Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: yui on November 21, 2007, 10:06:35 PM
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The reaction Co(NH3)5F2++H2O---Co(NH3)5(H2O)3+ + F- is acid catalyzed and proceeds according to the rate law
Rate= -d[Co(NH3)5F2+]/dt = k [Co(NH3)5F2+]a[H+]b
The times for half and for ¾ of the complex to react are given in the following table for the indicated temperatures and initial concentrations. Show what are the values of the exponents a and b must be.
[Co(NH3)5F2+]/M [H+]/M Temperature/ oC T ½ /h T ¾ /h
0.1 0.01 35 1 2
0.2 0.02 35 0.5 1
0.1 0.01 50 0.5 1
Is it accurate for me to say that there is a constant reduction of concentration from T ¾ to T ½ ? Hence a pseudo 1st order reaction is expected? Is that the purpose of putting this information up?
Comparing (1) and (3), when temperature increased by 15 oC, rate of reaction increased by ½.
Comparing (2) and (3), when [Co(NH3)5F2+] and [H+] were halved , the increased in temperature was cancelled out. Is it correct to have:
[0.5] a[0.5]b = 0.5 hence a+b=1, since acid is used as a catalyst, the order of reaction wrt to H+ is 0 and the order of reaction wrt to Co(NH3)5F2+ is 1?
However, comparing (1) and (2),
[0.1]a[0.01]b
_______________ = 1/ 0.5
[0.2]a[0.02]b
1
____= 2 ------ hence a+b = -1,
2a+b
How do I logic this out? Please help me. Thanks! ???
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Is that the purpose of putting this information up?
IMO its purpose is to show that successive half lives are constant.
This would suggest that the reaction is first order, so a+b=1.
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The half-life seems to be inversely proportional to the initial conc of Co(NH3)5F2 , which makes me think of a 2nd order rxn.
What perplexes me is not just the catalyst appearance in the rate law but also how T3/4 is greater than T1/2 . aren't you taking T3/4 to be the time when 3/4 [Co(NH3)5F2 ] remains i.e. 25% of the reactant is consumed ? how then does it take 1 hour for the reacant to reach half its value and 2 hours to reach its3/4 th?
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T3/4 is greater than T1/2 . aren't you taking T3/4 to be the time when 3/4 [Co(NH3)5F2 ] remains i.e. 25% of the reactant is consumed ? how then does it take 1 hour for the reacant to reach half its value and 2 hours to reach its3/4 th?
I think he meant 75% of reactant is consumed here:
The times for half and for ¾ of the complex to react are given
So, 2 half lives have elapsed. I may be wrong but I thought since the half life remains constant, this would indicate the reaction is first order.
Feel free to contradict me if I'm wrong though ;)
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Yes you're right it's logical to say at t=t3/4 , 1/4 of the reactant remains. I prolly was confused because of the usual convention used in chemical kinetics ... Anyway , how do you then conclude that the half-life is constant ? take the 1st two rows. At Temp = 35 degree C, the half-life doubles when the concentration is halfed. For a 1st order rxn , the half-life is indedepdent of the initial concentration. it is inversely proportional to the inital conc for a 2nd order rxn. Therfore , I think of a 2nd order rxn. It cant be 1st order where t{1/2} = f(T) alone is not a function of(C_o). Then take the 2nd and 3rd row , there's a difference of 15 C in temp , the conc is again halfed in the higher temp and we get a constant half-life in both cases. the difference in temp seems to compensate for the double initial initial concentration at 35 dgree C.
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Anyway , how do you then conclude that the half-life is constant ?
I guess I didn't word my post very well. I meant consecutive half lives are constant. So looking across rows, not down the columns.
For example, in the first case, T1/2=1 and (T3/4 - T1/2) is also 1. The fact that it doesn't change I thought would have indicated first order:
For 1st order: T1/2= ln2/k
So half lives are independant of concentration here - can't this be applied to consecutuve half lives here to show first order?
At Temp = 35 degree C, the half-life doubles when the concentration is halfed. For a 1st order rxn , the half-life is indedepdent of the initial concentration.
This seems to be true, and it seems to contradict my earlier reasoning :'(
Could you tell me what was wrong with it?
edit: poor spelling
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you're right , t3/4 = 2(t1/2) seems to hold in all cases ( a feature of a 1st order rxn ). I cant really explain why this is so when clearly the half-life seems inversely proportional to the initial conc in the 3rd case. Pretty much contradictory .
If this rxn occurs with H+ as a homogenous catalyst and if there exists a RDS , then according to the rate law given , it must be:
A + H+ ----> P This single step must be defining the rate.
H+ is the regenerated quickly via other fast steps and so it makes sense to say [H+] is approximately constant all the time. The reactant A is decreasing , so if u consider that and integrate , u'll get an equation that does not hold true with the data given. It beats me ... maybe there's we're somethng missing or perhaps the reaction mechanism is completely different .
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I've thought about this. The rxn could be 2nd order. a = 1 , b =1.
it makes perfect sense for me to assume that
A + H+ ----> P This single step must be defining the rate. [ H+] = const
Since t3/4 = 2{t1/2} then the order of A must be 1 as you said before.
-d[A] / dt = k[A][H+]^b
since [H+] does not change , this is like a pseudo 1st order rxn
integrate and all that , u should get this:
t1/2 = Ln2 / k [H+]^b
[H+] : initial conc of hydrogen ions
For the 1st row , t1/2 = 1 hour
t1/2 = Ln2 / k [0.01]^b = 1
Move on to the 2nd row. t1/2 = 0.5 hour . This is not because the conc of [A] is doubled but only because the conc of [H+] is doubled. In fact , the half-life of this reaction is independent of the initial conc of reactant A
t1/2 = Ln2 / k [0.02]^b = 0.5
take the ratios and solve ,
b = 1
From the 3rd row it can be deduced that the the new rate constant k' ( 50 C )= 2k ( 35 C ). k and k' can then be determined too by simple calculation ...
It seems logical to me. What do u think ?
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Of Course! Nice thinking, I agree with you. I overlooked the fact that H+ was only a catalyst and concentration doesn't change in each case, so the consecutive half lives indicate that A is first order. It gives no info on H+.
And as you suggest the order of H+ can then be found from comparing rows (since H+ isn't constant here) and then the rate constant can also be found.
Thanks Vant_Hoff for your clarification :)