a) At a certain temperature, the rate constant for this reaction is 5.09× 10–4 s–1. Calculate the half-life of cyclopropane at this temperature.
b) Given an initial cyclopropane concentration of 0.00190 M, calculate the concentration of cyclopropane that remains after 2.40 hours.
from the units of the rate constant it is clear that it is first order
units for first order reaction is s-1
relation between rate constant and half life is
t1/2 = 0.693 / k = 0.693 / 5.09× 10–4 s–1 = 1.36 x 103
part B
formula for first order is
k = (2.303 / t ) x log[A0/A]
A0 = intial concentration which is given as 0.0019 M
A = concentration after 2.4 h
t = 2.4 h convert in to sec = 2.4 x 60 = 144 min x 60 = 8640 s
k = rate constant = 5.09× 10–4 s–1
plug in these values
5.09× 10–4 s–1 = (2.303 / 8640 s ) x log[0.0019 / A]
log[0.0019 / A] = 1.909
0.0019 / A = 101.909
0.0019 / A = 81.20
A = 0.0019 / 81.20
A = 0.0000234 or
A = 2.34 x 10-5 M
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