[yui]

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!

[Padfoot]

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.

[Hunt]

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?

[Padfoot]

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 

[Hunt]

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.

[Padfoot]

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

[Hunt]

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 .

[Hunt]

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 ?

[Padfoot]

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