Hi !! We welcome you back to our lesson series where we evaluate a design problem of a process
to produce methanol from carbon dioxide and hydrogen.
In this second lesson, we are first going to interpret the design problem description
into a process flow sheet and then proceed to provide a step-by-step guide on how to
answer question one to question six.
The problem statement described that, the gas reactants comprising carbon dioxide and
hydrogen are fed into a CSTR at the specified conditions.
The reactants react to form methanol and water at complete conversion of 100%.
From the CSTR, the liquid product is then pumped using a centrifugal pump to be heated
in a shell and tube heat exchanger before being fed into a distillation column for separating
methanol from water.
In the distillation column, methanol of the required purity is produced as the top product
stream, while the wastewater is produced as the bottom stream.
This representation then gives the resulting process flow diagram.
Let us move on to start answering the questions.
Question-one deals with evaluating if the reaction will occur spontaneously at the given
conditions.
Let us first start by balancing the given reaction equation.
In order to form one molecule of methanol and one molecule of water, we will need six
hydrogen atoms.
This means that we will need to have three hydrogen molecules on the reactant side.
The carbon atoms already balance as there is one on the product side, and one on the
reactant side.
The oxygen atoms also balance as there is two on the product side, and two on the reactant
side.
Our reaction is now balanced, we can therefore proceed.
Remember that, from the second law of thermodynamics, some reactions occur spontaneously because
they give off energy in the form of heat, that is, the delta H of reaction is less than
zero.
Others reactions occur spontaneously because they lead to an increase in the disorder of
the system, meaning that, the delta S of the reaction is greater than zero.
Therefore, if a reaction is favoured by both delta H of reaction being less than zero,
and, delta S of the reaction is greater than zero, the reaction will definitely be spontaneous.
Also, if a reaction is not favoured by both delta H of reaction and delta S of the reaction,
meaning that delta H in NOT less than zero, and, delta S in NOT greater than zero, the
reaction will NOT be spontaneous.
The question that always arises is then, what happens when one of the potential driving
forces behind a chemical reaction is favourable and the other is not?
That is, what happens if delta H is less than zero and delta S is NOT greater than zero.
Or, what happens if delta H is NOT less than zero and delta S is greater than zero.
We can answer this question by defining a new quantity known as the Gibbs free energy
of the system, which reflects the balance between these forces.
The Gibbs free energy of a system, G, at any moment in time, is defined as the enthalpy
of the system, H, minus the product of the temperature, T, times the entropy of the system,
S. G equals H minus TS.
Therefore, the change in the Gibbs free energy of the system that occurs during a reaction,
delta G, is therefore equal to the change in the enthalpy of the system, delta H, minus
the change in the product of the temperature, T, times the entropy of the system, S.
For a reaction occurring at constant temperature, the Gibbs free energy can be calculated as
the delta H of reaction minus the delta S of reaction multiplied by temperature.
The change in the free energy of a system that occurs during a reaction can be measured
under any set of conditions.
If the data are collected under standard-state conditions, the result is the standard-state
free energy of reaction.
From this equation we can see that, delta G is negative for any reaction for which delta
H is negative and delta S is positive.
Therefore, delta G is negative for any reaction that is favoured by both the enthalpy and
entropy terms.
We can therefore conclude that any reaction for which delta G is negative should be favourable,
or spontaneous.
Conversely, delta G is positive for any reaction for which delta H is positive and delta S
is negative.
Any reaction for which delta G is positive is therefore unfavourable.
When a reaction is favoured by both enthalpy, delta H, and entropy, delta S, there is no
need to calculate the value of delta G to decide whether the reaction should proceed.
The same can be said for when reactions is neither favoured by enthalpy, delta H, nor
entropy, delta S. Free energy calculations become important for reactions favoured by
only one of these factors.
Reactions are classified as either exothermic when delta H is less than zero, or, endothermic
when delta H is greater than zero, on the basis of whether they give off or absorb heat.
Reactions can also be classified as exergonic when delta G is less than zero, or endergonic
when delta G is greater than zero, on the basis of whether the free energy of the system
decreases or increases during the reaction.
Let us now revisit our reaction with this knowledge.
Remember our reaction is that of 1 mole of carbon dioxide reacting with 3 moles of hydrogen
to form 1 mole of methanol and 1 mole of water.
We can calculate the standard-condition change of enthalpy of reaction from the standard
enthalpies of formation of the reaction compounds.
Similarly, the standard-condition change of entropy of reaction can be calculated from
the standard entropies of reaction components.
At standard-conditions we get the delta H of the reaction of -49.3 kilojoules per mole,
indicating that the reaction is exothermic and therefore favoured at this temperature.
However, we get the delta S of the reaction as -0.177 kilojoules per Kelvin per mole,
indicating that the reaction is not favoured by the entropy driving force.
This means that you will have to calculate the delta G of reaction to conclude if the
reaction is spontaneous or non-spontaneous.
You will have to calculate Gibbs free energy change of the reaction at the given reaction
temperature of 35 degrees Celsius.
Please go ahead and calculate delta G to answer whether the reaction is spontaneous or non-spontaneous
at the given conditions.
If we look at the literature, we find that, the thermodynamics of the reaction, as in
Carbon dioxide conversion and the selectivity towards methanol, is favoured by low temperatures
However, the kinetics of the reactions, which determine how fast the reaction is occurring,
is favoured by high temperatures The reaction is favoured by high pressure
due to the decrease in the number of moles, as we have four moles of reactants reacting
to form 2 moles of product.
Recent literature proposes a compromise in conditions at a T = 230°C and P = 50 bar
in order to optimise the yield and the rate of reaction.
We will take these conditions further as our newly suggested reaction conditions.
Let us now move on to the second question.
Question 2 asks if the reaction can be carried in a CSTR at the newly proposed conditions.
Remember we have suggested the new reaction conditions at a temperature of 230 degrees
Celcius, and pressure of 50 bar.
Before we answer this question, let us look at the three main types of reactors used for
continuous reaction processes.
The first type of reactor is the Continuous Stirred Tank Reactor, a CSTR.
The second type of reactor is a Plug Flow Reactor, also known as a PFR, or based on
the type catalyst used; it is frequently referred to as a packed bed reactor.
The third type of reactor is the Fluidized Bed Reactor, or sometimes referred to as a
Bubble Bed Reactor, if it comprises a gas-liquid reaction mixture.
The CSTR reactor is mainly used for liquid-phase reactions.
There are some uses where the CSTR is used for the gas-liquid reactions.
However, gas-gas reactions in a CSTR are extremely rare, as this type of reactor is not suitable
for this application.
A plug flow reactor is a preferred reactor for most phases, both homogeneous and heterogeneous
phases, that do not require back-mixing.
A fluidised bed reactor is an option for mixed-phase reactions, or homogeneous phases that are
solid catalysed.
Before we suggest which reactor is best suited for this reaction, let us check if the reactants
and products are liquids, or gases at the given reaction conditions of a temperature
of 230 degrees Celcius, and pressure of 50 bar.
The easiest way to do this is to use Chemcad simulation to check if these streams are gas,
liquid or mixed gas-liquid phase.
Okay!!
From Chemcad we can see that, at the given reaction conditions, the feed is gas and the
product stream is a gas-liquid mixed-phase stream.
For this type of reaction, a CSTR can work.
I am going to choose a PFR as in my opinion it is the best-suited reactor for this mixed-phase
reaction.
Particularly, because I know that this reaction will be catalysed by a solid catalyst.
The fluidised bed reactor would also be a suitable choice in a gas-liquid phase reaction
that is solid catalysed.
Proceeding on to question 3, we are required to evaluate the given reactor specifications.
We have been given the following information.
The volume of the reactor, V, which equals 800 litres
The flowrate of product, F, which equals 53 cubic meters per hour.
This flowrate of the product can be translated to calculate the flowrate of the feed, FO.
The rate of reaction, R, which equals 4 per minute
And also, we are given the conversion reactants to be 100%, meaning that, X equals 1
Let us look at the reactor design equations to see if we have all the information we need.
For a CSTR, the reactor sizing equation is as follows.
We have been given, the volume, the flowrate, the conversion, and the rate of reaction.
So the problem is over-specified and does not require us to calculate anything.
For a PFR, the reactor sizing equation is as follows.
Again here, we have been given, the volume, the flowrate, the conversion, and the rate
of reaction.
So the problem is over-specified and does not require us to calculate anything.
So, since the reactor design information is over-specified and does not require us to
do any calculation, we will therefore use a Conversion Reactor at 100% conversion, for
the simulation in Chemcad.
There is no need to use kinetics or equilibrium calculations.