Posted: **Fri Feb 03, 2017 12:36 am**

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

For this question, one can't assume that the N2H4 + H2 --> NH3 is balanced already. I ran into the same problem but then i realized that there's no way you could have that equation because the N2H4 reactant species has 2 Nitrogen atoms, so the product (NH3) must have a 2 coefficient in front of it.

Posted: **Fri Feb 03, 2017 5:24 am**

For number 9 why is delta S sys zero? shouldn't it be negative?

Posted: **Fri Feb 03, 2017 10:26 am**

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.

Sorry this is late and you've probably taken your quiz already. You have the correct formula, so if you plug in what they give you you will always have a positive value no matter what the temperature is. And for the reaction to be spontaneous DeltaG has to be negative. Hope this helps. :)

Posted: **Sun Feb 05, 2017 4:35 pm**

When do we get our quiz back??

Posted: **Wed Feb 08, 2017 2:35 pm**

This is about actual quiz 1...

It's about the last two parts of the questions about heat gained and lost.

During the winter, the water tank of a locomotive contains 100.0 x10^3 g of superheated steam at 155.0 C which cools to liquid water at 5 C. Assume no steam leaves the tank.

b) Calculate the heat gained or lost by the 100.0 x 10^3 g superheated steam during condensation. Assume standard pressure.

c) Calculate any other heat gained or lost and the total heat gained or lost by the steam as it goes from 155.0 to 5.0 C

I just don't know how to set it up because I am pretty sure I don't know what's going on conceptually.

It's about the last two parts of the questions about heat gained and lost.

During the winter, the water tank of a locomotive contains 100.0 x10^3 g of superheated steam at 155.0 C which cools to liquid water at 5 C. Assume no steam leaves the tank.

b) Calculate the heat gained or lost by the 100.0 x 10^3 g superheated steam during condensation. Assume standard pressure.

c) Calculate any other heat gained or lost and the total heat gained or lost by the steam as it goes from 155.0 to 5.0 C

I just don't know how to set it up because I am pretty sure I don't know what's going on conceptually.

Posted: **Thu Feb 09, 2017 12:58 pm**

Is there any chance we will get the answers to Quiz 1 to help study for the midterm?

Posted: **Mon Feb 13, 2017 4:29 pm**

For those looking for quiz 1 solutions, Wednesday's was posted here: viewtopic.php?f=160&t=18839

Posted: **Thu Feb 16, 2017 12:17 pm**

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.

Delta H is +

Delta S is -

Not possible for delta G to be negative at any temperature.

See page 37 in the course reader.

Posted: **Sat Mar 18, 2017 9:10 am**

When do we use Kelvin and when do we use C?

Posted: **Sat Mar 18, 2017 9:11 am**

How do you do Q6?