>>> Qn 1:
I am not given rate equation. I derive the rate equation from the
graph I plot. I plotted the graph and I realized I have 2nd-order
reaction. I still am clueless how to find rate constant. Because at
different points on the graph, I have different rate constants. ( I
find gradient at a particular point of the graph to get the rate in
moldm-3 )
Should rate constant for a 2nd-order reaction be the same?.
<<<
Rate constant, by definition, is a constant (ie. fixed value) at
any given temperature. In point of fact, Kc, Ksp, Kp, Ka, Kb, Kw,
etc, are all constants because k is a rate constant at any given
temperature. (Kinetics and Equilibria are
actually intimately related, although most JC students don't
appreciate this, because their JC teachers don't encourage them to
truly understand and appreciate Chemistry, but to just follow
dogmatic instructions for Chemistry.)
The decomposition of sodium azide to nitrogen gas involves the
formation of a stable diatomic nitrogen molecule from a
triatomic nitrogen azide ion, N3-. To do so would indeed require
(stoichiometry) 2 sodium azide ions coming together in a
redox reaction to form 3 nitrogen molecules and 2 sodium
atoms. And since in order to reduce sodium ions to sodium metals we
must involve lone electrons, we would expect photolytic
cleavage to form nitrogen free radicals are involved.
And since the highly reactive free radical must be involved in
the fast step, the formation of the free radical must be the slow
step, the rate determining step.
Therefore, the rate determining elementary step is indeed the
bimolecular reaction of 2 sodium azide formula units as the
reactants, and thus it is indeed a 2nd order reaction, and the rate
equation must be rate = k [NaN3]^2.
The above thought process, while not required at the JC level
when attempting Kinetics questions, is encouraged because it helps
you to see how all aspects of Chemistry are interconnected, and
helps you to truly understand, appreciate and enjoy Chemistry.
Which provides support that your graph, and your interpretation of
it, is indeed correct - it's a second order reaction.
For second order reactions, the half-life is not constant, but
can still be calculated by the formula t(1/2) = 1 / (k [A]0), where
the denominator is rate constant k multiplied by initial molarity
of reactant.
For all reactions, including 2nd order, the rate constant for
the reaction, k, can indeed be obtained by substituting any
pairs of values (rate and molarity of reactant) from the table
provided by the question, into the rate equation.
The perhaps-somewhat-annoying aspect of this question, is that
you're given volume of product instead of reactant. Therefore,
using stoichiometry of the overall equation, work out the volumes,
proportional to moles, hence proportional to molarity, of the
reactant. Hence, you can work out the rate of reaction based on how
fast the reactant is being used up, and together with the molarity
of reactant, substituted into the rate equation, gives you the
value of rate constant k.
>>>Q2 "to check for a constant half-life in a "[product]"
against "time" graph, you must use the y-axis intervals of 1/2a,
3/4a and 7/8a, in contrast to (if it were a usual [reactant]
against time graph) 1/2a, 1/4a, 1/8a."
Do you mean that a refers to the initial
concentration right from the start?
So I would need to find half-life from 1/2a to 1/4a,
another half-life from 3/4a to 3/8a
and so on... ?? <<<
Since your graph is based on product, a is the FINAL
molarity of product. The initial molarity, is of course zero.
(Doncha love these "product" instead of "reactant" sadistic kinetic
qns? Woot!)
Half life t1 = t2 = t3
t1 = 0 to 1/2 a
t2 = 1/2 a to 3/4 a
t3 = 3/4 a to 7/8 a