Question

Consider the reaction A → products at 282 K. The concentration of A was monitored over time and the data was analyzed by plotting. It was found that a plot of ln[A] vs time gave a straight line relationship. It was also observed that it took 20.7 s for the concentration of A to decrease from 0.689 M to 0.235 M. What is the half life for this reaction when [A]o = 0.689 M? t1/2 (in seconds)

Answer 1

This must be modelled as 1st order:

First = ln(C) vs. t

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

if x axis is "time" then the slope is "-k", and y-intercept is initial concentration C0. y-axis if ln(C) (natural logarithm of concentration)

from the data provided:

Cfinal = 0.235 M and Cinitial = 0.689 M takes, t = 20.7 s

substitute data to get K

ln(C) = ln(C0) - kt

ln(0.235) = ln(0.689) - k*(20.7)

k = (ln(0.235) -  ln(0.689) )/(-20.7)

k = 0.051964 1/s

now,

find the half life

for 1st order:

HL = ln(2)/k

HL = half life

HL = ln(2)/(0.051964) = 13.3389 seconds.