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Chapter 9: Rate Processes in Chemical Reactions

Chapter 9: Rate Processes in Chemical Reactions

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

Be the first to discuss this chapter!
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby Benjalyn » Sat Aug 08, 2015 1:07 am

The issue has been fixed.

Good luck with your studies :)
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Sun Oct 11, 2015 9:27 am

ilbe wrote:Queries with regards to Keq example in CHM9.8

In CHM 9.8 there is example of Keq as following
X(g) + 2Y(g) <-> Z(g)

in this case, can I use ICE method?
X(g) + 2Y(g) <-> Z(g)
I 5 12 0
C 5-x 12-2x x
E 5-4 12-8 4


The ICE method is used when there are unknowns in the concentrations which is not the case in this question.


ilbe wrote:Queries with regards to optimised web browser.

My appolosise to ask this kind of question in this thread. I was not able to find appropriate thread.
I used other computer and saw this thread can be 'edit' as the 'edit' button shows on the screen but my two laptops are not able to show 'edit' button.

What is optimised web browser? Please let me know. And you can move this thread to right place.

Thanks a lot.


In the instructions for each exam, Firefox is mentioned as the optimal browser.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Wed Nov 25, 2015 5:18 pm

1. Definitely not. 2 graphs can look identical, same curves, but that does not mean that the label of the axes must also be identical.

Gibbs free energy is a type of potential energy. BUT, when you see a potential energy diagram, then the difference between products and reactant is enthalpy: https://dr282zn36sxxg.cloudfront.net/da ... %2BIMAGE.1

If you are looking at a Gibbs free energy diagram, ONLY then is the difference between the products and reactants representing the change in Gibbs free energy: https://upload.wikimedia.org/wikibooks/ ... energy.JPG


2. That is not accurate. You cannot decide exothermic or endothermic only based on the sign of Gibbs free energy. You must take into account ALL terms in the relevant equation: ΔG = ΔH - T ΔS

It is important to see from the equation (ΔG = ΔH - T ΔS) that those terms should be clearly viewed as separate, important entities.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby laura.cp897117 » Fri Nov 27, 2015 7:24 pm

I have 2 questions:

[B]2) When the body’s intake of calcium is low, in order to maintain homeostasis, there is an increased renal absorption of calcium. Based on the information provided, which of the following changes in urine is most consistent with a decrease in calcium intake in a person with kidney stones?

A) Decreased oxalate ions
B) Increased solubility product
C) Decreased pH
D) Increased pH

Result: Your answer is wrong.
Your choice: A Correct choice: D

Explanation:
We learn from the question stem that a decrease in calcium intake is associated with increased renal absorption of calcium, which means there will be less calcium in the urine. Now we just apply Le Chatelier’s principle: Beginning with reaction (3), if calcium is low (i.e., being removed), the reaction shifts to the left, producing more oxalate ions; reaction (2) shifts to the left; and then reaction (1) shifts to the left. The net result of the latter two reactions is the net removal of hydrogen ions, which means that the pH increases. [B/]

Doesn't the question say that there is a DECREASE in calcium intake, therefore the [Ca+] in urine
would be higher?

[B] 7) The equilibrium constant of a reaction could be affected by which of the following?

A. the addition of a catalyst
B. an increase in temperature
C. a change in the initial concentrations of reactants
D. the reactants being converted mostly to product


Result: Your answer is incorrect.
Your choice: C Correct choice: B

Explanation: Answer choice B is correct as this is part of Le Chatelier’s principal. Moreover, changes in concentrations (answer choice C) or volume and/or pressure (answer D) can shift the position of an equilibrium constant without changing the equilibrium constant itself.

Adding a catalyst will decrease the activation energy; however, catalysts do not affect the position of an equilibrium as a catalyst will affect both the forward as well as the reverse reactions equally. A catalyst will only bring a system to equilibrium sooner and NOT affect the constant.

As for the temperature, changing the temperature of a system at equilibrium (i.e., an increase in temperature as for answer choice B), causes the equilibrium to shift because it changes the constant K. The enthalpy change of a reaction is the critical factor i.e., if it is exothermic, the heat is part of the products and is released and if it is endothermic, the heat is added and is part of the reactant side. So changing a temperature will affect the constant as the temperature will be part of the equation at equilibrium and added either to the reactant side (if endothermic) or product side (if exothermic). [B/]

Could you explain again how a change in temp causes the constant K to change? And also how this differs from a change in concentration does not cause the Keq to change? I assumed since Keq = [product]/[reactant], that the change in concentration will cause the Keq to change?
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Sat Nov 28, 2015 5:34 am

laura.cp897117 wrote:Doesn't the question say that there is a DECREASE in calcium intake, therefore the [Ca+] in urine would be higher?


I'm not following your reasoning. Here is what the explanation says on this issue, if it is not clear, please let me know: "We learn from the question stem that a decrease in calcium intake is associated with increased renal absorption of calcium, which means there will be less calcium in the urine". Logically if you eat less calcium, more calcium will be removed from urine in order for the body to maintain the same level of calcium (= homeostasis).

Just in case it's not clear: "intake" means the amount of food (or air) taken into the body. Intake does not refer to urine, nor any part of urine (this is called 'resorption').
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Sat Nov 28, 2015 6:03 am

laura.cp897117 wrote:Could you explain again how a change in temp causes the constant K to change? And also how this differs from a change in concentration does not cause the Keq to change?


Example
A + B <---> C + D + heat (exothermic)

Keq = [C][D] / [A][B]

If the temperature increases, it's like adding heat, so if the stress (added heat) is to the right side of the equation, then more A and B are made to relieve the stress. Therefore, given the equation for Keq above, the value of Keq must decrease as a result (because the relative magnitude of the denominator increases).


laura.cp897117 wrote: I assumed since Keq = [product]/[reactant], that the change in concentration will cause the Keq to change?


You are correct about the general structure for the equation for Keq. However, what do you mean by [product] and [reactant]? Those represent concentrations AT equilibrium! That's why Keq has "eq" for the equilibrium constant. Notice that answer choice C refers to "a change in the initial concentrations of reactants" which is not the same as referring to a change in concentration at equilibrium.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Tue Jan 26, 2016 5:57 pm

ilbe wrote:Queries with regards to sharpness of graphs of first, second order rxns.

Some sites show second order graph is sharper than first order graph and some site are opposite,
Which one is correct?

The site showing that second order graph is sharper:
https://www.chem.purdue.edu/gchelp/howt ... flife.html

The site showing that first order graph is sharper:
http://www.brynmawr.edu/chemistry/Chem/ ... flife.html

Please let me know.


1. You are asking an important question since it is assumed knowledge in order to understand ACER GAMSAT Practice Test 2, unit 2, especially question 9.

2. To get a perspective of the curves, see the diagram in CHM 9.2 in the GS book.

3. The 2 resources that you gave above are both correct, and both show the same result. You are doubting only because you may not have noticed that in the first resource https://www.chem.purdue.edu/gchelp/howt ... flife.html the x-axis is on a completely different scale. If you imagine the same scale, then all 3 resources above show the same general shapes with the first order graph descending more steeply than the second order graph.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby Neda » Fri Dec 29, 2017 7:25 pm

Could you please explain the following concept related to the Le Chatelier's principle?
"When there are different forms of a gaseous substance, an increase in total pressure favours the form with the greatest density, and a decrease in total pressure favours the form with the lowest density."
Thanks
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Sun Dec 31, 2017 6:18 am

Neda wrote:Could you please explain the following concept related to the Le Chatelier's principle?
"When there are different forms of a gaseous substance, an increase in total pressure favours the form with the greatest density, and a decrease in total pressure favours the form with the lowest density."
Thanks


There are many different ways to think about this, one way is as follows:

Density = (mass)/(volume)

Just be looking at the above equation, we can see that if mass is constant then lower volume means greater density.

How can you lower volume? Increase pressure!

(you can also prove the latter with common experience, or with the Boyle's relationship that PV = constant)
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby Neda » Sun Dec 31, 2017 8:05 am

Thanks for the information, but still it is unclear.
According the Le Chatelier principle, "If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change."
The question is:
Why an increase in total pressure favours the form with the greatest density? Why shouldn't it favour the form with the lower density which has greater volume to offset the change?
Regards
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Mon Jan 01, 2018 3:17 am

Neda wrote:Thanks for the information, but still it is unclear.
According the Le Chatelier principle, "If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change."
The question is:
Why an increase in total pressure favours the form with the greatest density? Why shouldn't it favour the form with the lower density which has greater volume to offset the change?
Regards


OK, let's write it out as a reaction at equilibrium so we can imagine what is really going on:

high density <--> low density

What does this mean in terms of volume?

low volume <--> high volume

Now, let's add pressure to the above reaction and it squishes high volume down to low volume which is high density.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby brad.willo2670 » Thu Aug 08, 2019 8:30 pm

9.6 Kinetic control vs thermodynamic control

The paragraph states 'suppose that C has the lowest Gibbs free energy (i.e. it is the most thermodynamically stable product)'.

Other than a measure of spontaneity, I'm trying to grasp the relationship between Gibbs and potential energy. I understand that gibbs is derived from enthalpy which in an of itself is derived from internal energy.
So what this passage is suggesting is that the lower the gibbs free energy, the lower the potential energy and therefore stability of the product. This is why the product is likely to form spontaneously.
Is that the general gist or do I have it blurred?

Thanks
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby goldstanda3269 » Fri Aug 09, 2019 10:11 pm

Thermodynamic free energy (one type being Gibbs free energy) is a subset of the total potential energy. They are similar but the differences are complex and far beyond this exam (https://physics.stackexchange.com/quest ... ing/146328).

The following similarity does pop up on GAMSAT, the curves of the 2 types of energy vs reaction progress (i.e. going from reactants to products):

https://ka-perseus-images.s3.amazonaws. ... e6577b.png (example of energy being released to the surroundings)

vs

https://cdn.kastatic.org/ka-perseus-ima ... c7f05f.svg (example of energy being absorbed from the surroundings)

Of course, for both, the lower the energy, the more stable.
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Re: Chapter 9: Rate Processes in Chemical Reactions

Postby brad.willo2670 » Tue Sep 03, 2019 6:11 pm

Differential vs Integrated rate laws:

To this point I have had minimal issue with these kinds of questions but something has just confused me regarding the graphing of these laws.

Example 1st order reaction:
Differential equation - rate = k(x)
Integrated equation - ln(X)t = -kt + ln(x)o

My question: is the differential equation ever graphed, or is it always the integrated equation? The integrated equation providing concentration vs time graph.
I have never seen 'rate' on an axis which is what came to confuse me. Does the differential equation simply provide a representation of the relationship of the concentration to the rate as opposed to something that would be directly graphed?

Thanks in advance!
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