In this problem we are considering the reverse water-gas shift reaction, and if you read
the problem statement, You will see that you are given some information about the forward
rate of reaction that is taking place when there is no product present, and eventually
we are asked to determine the initial rate of reaction for the reverse water gas shift
reaction when there is no CO or water present in that case. To do this we are going to need
to develop relationships between the forward and the reverse reaction. I am going to go ahead
and simplify this reaction of CO plus water to make CO2 and hydrogen. We will just call
it an A plus B going to C plus D reaction. Based on the given information, we are told
the form of the forward reaction is, and it is first order in both components A and B.
We can write it in terms of the concentration of component of A and B, I'll write it in terms
of the partial pressures which are really equivalent. We can convert between pressure
and concentration just dividing the pressure by RT, assuming ideal gas behavior. Which
is a good assumption under these conditions of low pressure. We are given in this problem
the rate of CO consumption under the case where we only have to worry about this forward
reaction. So I am going to designate this as being just the forward reaction. Alright,
and that rate is equal to 0.50 mol/min*g of catalyst, and that is equal to the forward
reaction rate constant, times each of the partial pressures, and they are each according
to the problem statement equal to 1 bar. So this allows us to very simply solve for the
forward reaction rate constant, and it is equal to 0.50 mol/(bar^2*min*g) of catalyst.
Now we want to try to use this information to determine something about the reverse rate
of the reaction. Here we anticipate that we are going to need to use an equilibrium relationship,
where we can write the total rate of net reaction as being equal to the forward reaction rate,
that we have the rate constant for, minus the reverse reaction rate. Then we know that
this at equilibrium has to be equal to 0. So therefore at equilibrium the reverse reaction
rate has to be equal to again this forward reaction rate. Alright we also know that we
can write an equilibrium relationship terms the equilibrium constant for this reaction,
and that equilibrium constant we can just ratio the partial pressures of the reactants
to the products, since the stoichiometry is 1 to 1 to 1 to 1. So the rate of the reverse
reaction if we combine these two equations will be dependent to the first order on the
partial pressures of C and D, and this is true at equilibrium. But we are also going
to use this expression for the rate of the reverse reaction away from equilibrium, because
it describes the rate going from products to reactants. It depends on the partial
pressure of our hydrogen and our CO2 for the reverse water/gas shift reaction. Alright,
so now we already know the value of Kf. We determined that earlier in the problem, and
we are given information on the partial pressures of C and D in the problem statement. So therefore
all we need to do is figure out what the value of Kc is, and Kc for an ideal gas, and for
a case where you have equal moles on the reactants and products as we do in this case. Kc we
can show is just equal to the equilibrium constant that is based on ratio of activities
instead of concentrations, and so that Kc is equal to the exponential of the negative
of the Gibbs free energy change of reaction, divided by RT. Now we are given data on this,
typically at 298 K if you look in various reference sources, and so we will first compute
this at 298 Kelvin even though the reaction conditions are at higher temperatures. So then we will
need to correct for the fact the temperature is not 298 Kelvin, and I should point out here
that earlier in the problem statement, that I should have defined our equilibrium constant initially
as being the pressure based equilibrium constant since I am working with partial pressures,
but as I mentioned earlier in the problem statement Kp and Kc are basically equivalent
here, because we are dealing with ideal gases and with the same number of moles on the reactant
and product side. We can show that Kp is equal to Kc. Alright so let's evaluate K at 298 K.
So when we evaluate this equilibrium constant we get the exponential, the negative of the
Gibbs free energy of the products minus the reactants, the Gibbs free energy of formation
of hydrogen is 0, since it is in its elemental state. Then in the denominator we simply divide
by RT. Where again we have to use the temperature of 298 K. The reason that we have to do that
is that these experimentally measured Gibbs free energies of formation were also measured
at 298 K. So we have to use that same temperature here. If we solve for the value of this equilibrium
constant at 298 K, we get a value around 103,000. Now of course we want to solve for the equilibrium
constant at the higher temperature, the actual reaction temperature in this case of 600 K,
and so to do that we will have to use the Van't Hoff relationship. The Van't Hoff relationship
uses the heat of reaction, the enthalpy of reaction. I am expressing this in terms of
the negative enthalpy of reaction that we are going to use in the expression, and so
it is going to be the enthalpy of the products, the negative of it, again the product hydrogen
has an enthalpy of formation of 0, because it is in its elemental state, and then we
subtract the energy of the reactants. Again we can look up in the back of a thermo book to give us
an overall negative heat of reaction of around 41 kJ/mol. So in other words since the negative
heat of reaction is positive it is an exothermic reaction, and so we are going to expect this
value of the equilibrium constant to decrease as we increase the temperature. Now we can
write up the Van't Hoff relation for the K at about a value of 600 Kelvin, and it is equal
to a K at 298 times this exponential term. Where we are taking the reciprocal of temperature
difference between our 2 conditions. Alright, what this Van't Hoff equation assumes is
that the heat of reaction is approximately constant over the temperature range between
298 Kelvin and 600 Kelvin, which turns out to be a reasonable assumption in this case. You can use a more
complex form involving the heat capacities of the different gases for more accuracy as we
will see in a future screencast. We have already solve for the value of K, 298 Kelvin, above. We have
already solve for the negative heat of reaction that we can plug in. R we are going to again
use 0.008314 kJ/mol*K, and therefore we can solve for our equilibrium constant at 600
Kelvin, and we find that is has dropped why down to a value of around 24, and now we have what
we need in order to solve for our final value for the reverse reaction rate. Remember the
only thing that we were missing was this value of Kc, and now we have it. So if we just plug
in the necessary values according to the problem statement we get the following rate
expression for the reverse reaction rate, where again we plug in our k forward our equilibrium
constant and the partial pressures of CO2 and hydrogen that are given to be 2 bar and
half a bar, and ultimately solve for the initial rate of the reverse reaction under those conditions
of being around 0.02 mol/min. So we can see that the given information for the initial
rate for the forward reaction when we combine that with the ability to solve for the equilibrium
constant for that reaction at our temperature at 600 Kelvin enables use to solve for the initial
rate of the reverse reaction.