October 21, 2019, 04:31:51 pm

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Bri MT

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« Reply #8160 on: October 05, 2019, 06:49:48 pm »
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Hi Guys,

Why doesn't a change in pressure affect the equilibrium constant (Kc)?

The equilibrium constant is defined as the value of the concentration fraction when equilibrium is reached. It indicates the extent of reaction (i.e. how far a reaction will proceed towards the products).

Let's say a decrease in pressure causes a net forward reaction due to Le Chatelier's principle (as there are more particles of products than there are reactants). Wouldn't this increase the yield of products AND increase the value of the equilibrium constant (Kc)? Because an increase in yield = increase in products = reaction proceeds more towards products (i.e. increased extent of reaction)?

I might sound confusing here, however remember that equilibrium constant is different from equilibrium position. Changes in pressure affect the equilibrium position but not the equilibrium constant (Kc). And why they do not affect both, please help me out haha. The book states that the equilibrium constant (Kc) is altered ONLY by changes in temperature and no other factor(s).

When you look at the equation, remember that the concentrations are to the power of the co-efficients.
Let's say you had 2A <=> B
Originally you had [ B]/[A]^2 =K
Now you double the volume which means the concentration of each drops by half. If we just divide by half:
([ B]/2)/([A]/2)^2 =K
[ B]/([A]/2)^2 = 2K
[ B]/[A]^2 * 2^2 = 2K
[ B]/[A]^2 = .5K  <- but this can't be true

For the equilbibrium constant to stay the same when there's a change in volume & uneven amounts of moles on each side,  the position of equilibrium has to shift.

Hope this helps

Edit: oh no the forums thought I was trying to bold whenever I indicated the concentration of B. Hopefully that's fixed now
« Last Edit: October 12, 2019, 03:05:50 pm by Bri MT »
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Erutepa

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« Reply #8161 on: October 05, 2019, 07:46:18 pm »
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Hi Guys,

Why doesn't a change in pressure affect the equilibrium constant (Kc)?

The equilibrium constant is defined as the value of the concentration fraction when equilibrium is reached. It indicates the extent of reaction (i.e. how far a reaction will proceed towards the products).

Let's say a decrease in pressure causes a net forward reaction due to Le Chatelier's principle (as there are more particles of products than there are reactants). Wouldn't this increase the yield of products AND increase the value of the equilibrium constant (Kc)? Because an increase in yield = increase in products = reaction proceeds more towards products (i.e. increased extent of reaction)?

I might sound confusing here, however remember that equilibrium constant is different from equilibrium position. Changes in pressure affect the equilibrium position but not the equilibrium constant (Kc). And why they do not affect both, please help me out haha. The book states that the equilibrium constant (Kc) is altered ONLY by changes in temperature and no other factor(s).
I know Bri MT beat me to it, but I thought I would add my inferior and probably convoluted 2 cents

I also had trouble wrapping my head around this and this reasoning here helped me understand a bit.

As we know, If the pressure of the system is increased, the system will partially oppose this change favouring the reaction that produces less moles of gas, such to decrease the pressure.
If we regard the equilibrium equation:

Initially, the Kc of this reaction would be given as

Making up random concentration values for an established equilibrium where

Thus

Now, if we were to double the pressure, the reaction would no longer be at equilibrium, and the reaction quotient would no longer equal the equilibrium constant as shown:

As we also know, the system will partially oppose this increase in pressure by favouring the forward reaction, thus increasing the concentration of N2 O4(g) and decreasing the concentration of NO2(g). This would occur to the extent that equilibrium is re-attained as the reaction quotient becomes equal to the initial equilibrium constant.
Thus, a change in pressure can be said to change the equilibrium position, however, does not change the equilibrium constant since, as it partially opposes the change in pressure, the reaction quotient becomes equal to the initial equilibrium constant.
It is important to now, however, that any of this only occurs when there is a difference in the molarity of gasses on each side of the reaction.

However, if we regard an increase in temperature for the reaction:

The system is no longer at equilibrium as it will favour the reverse reaction to decrease the temperature. However initially, the reaction quotient is unchanged as the change in temperature does not initially affect the concentrations of the products and reactants. Thus, as the reverse reaction is favoured and equilibrium is reattained, the equilibrium concentrations are changed thus changing equilibrium constant.

Hopefully, this logic is sound, although, please do point out anything nonsensical.
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« Reply #8162 on: October 08, 2019, 07:58:04 pm »
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I don't know which thread to ask this but I have major worries for my chem study score.
I am rank 4 for unit 3 and rank 2 for unit 4. Rank 1 is the same person for both units but i fear that she may not perform at her best due to burning out. So let's say i get around 80-85 for the exam and she gets around 70, by how much will my study score get effected. Im trying to get into my 40s so i want to know how much i will be affected.

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« Reply #8163 on: October 09, 2019, 09:37:54 pm »
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Hi, I'm wondering what exam scores are needed to get a 43, 45 and/or 47 study score in Chem on the final exam. Thanks!

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« Reply #8164 on: October 10, 2019, 12:44:05 am »
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Hi, I'm wondering what exam scores are needed to get a 43, 45 and/or 47 study score in Chem on the final exam. Thanks!

Assuming you have at least an A SAC average with a decently strong cohort I would say getting a 43, 45 or 47 will require somewhere around 102/120, 107/120 and 111/120 respectively. Note that this would also be on an easy exam (e.g. 2017).
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EllingtonFeint

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« Reply #8165 on: October 10, 2019, 09:58:11 am »
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Hey,
I'm sure this is probably answered in the study design somewhere but I can't find it...
Should I memorise the structures and formulas of glucose and fructose for the exam??
Thank you
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Erutepa

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« Reply #8166 on: October 10, 2019, 10:15:04 am »
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Hey,
I'm sure this is probably answered in the study design somewhere but I can't find it...
Should I memorise the structures and formulas of glucose and fructose for the exam??
Thank you
No, you don't need to memorise these. Both of these molecules are in the data booklet given to you in the exam .
you can find the data booklet here: https://www.vcaa.vic.edu.au/Documents/exams/chemistry/chemdata-w.pdf
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EllingtonFeint

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« Reply #8167 on: October 10, 2019, 11:47:03 am »
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Hey,
I've been asked to compare the density of starch and cellulose.
I was just looking at the structural formula of the two molecules and it looks like cellulose has more of a 'kink' in its polymers (?) than starch so that it doesn't pack as closely but I'm not quite sure if this reasoning is adequate (or even true).
I'm just curious as to whether or not my idea is correct and which molecule would be more dense.
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galaxy21

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« Reply #8168 on: October 10, 2019, 11:59:36 am »
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Do we need to know gravimetric analysis for the exam? It's in some of the past papers (from the old study design) but I can't see anything in the current study design (other than in units 1/2).
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jnlfs2010

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« Reply #8169 on: October 10, 2019, 01:05:20 pm »
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Do we need to know gravimetric analysis for the exam? It's in some of the past papers (from the old study design) but I can't see anything in the current study design (other than in units 1/2).

Gravimetric analysis is in units 1/2 but no longer in units 3/4 for Chemistry. So is atomic Absorption spectroscopy
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Erutepa

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« Reply #8170 on: October 10, 2019, 01:10:01 pm »
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Hey,
I've been asked to compare the density of starch and cellulose.
I was just looking at the structural formula of the two molecules and it looks like cellulose has more of a 'kink' in its polymers (?) than starch so that it doesn't pack as closely but I'm not quite sure if this reasoning is adequate (or even true).
I'm just curious as to whether or not my idea is correct and which molecule would be more dense.
Starch consists of both Amylose and Amylopectin. Amylopectin has branches every ~20 units and Amylose forms a helical structure. As such, both starch molecules are less closely packed then cellulose which is not branched and does not form a helix, allowing cellulose to more closely pack together. As such, starch is less dense than cellulose.
This little summary table from bioninja might be able to help with visualising the differences:
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« Reply #8171 on: October 12, 2019, 10:31:10 am »
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Hi AN

Answers are also attached, I particularly don't get part a and b.

Erutepa

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« Reply #8172 on: October 12, 2019, 10:36:17 am »
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Hi AN

Answers are also attached, I particularly don't get part a and b.
It seems you have only attached the answers, not the questions.

What parts of a and b are you specifically confused about?
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« Reply #8173 on: October 12, 2019, 10:44:30 am »
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Sry my bad, Q is attached.

I don't understand how you get copper in the equation.

thanks