This is an exam problem that deals with the kinetics of an elementary, reversible reaction,
and so as we read the problem statement, because it is elementary and reversible, then we know
that we can write down the rate of consumption of A as being equal to a forward rate constant,
which I'll call k_f, times the concentrations of the reactants raised to their stoichiometric
coefficients, which in this case means to the first power, minus 1 over an equilibrium
constant times the concentrations of the products raised to their stoichiometric coefficients.
In the problem statement, we're given information about the forward rate constant, k_f.
In part A, we're asked to determine the activation energy for that forward rate constant.
If we're determining an activation energy, then we can use the Arrhenius-type expression
to do that, and so one form of the Arrhenius expression is that if we look at k forward
for the higher temperature over k forward for the lower temperature, and I will give
those superscripts 2 and 1 respectively, we know that k forward with the superscript 2
has a rate constant of 4.3 liters per mole second, and for k forward in the lower temperature
case, it's 2.0 liters per mole second.
I'm not bothering to write down the units here because they're the same units in the
numerator and denominator and they'll cancel.
And then one of our Arrhenius expressions is that this is equal to the exponential of
the forward activation energy, Ef over R, times 1 over T1 minus 1 over T2.
We can then substitute in for T1 and T2.
T1 will be equal to 318 Kelvin, because we need to use absolute temperature.
T2 is going to be equal to 348 Kelvin.
In this case I'll use units of kilojoules per mole Kelvin.
If we plug in those, our only unknown is Ef, and we can solve that most easily by taking
the log of both sides, but I won't show those details.
And so we get that Ef is a little over 23 kilojoules per mole.
We've solved for the activation energy for the forward rate constant.
In part B, we're told that there's an equimolar mixture of A and BC that's used as the feed,
and then we achieve an equilibrium conversion of 75%.
The question is, what will be the k for the reverse reaction at the lower temperature?
And so we need to solve for k_r under condition 1, which is the lower temperature of 318 Kelvin,
or 45 degrees C. To solve for this we'll realize that this would be equal to k forward for
that lower temperature divided by Kc at that lower temperature, where Kc is the equilibrium
constant associated with this reaction, the concentration based equilibrium constant,
and that concentration based equilibrium constant is equal to the following, where the superscript
e denotes that we're at equilibrium.
And then we can assume for the moment that we're dealing with a constant volume type
of reaction, so in that case, we can substitute in for the product concentrations and the
reactant concentrations, okay, and so for example the concentration of AB at equilibrium
will be equal to the concentration of A that we started with times the equilibrium conversion,
and the same will be true for the concentration of C, and that's because there's a 1 to 1
stoichiometry, so for every mole of A that we convert, we produce 1 mole of both AB and
C. And similarly, if we plug in for the concentration of A assuming constant volume, then we get
a C_A0 times 1 minus Xe based on the definition of conversion, alright so this is the equilibrium
condition, and then it's an equimolar feed, and so the concentration of BC is always equal
to the concentration of A. So we can put squares here, the C_A0 terms go away, and then we
can plug in for Xe that is just equal to 0.75.
And so ultimately we can solve for Kc being equal to 9.
We just do a straight substitution of, we already know the value of k_f as given in
the problem statement, and so that means we can solve for k_r at this lower temperature
of being equal to 0.22 liters per mole second.
And if we go up and look at part C, we need to estimate the reverse rate constant at 75
degrees C, and to do this we basically have to know the activation energy for the reverse
rate constant.
The key piece of information that we need comes in this part of the problem statement,
where we're given that the reaction is exothermic by 40 kilojoules per mole of A reacted, and
we know the activation energy for the forward rate constant, and so what we can do for part
C is realize that if we're looking at this exothermic reaction on a potential energy
diagram, that if this is an exothermic reaction, then the products are at a lower energy than
the reactants, and that the activation energy for the reverse reaction should therefore
be equal to the activation energy for the forward reaction plus the amount of heat that's
released from the exothermic reaction.
And it can be hard to remember to include this negative sign, but we can always, if
we're looking at this in terms of the reactants and products here, then we can always recall
what the signs should be by just plugging in the numbers.
So we've already solved for the forward activation energy as 23.5 kilojoules per mole, and we
were given in the problem statement that we have an exothermic reaction by 40 kilojoules
per mole, and we know what the signs should be here because we know that the barrier should
be higher in the reverse direction for this exothermic reaction, and therefore we need
to make a larger number and add those two numbers together.
Alright, and so if we do that, we get 63.5 kilojoules per mole, and now it's just a matter
of, since we have the reverse rate constant at the lower temperature and we have the activation
energy, it's just a matter of using the Arrhenius expression in order to solve for the reverse
rate constant at the higher temperature.
And so I'll rewrite that out now, though we've already used it in part A, and so the Arrhenius
expression is shown here, where we make use again of our new activation energy for the
reverse reaction, and the rate constant at lower temperature, at 45 degrees C, for the
reverse reaction.
We multiply inside the exponential by the reciprocal of the temperature difference,
and we get this value for the reverse rate constant at the higher temperature of 1.74.
What we notice, and is a good check for us, is that the reverse rate constant goes up
by a much greater factor as we increase the temperature than the forward rate constant,
so roughly a factor of 8 increase as we increase the reaction temperature by 30 degrees C.
We see that instead for the forward rate constant that we only get a little more than a doubling
of the rate constant as we increase the temperature, which makes sense, that the reverse reaction
has a higher activation energy, and so it's going to be more and more favored as we approach
higher temperature for this exothermic reaction.
And that's also consistent with our picture for equilibrium, that the value of Kc for
this exothermic reaction should go down as we increase the temperature, so that we're
favoring the reverse reaction, relative to the forward.