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Chapter 8: Enthalpy and Thermochemistry

Chapter 8: Enthalpy and Thermochemistry

Postby admin » Fri Aug 27, 2010 12:31 am

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Re: Chapter 8: Enthalpy and Thermochemistry

Postby Tmeredith7702 » Sat Jun 21, 2014 3:13 am

Re: Q10 for chapter 8 of general chemistry

The answers says that R is greatest for D. But, R = thickness / k. D's thickness is 0.7 and k is 0.04. Hence R = 0.7/0.04 = 17.5. But A's thickness is 75 and k is 0.05, hence R = 75/0.05 = 1,500. Why isn't A therefore the right answer?
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Sun Jun 22, 2014 4:47 am

With the information that you had, you definitely got the answer correct.

There was one dot missing (i.e. a typo). A's thickness of 75 was meant to be .75, hence R = 15 and D is the greatest at 17.5. The question has been updated. We also took the opportunity to add one question to the end of that passage/page which requires the use of dimensional analysis (a popular GAMSAT subtopic) which we think you should try. Good luck!
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby ElleRT » Sat Aug 22, 2015 12:21 am

Hi,

For page 103 of chem. I was wondering for the second example why do you flip teh H2O (l) --> H20 (g) round?

Thanks
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby ElleRT » Sat Aug 22, 2015 12:52 am

Hi,

Just wondering at the start of the chapter it says about understand the area under curve of PV diagram but I can't find any diagram in this chapter so I'm not sure where to look.

Thanks
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Sat Aug 22, 2015 7:25 am

ElleRT wrote:Hi,

For page 103 of chem. I was wondering for the second example why do you flip teh H2O (l) --> H20 (g) round?

Thanks


It all begins with the first reaction R1. The first reaction R1 is "flipped" so that B2O3 gets on the right side which is the desirable result (i.e. that is where B2O3 is in the desired summary reaction R). But when R1 is switched, H2O(g) in R1 is now on the right side. But there is no H2O(g) in reaction R so we need a way to get rid of H2O(g). The only way to do so is to have the same amount of H2O(g) on the OTHER side of the reaction so that all H2O(g) cancels. So R2 is "flipped" so that H2O(g) goes to the left side, also with a 3 in front, so all H2O(g) cancels.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Sat Aug 22, 2015 7:34 am

ElleRT wrote:Hi,

Just wondering at the start of the chapter it says about understand the area under curve of PV diagram but I can't find any diagram in this chapter so I'm not sure where to look.

Thanks


Well, I did not want to give away the surprise but let's say the matter will be clarified after completing your chapter review practice questions for this chapter!
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A question about 22) in Q & A

Postby li.zhang96890 » Thu Nov 05, 2015 2:15 pm

"22) When the reaction N2(g) + 3H2(g) <--> 2NH3(g) is at equilibrium, it is "far to the right." The forward reaction must be:

A) ectoplasmic.
B) exothermic.
C) endothermic.
D) endergonic.

Your Answer: B
Correct Answer: B

Explanation: ΔG = ΔH - T(ΔS)
Since the reaction is spontaneous, ΔG < 0.
Since ΔS < 0, -T(ΔS) > 0.
Thus, for ΔG to be strictly less than zero, ΔH must be strictly less than zero, which means that the reaction is exothermic."

My question is how do you get "ΔS < 0" and "ΔG < 0"? The question says the reaction is at equilibium, so ΔG = ΔH - T(ΔS)= 0. And another question is that I cannot interpret any information from " it is far to the right". So could you tell me what can I get from this condition?
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A question about 21) in Q & A

Postby li.zhang96890 » Thu Nov 05, 2015 2:35 pm

"21) In the reaction N2(g) + 3H2(g) <--> 2NH3(g), the entropy among the molecules involved:

A) increases.
B) remains the same.
C) decreases.
D) cannot be determined with the information given.

Your Answer: A
Correct Answer: C

Explanation: In this reaction, four molecules react to form two molecules, thus increasing the order and decreasing the entropy."

According to the second law of thermodynamics, entropy of an isolated system will never decreases, so I chose A. Is it correct that for any reaction, entropy of the system always increases?
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Re: A question about 22) in Q & A

Postby goldstanda3269 » Fri Nov 06, 2015 12:47 am

li.zhang96890 wrote:"22) When the reaction N2(g) + 3H2(g) <--> 2NH3(g) is at equilibrium, it is "far to the right."

My question is how do you get "ΔS < 0" and "ΔG < 0"? The question says the reaction is at equilibium, so ΔG = ΔH - T(ΔS)= 0. And another question is that I cannot interpret any information from " it is far to the right". So could you tell me what can I get from this condition?


"Far to the right" means that there is far more product(s) than reactant(s). There are many interpretations depending on the situation, for example: 1) The reaction moves to completion (i.e. from left to right); 2) the equilibrium constant K must be greater than 1, which means that Gibbs free energy ΔG is negative (ΔG = -RT ln K; CHM 9.10).

It is very important for the GAMSAT to quickly determine the sign from entropy (randomness) which is negative when decreasing and positive when increasing. The reaction provided goes from 4 moles down to 2 moles, from 2 molecules to 1 molecule, on both accounts, it is getting more organized, LESS random so ΔS < 0.

"Is it correct that for any reaction, entropy of the system always increases?" is saying 2 things at the same time which are in conflict: the reaction vs. the system. In a reaction, entropy can go up or down. In the system, entropy always increases. When you see a reaction, you are looking at it in isolation, we are not being provided additional information (where is it, what else is happening, what is giving energy to the reaction or taking energy from it) and so the entropy of the reaction must be determined since it can be going up or down.
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Re: after reading the reply about Q22, I still have a question

Postby li.zhang96890 » Sun Nov 08, 2015 12:40 pm

At first, thank you for your reply. In the video of this chapter and on the book CHEM 8.10 Free Energy, it all mentioned that the reaction is at a state of equilibium if ΔG= 0. So I am confused because this contradicts your conclusion which is ΔG < 0. Honestly, I think more products than reactants does not necessarily mean K must be greater than 1 since no details of concentration of reactants and products are given in the question. If, for example, [NH3] = 16M, [N2] = 4 M and [H2]= 4M, we can say the reaction is far to the right. But in this case, Keq= [NH3]2/[N2][H2]3 = 1. Then, according to ΔG = -RT ln K, ΔG= 0.

What do you think?
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Mon Nov 09, 2015 11:15 am

"Far to the right" (or left) is not a clear, scientific statement. Despite this, it is an expression used relatively often is chemistry and biochemistry, and its purpose, due to the word "far" is to consider something extreme. You have described a situation where 2 reactants with a total of a molarity of 8 produces a product with a molarity of 16, whereas, "far" would certainly include a 10 times difference, or 100 times, or 1000 times, rather than double or 4 times.

Of course, there is context here: saying "far" and then having limited answers within the context of considering extreme values.

"the reaction is at a state of equilibium if ΔG= 0"
- True, by definition.

OK, let's ignore the explanation given to the question and start over while strictly considering the issue of equilibrium.

ΔG = ΔH - T(ΔS)
0 = ΔH - T(ΔS)
ΔH = T(ΔS)
Keeping in mind that T in kelvin must always be positive, this means that ΔH and ΔS must always have the same sign. So now we just need to determine: What is the sign for ΔS? Well, as mentioned previously, entropy is clearly decreasing in the reaction provided (going to the right), meaning that ΔS must be negative, meaning ΔH must be negative, meaning the reaction is exothermic.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Tue Dec 29, 2015 4:45 am

ilbe wrote:Queries with regards to Q12.

1.Spherical structure is the minimum area under the same volume condition?
2. Can I ask that would be the maximum area for the same volume?


1. Yes.

2. Not GAMSAT assumed knowledge. If you want to explore it: https://en.wikipedia.org/wiki/Surface-a ... lume_ratio
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby Sheri » Tue Mar 08, 2016 10:47 pm

Hi GS,

Could you explain q12?
I don't get the explanation there. "Thus, when competing with shapes of equal volume, it will have a smaller surface area than any of the rest. Minimizing the area translates into reducing the size of the numerator while, by hypothesis, the denominator stays the same. This should produce a decrease in Q – an increase in energy efficiency."
which equation is the question/answer referring to when stating numerator and denominator?
Thanks!
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