3. A certain reaction has the following general form: 2A > B Concentration vs time data were collected for this reaction, at 50°C and an initial concentration of 0.0200 M. It is determined that a plot of ln[A] vs. time resulted in a straight line with a slope value of - 2.97 X 10-2 min-1.
A. Write the rate law.
B. Write the integrated rate law
C. What is k for this reaction (or what is the rate constant for this reaction)?
D. How much time is required for the concentration of A to decrease to 2.50 X 10-3M?
As per the data given, the slope of lnA vs time is giving straight line suggests the reaction to be first order. since the order (n) and rate (-dA/dt) are related as -dA/dt=k[A]n, n is order, K is rate constant for 1st order, n=1
aand hence -dA/dt= K[A], and integration of equation of the equation with [A]=[A]0 at t= 0 and [A]=[A] at t=t is
[A]= [A]o*exp(-Kt), (!)
or lnA=ln[A]0-Kt (2), so a plot of lnA vs time (t) gives straight line whose slope is -K. hence from the given data, K=2.97*10-2/min
given [A]0=0.02 and [A]=2.5*10-3M
from Eq.2, ln(2.5*10-3)= ln(0.02)-2.97*10-2t, time, t= 70 min
the rate law is -dA/dt= 2.97*10-2t
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