Chemistry P41 How was it?

[Marks&Spencers]

First of all, the relative values of Kc between reactions with different orders cannot be compared by just matching the numbers, since the units differs!. I.E., Kc, of esterification for example has no units, while the Kc in that question has a unit. This means that you cannot say in that situation whether it is high or low. As the Kc value with that unit would usually be in the order of 10^12 for example, which is higher than 10^9 that I got.

Secondly, in the chemistry course book (if you have it), it talks about how unlikely the reaction of Ecell below 0.30V to occur if it is not done under standard conditions. This means this value is low, when coming to talk about E° values.

Here is an example, Sn4+ + 2e- <---------------------> Sn4+ E°= +0.15V
You have been told that Sn+4 is slightly stable than +2 which means the position of equilibrium lies a little bit on the left. The value of that E° is close to 0.23V , so similar situation.

Well what about the values which you quoted for the concentrations of the products? They were much higher than those of the reactants, like they were in the range of 10^-2, while those of the products were in the range of 10^-4. Doesn't that mean there are more products, so the forward reaction is favoured?

 

[Starry_night]

No man, I'm pretty sure they asked for the value of k, and not the units. Or may be both! I didn't find the units!! !

NOT SURE THOUGH

I'm pretty sure it was just to find the value of k. I thought it was weird that they didin't ask for units as well because they usually do, so I re-read the question and it only said 'find the value of k'.

 

[iKhaled]

man they asked for unit of K

noo they didn't ask for the unit! they asked for the value

 

[Jaf]

why not the half life bro?
arent we taught that? and even its in the cie course book isnt it? like for finding the concentration of the of ethey etanoate we calculate the successive half life if they are the same then its of the order=1 and when i did i got the same half life hence it was the 1st order
now for order of hcl i drew tangent on the initial points of both the graphs and then calculated the gradient(rate) and it was doubled in the second experiment when higher concentration(which was also double) now it shows hcl rate=1
so over rate=k(hcl)(ehyl etanoate)
iam not sure if iam right but this is the way i did. tell me if iam right becuase these type of question have come n the past papers seperately this time they gave together i think they came before in paper before 2007 if i remember right.

Well the method COULD be used for the first part since it was more or less a 'prove' question. We knew the order was 1. So no problem there. It's the second part I'm getting worked up over. Oh well...

 

[Jaf]

man they asked for unit of K

LOL no. They only asked for the 'k' value. No units. I still gave the units though.

 

[vivo990]

Well what about the values which you quoted for the concentrations of the products? They were much higher than those of the reactants, like they were in the range of 10^-2, while those of the products were in the range of 10^-4. Doesn't that mean there are more products, so the forward reaction is favoured?

First of all, the question said to calculate the E⁰ value of the reaction, then HENCE, use that value to suggest whether the position of equilibrium is on the right or the left.
Secondly, regarding the concentration values, the reaction might being held under non standard conditions, so position of equilibrium is affected. Thus, you have to follow the instructions that is given in the question. Plus, you can make use of information before this part, but not after it.

 

[Jaf]

First of all, the question said to calculate the E⁰ value of the reaction, then HENCE, use that value to suggest whether the position of equilibrium is on the right or the left.
Secondly, regarding the concentration values, the reaction might being held under non standard conditions, so position of equilibrium is affected. Thus, you have to follow the instructions that is given in the question. Plus, you can make use of information before this part, but not after it.

Wait, I'm a bit lost too. What non-standard conditions? :S I do remember the equation mentioning something but only very faintly.

 

[Jaf]

Secondly, in the chemistry course book (if you have it), it talks about how unlikely the reaction of Ecell below 0.30V to occur if it is not done under standard conditions. This means this value is low, when coming to talk about E° values.

If that's what you read, then you completely missed the point that was being made. It doesn't say at all that if the electrode potential of cell is less than 0.3, the reaction wouldn't occur. It says if the DIFFERENCE of the E°cell between the reaction under standard conditions and the reaction under non-standard condition is greater than 0.3, only THEN is the reaction unfeasible. Go read it again.

Here is an example, Sn4+ + 2e- <---------------------> Sn2+ E°= +0.15V
You have been told that Sn+4 is slightly stable than +2 which means the position of equilibrium lies a little bit on the left. The value of that E° is close to 0.23V , so similar situation.

What in the worlllddd. How do you come to the conclusion that Sn4+ ion is MORE stable that the Sn2+ ion? The +2 oxidation states are MORE stable down the group. Sn2+ is the second last element. The +4 oxidation states get LESS stable down the group. Sn4+ is almost certainly less stable Sn2+ since it's so low in the group. In fact the equation you quoted just justifies all this. I'm sorry you're making no sense lol

 

[vivo990]

If that's what you read, then you completely missed the point that was being made. It doesn't say at all that if the electrode potential of cell is less than 0.3, the reaction wouldn't occur. It says if the DIFFERENCE of the E°cell between the reaction under standard conditions and the reaction under non-standard condition is greater than 0.3, only THEN is the reaction unfeasible. Go read it again.

Yes, I agree. The reaction given might not occur if the condition is not standard. But why do you think? Well, the E value is low. and it should be interpenetrated that 0.3V is the minimum standard in which we can " predict with confidence that the reaction might occur even if the conditions are not standard (this is given in the book)"

Ok: now we can stretch little bit our understanding to apply some of the concept in equilibrium reactions and their E value ( since we were dealing with complete reaction the whole course, right?). If this question come in the exam ( and it actually came), then I have to use the E value of the reaction to say whether the equilibrium will lie to the right or the left. But to what should I compare that value with? The only standard which I know is the 0.3V (according to what is mentioned in the course book), then the only way to solve that question using my LIMITED A-Level knowledge is to compare my answer (+0.23V) with the standard value that I know (and expected to know).

What in the worlllddd. How do you come to the conclusion that Sn4+ ion is MORE stable that the Sn2+ ion? The +2 oxidation states are MORE stable down the group. Sn2+ is the second last element. The +4 oxidation states get LESS stable down the group. Sn4+ is almost certainly less stable Sn2+ since it's so low in the group. In fact the equation you quoted just justifies all this. I'm sorry you're making no sense lol

Go read the Chemistry course book, in Group 4 chapter. It exciplicitly said that +4 is slightly more stable than plus two. I did not deduce that. It's a FACT stated in the book. Ask your teacher even.

 

[Jaf]

Yes, I agree. The reaction given might not occur if the condition is not standard. But why do you think? Well, the E value is low.

and it should be interpenetrated that 0.3V is the minimum standard in which we can " predict with confidence that the reaction might occur even if the conditions are not standard (this is given in the book)"
If this question come in the exam ( and it actually came), then I have to use the E value of the reaction to say whether the equilibrium will lie to the right or the left. But to what should I compare that value with? The only standard which I know is the 0.3V (according to what is mentioned in the course book), then the only way to solve that question using my LIMITED A-Level knowledge is to compare my answer (+0.23V) with the standard value that I know (and expected to know).

You're missing the point, again. +0.3V is NOT a standard value you can use. It is the DIFFERENCE between two values. That's not the same thing. Since we had nothing to compare our potential with (as you say), we can't say if the difference is more or less than 0.3V. You're using the 0.3V value in an incorrect way. So keeping in mind that ALL positive Ecells are feasible, we say the equilibrium lies towards products. This is a rule. You can not deny it. Just google 'positive cell potential' and you will have results telling you that whenever the cell potential is positive, the reaction is spontaneous. There's no mention of the magnitude of the positive voltage. Read this (the guy explains in very well):
http://wiki.answers.com/Q/How_can_you_tell_if_a_redox_reaction_is_spontaneous_or_not

Go read the Chemistry course book, in Group 4 chapter. It exciplicitly said that +4 is slightly more stable than plus two. I did not deduce that. It's a FACT stated in the book. Ask your teacher even.

Okay, agreed. But pay attention to the 'slight'. The electrode potential (which the question was about) suggests otherwise.

If you still don't get it, then lets agree to disagree.

 

[iKhaled]

Jaf what was ur value of Kc ?

 

[Jaf]

Jaf what was ur value of Kc ?

2.5 x 10^9

Uptil now I haven't seen a good reason why this shouldn't be the answer.

 

[vivo990]

You're missing the point, again. +0.3V is NOT a standard value you can use. It is the DIFFERENCE between two values. That's not the same thing. Since we had nothing to compare our potential with (as you say), we can't say if the difference is more or less than 0.3V. You're using the 0.3V value in an incorrect way. So keeping in mind that ALL positive Ecells are feasible, we say the equilibrium lies towards products. This is a rule. You can not deny it. Just google 'positive cell potential' and you will have results telling you that whenever the cell potential is positive, the reaction is spontaneous. There's no mention of the magnitude of the positive voltage. Read this (the guy explains in very well):
http://wiki.answers.com/Q/How_can_you_tell_if_a_redox_reaction_is_spontaneous_or_not

I'm not missing any point. All of what you have said was taught to me, plus it is in the book. I guess you're not giving a second thought about what did I say.

Okay, agreed. But pay attention to the 'slight'.

Even if it is by just a micrometer to the left, it is still more to the left.

The electrode potential (which the question was about) suggests otherwise.

From where did you get that? E value for the forward reaction is +0.23V

If you still don't get it, then lets agree to disagree.

I guess you're right in that one. Still I didn't find your reasoning satisfactory, so let's agree to disagree.

 

[Mairaxo]

Well, many people I've seen got 25. Still trying to figure out what reasoning did they approach.
My Kc was 2.5x10^9. Did anyone get that value also? Or did anyone know somebody got that value?

me 2!!

 

[Mairaxo]

heyy dont worry man! m cnsidered a genius and yeah even i got the same value so ur right

Ok can you please explain to me how you got that 25? You seem like a smart person and I didn't get that answer so now I'm kind of freaking out. I got 2.5 x 10^9. I know it's a big number but I checked it a million times and this is how I got it. I did it the same way another poster on this thread did it.

t

 

[hana.lula.nur]

About the rate of reaction question... I just didn't think half-life was the method to do it. (at our level, especially using only a graph; you would need differentiation if it were a second order reaction and since we didn't know, we shouldn't use half life)
I drew a tangent for the first curve at the first point (0,0.2). Found the gradient which = rate. Then I drew a tangent at half the conc of ethyl ethanoate and found the rate again. The ratio of concentrations was 1/2 and so was the ratio of rate of reaction. So order = 1.

I drew a third tangent at the same point (0,0.2) but this time for the curve I drew. I showed the gradient to be half of the gradient of my very first tangent (which was for the first curve, and drawn at the same point). The ratio was 0.7 something for the rate (and I'm pretty sure it was because the inaccuracy of the tangents I drew). I stated that was approximately 0.5, so order is 1.

I haven't seen anyone mention this method, so I'm getting a bit worried now... :/

THAT IS EXACTLY HOW I DID IT! Thank youuu so much for reassuring me!!

 

[Mairaxo]

THAT IS EXACTLY HOW I DID IT! Thank youuu so much for reassuring me!!

yup so did i

 

[Jaf]

I'm not missing any point. All of what you have said was taught to me, plus it is in the book. I guess you're not giving a second thought about what did I say.

Even if it is by just a micrometer to the left, it is still more to the left.



From where did you get that? E value for the forward reaction is +0.23V


I guess you're right in that one. Still I didn't find your reasoning satisfactory, so let's agree to disagree.

The ms is out. ^_^ http://olevel.sourceforge.net/papers/9701/9701_s13_ms_41.pdf