-
Practice your interview with the new SDN AI Interview Coach. Choose a school, answer by voice or typing, and receive a personalized feedback report. Available now to all SDN members. Try the AI Interview Coach.
Rate RXN--Need your Help
Started by Tooth09
Advertisement - Members don't see this ad
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
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
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.
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..
