the relation between rate of decomposition of H2O2 ( -d[H2O2]/dt) vs rate consta
ID: 700590 • Letter: T
Question
the relation between rate of decomposition of H2O2 ( -d[H2O2]/dt) vs rate constant K and order (n) is given by
-d[H2O2]/dt= K[H2O2]n,
at t=0 , [H2O2]= [H2O2]o= initial concentration and at t=t, [H2O2]= [H2O2], concentration of H2O2 at any time
are the boumdary conditions
when n= 0, the reaction is zero order and the integrated expresion becomes [H2O2] = [H2O2[0-Kt, so a plot of [H2O2] vs time gives straight line, with a slope of -K, if the reaction is zero order.
when n=1 , the reaction is 1st order and the integrated expression becomes ln [H2O2] =ln[H2O2]o-Kt, so a plot of ln[H2O2] vs time gives straight line with a slope of -K for 1st order reaction
when n=2, the reaction is second order and integrated expression becomes 1/[H2O2]= 1/[H2O2]o+ Kt, so a plot of 1/[H2O2] vs t gives straight line whose slope is K for second order reaction.
the plots are shown in the following.
from the plots, it can be concluded that the reaction is 1st order with highest R2 value.
hence the equation of best fit is ln[H2O2] =ln[H2O2]o-Kt
to identify the constant K from the equation of best fit, the equatino is rewritten as
ln[H2O2] = ln [H2O2]o- 100*K(*t/100)
hence the slope will be 100K.
100K= 0.083 and K=0.083*10-2 /sec
the equation of best fit is ln[H2O2] = ln[H2O2]o-0.083*10-2*t
at 4000 second, ln[H2O2] =0 -0.083*10-2*4000
[H2O2] =0.036M, rate law is -d[H2O2]/dt= 0.083*10-2 [H2O2]
Explanation / Answer
9. The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature Time (s) 0 120 ± 1 300 ± 1 600 ± 1 1200 ± 1 1800 ± 1 2400 ± 1 3000 ± 1 3600 ± 1 [H,O,] (mol/L) 1.00 0.91 0.78 0.59 0.37 0.22 0.13 0.082 0.050 3 points Assuming that determine rate law, the integrated rate law, and the value of the rate constant. Also calculate [H2O2] at 4000. seconds after the start of the reaction
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