Half Life and Rate Laws

[Bill Nye Guy]

Hey everyone i was just reviewing chemistry for a class i am going to take next semester and i came upon some questions i am trying to solve which i think might be right but i am not sure.

 

1.) The rate constant for the decomposition of N205 at 70 celisus is 6.82x10-3s-1 N2O5(g)-->2NO2(g) + O2(g)

 

a.) what is the rate law of this reaction? be sure to include the value of the rate constant in this expression.

 

*my answer* rate=k[NO2]2[O2] since NO2 is 2 and O2 is 1 overall the rate law is 3. I said the rate constant is 6.82x10-3s-1 since that is what it was for N2O5.

 

b.) what is the half life of this reaction if the initial concentration of N2O5 is 0.0125M?

 

*my answer* I believe t1/2 is equal to 0.693/k so its 0.693/0.0125.

 

c.) what concentration of N2O5 will remain after 2.5 MINUTES if the initial concentration is 0.0125M?

 

*my answer* 150(1/2)=0.693/k

 

 

 

 

I just want to know if i am doing this right and if not what am i doing wrong and what are the right answers. If i do well on this or if i understand it, then i think i will take the class, which apparently deals a lot with this kinda stuff. Thanks again for your help everyone.

 

Peace.

[woelen]

I don't think you are correct. The terms, as you use them give me the impression that the behavior is simply negative exponential, first order. Only the concentration of N2O5 matters, this is not an equilibrium reaction with low-rate back and forward reactions:

 

Question 1:

 

d[N2O5]/dt = -k * [N2O5], where k = 6.82*10^(-3)

 

Question 2:

 

You need to solve the rate equation, as given in question 1. The solution of this is

 

[N2O5](t) = [N2O5](0) * exp(-t*k), Here [N2O5](t) is the concentration of N2O5 at time t, and [N2O5](0) is the initial concentration. With exp() I mean the exponential function.

 

Now you want to determine the half life, when the initial concentration is 0.0125 M. Because this is a linear equation, half life does not depend on initial concentration. You simply have to determine the time t, for which exp(-k*t) = 0.5.

 

This is the case for k*t = ln(2) --> t = ln(2) / k = 101.6 (s)

 

Question 3:

 

2.5 minutes is 150 seconds. So, using the solution of the equation, given in question (2), one finds

[N2O5](150) = [N2O5](0)*exp(-150*0.00682) = 0.0125*exp(-1.023) = 0.00449 M