homework solution:
exp 1: 0.26 = k [1]x[0.25]y
exp 2: 0.52 = k [1]x[0.5]y
}
exp 4: 4.16 = k [2]x[1]
exp 5: 9.36 = k [3]x[1]
}x=2
exp 1: 0.26 = k [1]2[0.25]1
}k=1
exp 3: N = 1 [1]2[1]
}N=1
exp 6: 16.64 = [4]2[M]
}M=1
y=1
29-1
First Order Reactions
e.g., A
First order: Rate = k[A]
products
-d[A]/dt = k[A] Y -d[A]/[A] = k dt
A
-I d[A]/[A] =
A0
t
0
Ik dt
ln [A]
[A0]
ln [A] = -kt + ln[A0]
y
= mx + b
t
ln [A0]/[A] = kt
slope = -k
29-2
First Order: summary and useful equations
Reaction rate = k[A] or -d[A]/dt = k[A]
[A] = [Ao] e-kt
p 572: Ebbing
or Rn[A] = Rn[Ao] - kt
ln{[Ao]/[A]} = kt or ln[Ao] - ln[A] = kt (integrated form of rate
law: from integral calculus)
[A] = [Ao] e-kt
First order rate constants have the units (time)-1.
half-life of any reaction is the time taken for the concentration of a reactant
to decline to half its original value so that at time t½, [A]= [Ao]/2.
Substitute [A] = [Ao]/2 into the 1st order equation:
ln{[Ao]/[Ao/2]} = kt½
t½ = (ln 2)/k = 0.693/k
The half-life of a first order reaction is independent of the initial
concentration of the reactant. (see later 14C radioactive dating)
29-3
example
For the 1st order decomposition of N2O5 at a given temperature, it was found that
[N2O5] was 0.0300 mol L-1 after 10.0 min and 0.0150 mol L-1 after 30.0 min.
Calculate the rate constant k and the original N2O5 concentration when the
reaction was started.
ln[A] = ln[A0] - kt; so, ln [A0] -ln [A] = kt
ln 0.03 - ln 0.015 = 0.7 = kt = k x 20
0.7/20 = k = 0.035 min -1
ln [A]
ln[A] = ln[A0] - kt
ln[A0] = ln[A] + kt
t
[A0] is now at time =0
so: ln[A0] = ln[0.03] + 10x 0.035
29-4
= -3.156
[A0] = 0.042 molL-1
radioactive carbon dating
14N:
14C:6p8n
7p7n
Carbon exists as 3 isotopes 12C, 13C and 14C
radioactive
14C
decays at 14 disintegrations /minute in a living sample
once a sample dies no more 14C
can be incorporated
so gradually the amount of 14C decreases:
so number of disintegrations
seen decreases --eventually to zero (none left)
by measuring how much is left we can tell when
something died -- thus how old it is
half life: time for half of it to decay away
is 5.73 x 103 years
how do we know that ...and how do we date objects?
29-5
radioactive decay is a first order process : if you plot ln [14C] vs t you get a
straight line
lets work out the time for the [14C] to go to one
half its initial value
you get this by measuring the
number of disintegrations
at this point [14C] = [14C]0
ln
[14C]0
- ln
[14C]
ln[14C]
= kt
slope = -k
ln ([14C]0 / [14C]) = kt
also see page 29-3
ln 2 = kt 1/2
kt 1/2 = ln 2 = 0.693
t
example
29-6
example
a carbon sample from the dead sea scrolls had a disintegration rate of
11 dis./min. How old are the scrolls?
t 1/2 of 14 C is 5.73 x 103 years ( you get this from decay rate)
so: k = 0.693/ 5.73 x 103 = 1.21 x 10-4 year -1
ln 14/11 = 1.21 x 10 -4 t
t = 2 x 10 3 years
about 2000 years old!
disintegrations for a "fresh" sample: 14 is nothing to do with
14C
29-7
another example
14C
undergoes 1st order radioactive decay with a half life of 5730 years. If
20% of the original 14C is present (left) in a sample then how old is it?
for 1st order ln [A] = ln [A]0 -kt
ln 100/20 = 1.21 x 10 -4 t
t = 1.609/ 1.21 x 10 -4 = 13301 years
shroud of Turin: FAKE!!
29-8