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Author Topic: VCE Chemistry Question Thread  (Read 2913251 times)  Share 

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HighTide

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Re: VCE Chemistry Question Thread
« Reply #3870 on: August 01, 2015, 11:09:16 pm »
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Few questions about equilibria:

1. Predict the effect of the changes on the position of each equilibrium: Removal of from the equilibrium:

2. For the exothermic reaction: , how would you alter the temperature of the equilibrium in order to product a net forward reaction? I understand that because the forward reaction is exothermic, but what has this to do with decreasing the temperature to produce the forward reaction?

3. An equilibrium mixture containing: The volume of the container was increased at constant temperature and a new equilibrium was established. How would the concentration of be altered? I would say that because the pressure is being decreased, the more particles (moles) is favourable, and hence the forward reaction would increase and so the production of would increase. That being said, although the particles of nitrogen dioxide are being relatively increased, the volume is also increasing so does this actually mean that the concentration is increasing ?

Cheers.
1. If I had a solution of ethanoic acid, hydronium and acetate ions and I removed acetate, the concentration of acetate would decrease. Now imagine the backward reaction. Think of the backward reaction as any normal reaction. So if I removed the acetate ions, I remove reactants, and if I removed reactants, I reduce the chance of successful collisions occurring and so reaction rate decreases. So if the backward reaction occurs less frequently, more reactants will be converted to products, causing the forward reaction to occur to a greater extent --> Shifts in favour of forward reaction.
2. Reaction is exothermic. So, backward reaction is endothermic. Adding temperature= adding energy. Since the reaction is exothermic, energy is released, and so adding energy has no effect on the exothermic reaction other than increasing reaction rate.

On the other hand, when the reaction is endothermic, adding energy facilitates the reaction.So if I increased the temperature, the reaction has more energy to absorb from the environment, thereby causing the backward reaction to occur more frequently.Now, if I wanted to increase the yield of the forward reaction, I need to prevent the backward reaction from occurring. One way I can do this is by decreasing the temperature to prevent the backward reaction from occurring (although it may affect reaction rate).

3. Increase volume--> Decrease pressure-->  Initial decrease in concentration--> Forward reaction occurs to greater extent due to particles--> Gradual increase in nitrogen dioxide concentration, gradual decrease in dinitrogen tetroxide concentration.

I'm confused about changing the equilibrium position. My text says that by adding reactant (or imposing any other change excluding temperature) a new equilibrium is established. Does this mean there is a new value of K?
No. K value is only changed by temperature. Adding reactants or products will partially offset the dynamic equilibrium. The position of equilibrium will thereby shift to counteract the change and create a new equilibrium with different concentrations of products and reactants but same K value. 
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thushan

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Re: VCE Chemistry Question Thread
« Reply #3871 on: August 02, 2015, 04:09:29 am »
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I'm confused about changing the equilibrium position. My text says that by adding reactant (or imposing any other change excluding temperature) a new equilibrium is established. Does this mean there is a new value of K?

At time zero, Q (reaction quotient) is equal to K (equilibrium constant). When you add a reactant, you increase its concentration; this decreases the value of Q. Therefore, Q is no longer equal to K (Q < K) and therefore the system is not at equilibrium. The system will favour the forward reaction (in this case), which will subsequently increase Q back up to K. When Q = K, the system is at equilibrium again.

Note that K has never changed here; it's Q that changes.
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cosine

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Re: VCE Chemistry Question Thread
« Reply #3872 on: August 02, 2015, 08:14:33 am »
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At time zero, Q (reaction quotient) is equal to K (equilibrium constant). When you add a reactant, you increase its concentration; this decreases the value of Q. Therefore, Q is no longer equal to K (Q < K) and therefore the system is not at equilibrium. The system will favour the forward reaction (in this case), which will subsequently increase Q back up to K. When Q = K, the system is at equilibrium again.

Note that K has never changed here; it's Q that changes.

But why does not K change? Doesn't the overall concentration of the products increase? Does not K indicate how much product is produced in a reaction at equilibrium?
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cosine

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Re: VCE Chemistry Question Thread
« Reply #3873 on: August 02, 2015, 11:07:29 am »
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(assuming temperature remains the same) Why is it that even if you change the number of mol per reactant and product, the equilibrium constant will still remain the same? Yes sure I get it's a constant, and hence it remains the same, but why?

When you have K =50M I know that this means that the forward reaction have occurred to a greater extent and the products have a higher concentration than the reactants, but what does the Molarity (units) actually mean? And what does the 50 mean? Like what would be the difference between K=25 and K=50?

Cheers.
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Redoxify

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Re: VCE Chemistry Question Thread
« Reply #3874 on: August 02, 2015, 11:52:17 am »
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(assuming temperature remains the same) Why is it that even if you change the number of mol per reactant and product, the equilibrium constant will still remain the same? Yes sure I get it's a constant, and hence it remains the same, but why?

When you have K =50M I know that this means that the forward reaction have occurred to a greater extent and the products have a higher concentration than the reactants, but what does the Molarity (units) actually mean? And what does the 50 mean? Like what would be the difference between K=25 and K=50?

Cheers.

To answer the first part of the question,
The reason why equilibrium stays the same is related to Le chatliers principle, where either the forward or backward net reaction will occur to maintain equilibrium
hence if you remove the products, to maintain equilibrium the equation shifts to the right, thus decreasing the concentration of the reactants :)
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Acid

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Re: VCE Chemistry Question Thread
« Reply #3875 on: August 02, 2015, 04:10:41 pm »
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Someone please help!

This is an acid-base equilibrium question

A solution is prepared by adding 0.500 mol of methanoic acid, HCOOH, and 0.100 mol of sodium methanoate, HCOONa, to 2.00 L of water. Determine the concentration of hydronium ions in this solution. In you calculations assume that
-the equilibrium concentration of HCOOH is approximately equal to the initial concentration of HCOOH.
-the equilibrium concentration of HCOO- is approximately equal to the initial concentration of HCOO-.

The answer is 9.0*10^-4 M

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Acid

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Re: VCE Chemistry Question Thread
« Reply #3876 on: August 02, 2015, 04:38:58 pm »
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But why does not K change? Doesn't the overall concentration of the products increase? Does not K indicate how much product is produced in a reaction at equilibrium?

In case of pressure and concentration changes the system will readjust itself to satisfy the K value. K is only affected by temperature. Whenever the forward reaction is favoured due to temperature change the K value will be larger, and whenever the backward reaction is favoured due to temperature change the K value will be smaller. 
K does not change, the disturbances in the equilibrium cause the system to respond in a way where either the forward or the backward reaction is favoured. K value is indicative of the yield of a reaction.  :)
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cosine

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Re: VCE Chemistry Question Thread
« Reply #3877 on: August 02, 2015, 04:59:19 pm »
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In case of pressure and concentration changes the system will readjust itself to satisfy the K value. K is only affected by temperature. Whenever the forward reaction is favoured due to temperature change the K value will be larger, and whenever the backward reaction is favoured due to temperature change the K value will be smaller. 
K does not change, the disturbances in the equilibrium cause the system to respond in a way where either the . K value is indicative of the yield of a reaction:)

But if the forward reaction is favoured, then the products will increase. If K = yield of reaction, then the products are increasing, so why does K not increase too?
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Acid

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Re: VCE Chemistry Question Thread
« Reply #3878 on: August 02, 2015, 05:21:10 pm »
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But if the forward reaction is favoured, then the products will increase. If K = yield of reaction, then the products are increasing, so why does K not increase too?

Because it is increasing the concentration fraction at the time equilibrium is disturbed, but not the concentration fraction at equilibrium.
It will try to restore the system back to K at equilibrium. So basically, Q (reaction quotient) will change, but not K at equilibrium. Q will try to go back to being K at equilibrium. I hope I make sense? haha
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cosine

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Re: VCE Chemistry Question Thread
« Reply #3879 on: August 02, 2015, 05:51:59 pm »
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Because it is increasing the concentration fraction at the time equilibrium is disturbed, but not the concentration fraction at equilibrium.
It will try to restore the system back to K at equilibrium. So basically, Q (reaction quotient) will change, but not K at equilibrium. Q will try to go back to being K at equilibrium. I hope I make sense? haha

Sorry but that makes no sense to me at all :(

Question:

If you added water to a solution containing
i). How would the initial concentration of the reactants be altered? I would say that because we have added water, we have diluted the solution and hence the initial concentration of the ions will be decreased. Is this true?
ii). Describe the equilibrium shift. Wouldn't there be a net backwards shift because there are more mol particles in the reactants side?
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Acid

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Re: VCE Chemistry Question Thread
« Reply #3880 on: August 02, 2015, 06:21:20 pm »
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Sorry but that makes no sense to me at all :(

Question:

If you added water to a solution containing
i). How would the initial concentration of the reactants be altered? I would say that because we have added water, we have diluted the solution and hence the initial concentration of the ions will be decreased. Is this true?
ii). Describe the equilibrium shift. Wouldn't there be a net backwards shift because there are more mol particles in the reactants side?
Sorry :(
i) Yep, the initial concentration would decrease.
ii) There will be a net backward reaction because dilution causes the reaction to move towards higher number of particles. Since the left side of the reaction has more particles, backwards reaction will be favoured.

Is this for a practical? My school did this exact practical a few days ago.
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cosine

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Re: VCE Chemistry Question Thread
« Reply #3881 on: August 02, 2015, 06:30:21 pm »
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Sorry :(
i) Yep, the initial concentration would decrease.
ii) There will be a net backward reaction because dilution causes the reaction to move towards higher number of particles. Since the left side of the reaction has more particles, backwards reaction will be favoured.

Is this for a practical? My school did this exact practical a few days ago.
Yah thank you!

Also, how would the graph be altered? Like would there be an immediate drop in all three of the particles on the graph? And then as the back reaction is adamant, the reactants will slowly increase and the product will decrease further, and then the new equilibrium is reached?
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paper-back

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Re: VCE Chemistry Question Thread
« Reply #3882 on: August 02, 2015, 06:36:24 pm »
+1
Someone please help!

This is an acid-base equilibrium question

A solution is prepared by adding 0.500 mol of methanoic acid, HCOOH, and 0.100 mol of sodium methanoate, HCOONa, to 2.00 L of water. Determine the concentration of hydronium ions in this solution. In you calculations assume that
-the equilibrium concentration of HCOOH is approximately equal to the initial concentration of HCOOH.
-the equilibrium concentration of HCOO- is approximately equal to the initial concentration of HCOO-.

The answer is 9.0*10^-4 M

This was my working out; Please correct me if I'm wrong anywhere

We have two reactions here:
(1) CHOOH + H2O <-> CHOO- + H3O
(2) CHOONa -> CHOO- + Na+

Assuming the second reaction occurs to completion then the concentration of CHOO- is 0.05M
The first reaction has a K value of 1.8x10^-4
The information states that the "the equilibrium concentration of HCOOH is approximately equal to the initial concentration of HCOOH", this suggests that reaction 1 has only proceeded to a minute extent. If this is true, then little CHOO- would have formed from reaction 1, and hence all CHOO- that has formed can be assumed to be as a result of reaction 2

Therefore, the concentration of CHOOH is approximately 0.25
And the concentration of CHOO- is approximately 0.05
The first reaction has a K value of 1.8x10^-4 - data  booklet

Therefore, K=([CHOO-]x[H3O])/CHOOH
(1.8x10^-4)=0.05 x [H3O]/0.25
Solve for H3O and you get 9x10^-4
« Last Edit: August 02, 2015, 06:42:52 pm by paper-back »

Acid

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Re: VCE Chemistry Question Thread
« Reply #3883 on: August 02, 2015, 06:42:30 pm »
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This was my working out; Please correct me if I'm wrong anywhere

We have two reactions here:
(1) CHOOH + H2O <-> CHOO- + H3O
(2) CHOONa -> CHOO- + Na+

Assuming the second reaction occurs to completion then the concentration of CHOO- is 0.05M
The first reaction has a K value of 1.8x10^-4
The information states that the "the equilibrium concentration of HCOOH is approximately equal to the initial concentration of HCOOH", this suggests that reaction 1 has only proceeded to a minute extent. If this is true, then little CHOO- would have formed from CHOO-, and hence all CHOO- that has formed can be assumed to be as a result of reaction 2

Therefore, the concentration of CHOOH is approximately 0.25
And the concentration of CHOO- is approximately 0.05
The first reaction has a K value of 1.8x10^-4 - data  booklet

Therefore, K=([CHOO-]x[H3O])/CHOOH
(1.8x10^-4)=0.05 x [H3O]/0.25
Solve for H3O and you get 9x10^-4

Thanks a lot! Makes sense :)
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Acid

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Re: VCE Chemistry Question Thread
« Reply #3884 on: August 02, 2015, 06:43:30 pm »
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Yah thank you!

Also, how would the graph be altered? Like would there be an immediate drop in all three of the particles on the graph? And then as the back reaction is adamant, the reactants will slowly increase and the product will decrease further, and then the new equilibrium is reached?

Yep, that's correct.
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