Question

Plot the natural log of the concentration of the ozone Versus time. Since we get a straight line, that mean., that we can consider an integrated rate law for this reaction to be given by the following: ln([O_3]/[O_3]_o) = -kt How long does it take for half the ozone to decompose, and what is the concentration of ozone at that time? What is the initial reaction rate of the ozone decomposition reaction? Provide both the magnitude and the units. What is the rate of the reaction after 1.93 hours have passed? Why does the reaction rate change with time? What is the order of the reaction with respect to ozone? Consider that the reaction rate is independent of all other species. What is the overall order of the reaction? If the concentration of the ozone were doubled, what would happen to the reaction rate? What parameter in the integrated rate law determines the slope of the straight line in the graph?

Answer 1

The plot of ln[O3]/[O2] vs time is drawn and shown below. Time scale is taken as t*10-2

The rate constant from the slope is 0.054/100 =0.00054/sec

the equation of best fit is ln [O3]/[O3]o = -0.00054*t

Half life is the time required for the concentration to drop to 50% of its initial value.

ln 1/2 = -0.00054*t

t= 0.693/0.00054= 1283 seconds

from the best fit ln [O3]/[O3]o= -0.00054*t

[O3]/[O3]o =exp(-0.00054t)

d[O3]dt= -[O3]0 exp(-0.00054t)*0.00054

at t=0 -d[O3]/dt = 90*exp(-0.00054*0)*0.00054= 0.0486 uM/s

after 1.93 hrs, =1.93*60 minutes*60 seconds =6948 seconds

-d[O3]/dt = 90* exp(-0.00054*6948)* 0.00054=0.0011 uM/sec

the rate -dCA/dt= KCO3= K[O3]0*(1-XA)

hecne as the conversion changes the reaction rate also changes