Question

The decomposition of XY is second order in XYand has a rate constant of 7.20

Answer 1

Answer:  Directly below is a rigorous mathematical derivation, but at the bottom is a "simpler" method that seems to get different answers - you should check with your teacher to see why they aren't consistent.

First approach:

You really need to derive the equation you need to use, by using the definition of a 2nd order reaction:

d[XY]/dt = -k*[XY]^2

You need to integrate this to change it from a differential equation into an algebraic equation.

First rearrange the equation to separate variables in preparation for integration:

d[XY]/[XY]^2 = -k*dt

Now integrate both sides of the equation from [XY]=[XY0] to [XY]=[XY] and t=0 to t=t:

-1/[XY] = -k*t (evaluate this over above limits)

=> -1/[XY] - {-1/[XY0]} = -k*t - 0

=> 1/[XY0] - 1/[XY] = -k*t, or

1/[XY] - 1/[XY0] = k*t

This is the equation you should use. Since you were already given the rate constant, I'm not sure why they gave you the half-life information - you don't seem to need it.

For part 1): you can solve for t:

t={1/[XY] - 1/[XY0]}/k

Since you know [XY]= 0.125*[XY0]:

t= {1/(0.125*[XY0]) - 1/[XY0]}/k

= (1/0.125-1/1)/([XY0]*k)

= 7/(0.100M)/(7.08x10^(-3))

= 9887 seconds

For part 2) you have:

t={1/[XY] - 1/[XY0]}/k

={1/[6.60x10^(-2)] - 1/[0.150]}/(7.08x10^(-3))

=1198 seconds


For part 3) it is unclear - you've given information about a specific chemical (maybe that's what XY is?). I'll let you finish that part.

Note: There seems to be some conflict between these answers and the half-life information you were given (not consistent). If you're not at the math level required for the above explanation, your teacher may have wanted you to simply use the half-life information as follows:

Second Approach:

From the definition of half-life, the concentration is halved after every half-life.

For part a) a final concentration of 12.5% the original concentration represents (0.50)^3 times the original concentration or 3 half-lives worth of decay. From that you could simply calculate t as

t= 3*(half-life) = 3*(1410 s) = 4230 seconds (Note this conflicts with the original calculation of part 1) above)

For part b)

((6.6x10^(-02))/(0.150)) = 0.44 = (0.5)^x, where x=number of half-lives.

Solve for x: x=ln(0.44)/ln(0.5) = 1.184 half-lives

so t=(1.184)*(1410s) = 1670seconds (again different from original calculation).