Chemistry P41 How was it?

[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