Download Chemical Kinetics Problem Set: First Order Reactions and Rate Coefficients and more Lecture notes Chemical Kinetics in PDF only on Docsity!
Problem Set #1: Chemical Kinetics
14.7 a)
At the time when [A] = 0.3580 M, rate = 1.76 X 10 -5^ Ms-
b) To do this we need to assume the rate of reaction is constant over the time interval.
Therefore, Ä[A] = rate X Ät = (1.76X10 -5^ Ms )(60 s) = 1.06X10-1^ -3M
Hence [A] = 0.3580 - 0.00106 = 0.3569 M
Since [A] has not changed very much during the 60 sec, the assumption is valid.
c) When is [A] = 0.3500?
Make the same assumption about the Rate as in part b).
14.12 Rate = k[A] [B]p^ q
Use experiments1 and 2 to remove any dependence on B.
0.323 = (0.33)p p = 1
There is no simple way to remove the dependence on [A], so use experiments 1 and 3.
12.38 = 6X2q q = 1
Calculate the rate coefficient from experiment 1
Rate = k[A][B]
14.14 Rate = k[A] [B] [C]p^ q^ r
a) Using experiments1 and 2
2 = 2p p = 1
similarly,
4 = 2q q = 2
and
16 = 2 X2 X(1/2)p^ q^ r 16 = 2X4X(1/2)r r = -
b)
14.53 a) Plot ln k versus 1/T:
b) A linear regression of the above plot gives a slope of -13518 K.
But the slope = -E /R a
E a = -slopeR = 135188.314 = 112 kJ mol
c) You can read the 40 C value off the plot or use the equation and the 0 C valueo^ o for k.
k 2 = 3.05X10 s -3 -
The units of the rate constant indicate that this is a first order reaction. (It is also mentioned that this is a first order reaction in the “Integrative Example” on page 643.)
14.70 Since O 3 is the only reagent, there is really only one possible reaction for a fast, reversible first step:
Then there is a slow step, most likely:
The stoichiometry for this mechanism is consistent with the reaction under study.
The slow step determines the rate law
Rate = k [O ][O] 3 3
We need to use the steady state approximation to obtain [O] (i.e. the concentration of the intermediate).
rate of production of O = -(rate of consumption of O)
k [O ] = k [O ][O] + k [O ][O] 1 3 2 2 3 3
Therefore,
Substitute this expression for [O] into the rate law:
But, since the last step of the mechanism is slow:
k [O ][O] << k [O ][O], or k [O ] << k [O ] 3 3 2 2 3 3 2 2
Therefore,
which is consistent with the observed rate law if k = k k /k. 1 3 2
