Rate RXN--Need your Help

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Tooth09

Full Member
10+ Year Member
15+ Year Member
Joined
Apr 2, 2008
Messages
15
Reaction score
0
Exp.[CoBr(NH3)52+][OH-]initial rate (M/sec)13.0 x 10-23.0 x 10-21.37 x 10-326.0 x 10-23.0 x 10-22.74 x 10-339.0 x 10-29.0 x 10-21.23 x 10-2

Find Rate law expression?? I can determine for CoBr. but Cant really find for OH-. Please help..Thanks

Members don't see this ad.
 
I can ot make sense of your post can you repost with apporiate columns.....thanks
 
Members don't see this ad :)
XP CoBr OH rate

1. 3.0 x 10-2 3.0x10-2 1.37x10-3
2. 6.0x10-2 3.0x10-2 2.74x10-3
3. 9.0x10-2 9.0x10-2 1.23x10-2

what is the rate law expression? How do you determine for OH-..please explain thanks..
 
The rate law would be rate=(1.5)[CoBr][OH]

To find CoBr:
from experiment 1 to 2 we double [CoBr], and our rate doubles. therefore 1st order.

To find OH:
when going from experiment 1 to 3, our [CoBr] increases by 3, but our rate increases by 9. Since we tripled our [OH] as well, we see that the product tripled making [OH] first order.

Now plug values into our equation

rate=k[CoBr][OH]
1.37*10^-3=k[3.0*10^-2][3.0*10^-2]
k=1.5

rate=(1.5)[CoBr][OH]

correct me if im wrong
 
Drgreen..Since 1 to 3.. both OH and CoBr increase 3 times and rate increase 9 times..so, it should be second order???
 
Yes you are correct! the overall reaction is second order.
the rate of CoBr is first order, the rate of OH is first order... 1+1=2
1.5 represents k, our rate constant, not to be confused with overall rxn order.

to put it simply, when you double the concentration of CoBr, you double the rate of the reaction, when you double the concentration of OH you double the rate of the reaction, but when you double the concentration of both CoBr AND OH, the reaction rate quadruples.
 
Drgreen... I know that the overall order of the rxn is 2.. I am a little bit confuse how to determine the rate order for OH only.. From exp 1 and 3-- the [OH] is triple (3times) and the rate rxn is 9 times which lead to 3^y = 9 therefore y =2; then [OH]^2 or second???? please explain..
 
Top