In chapter 2 we saw that if we had –rA as a function of X, [–rA= f(X)] we could size many reactors and reactor sequences and systems.
How do we obtain –rA = f(X)?
We do this in two steps
1. Part 1 Rate Law – Find the rate as a function of concentration,
–rA = k fn (CA, CB …)
2. Part 2 Stoichiometry – Find the concentration as a function of conversion
CA = g(X)Combine Part 1 and Part 2 to get -rA=f(X)
A rate law describes the behavior of a reaction. The rate of a reaction is a
function of temperature (through the rate constant) and concentration.
|Relative Rates of Reaction (p. 90)||top|
aA + bBcC + dD
|Power Law Model||top|
|Rate Constant, k||top|
k is the specific reaction rate (constant) and is given by the Arrhenius
E = activation energy (cal/mol)
A, and k, depend on overall
The larger the activation energy, the more temperature sensitive k and thus the reaction rate.
Why is there an Activation Energy?
(1) the molecules need energy to distort or stretch their bonds in order to break them and to thus form new bonds
(2) as the reacting molecules come close together they must overcome both steric and electron repulsion forces in order to react
In our development of collision theory we assumed all molecules had the same average energy. However, all the molecules don’t have the same energy, rather there is distribution of energies where some molecules have more energy than others. The distraction function f(E,T) describes this distribution of the energies of the molecules. The distribution function is read in conjunction with dE
One such distribution of energies is in the following figure.
By increasing the temperature we increase the kinetic energy of the reactant molecules which can in turn be transferred to internal energy to increase the stretching and bending of the bonds causing them to be in an activated state, vulnerable to bond breaking and reaction.
We see that as the temperature is increased we have greater number of molecules have energies E A or greater and hence the reaction rate will be greater.
You can tell the overall reaction order by the units of k
|CA||-rA||Reaction Order||Rate Law||k|
|(mol/dm3)||(mol/dm3*s)||zero||-rA = k||(mol/dm3*s)|
|(mol/dm3)||(mol/dm3*s)||1st||-rA = kCA||s-1|
|(mol/dm3)||(mol/dm3*s)||2nd||-rA = kCA2||(dm3/mol*s)|
The activation energy is a measure of the minimum energy a that the reacting molecules must have in order for the reaction to occur.
The reaction of AB + C to form A + BC is shown above along the reaction coordinate. One way to think of the reaction coordinate is the linear distance between the AB molecule for a fixed linear distance between the AC molecule. At the start of the reaction the AB distance is small and the BC distance is large. As the reaction proceeds, the A–C distance remains fixed but B moves away for A and closer to C and the energy of system increases. At the top of the barrier molecule B is equal distance from A and C. But as it crosses the barrier it moved close to C to form the BC molecule and the A molecule alone.
The following movie was made by the students of Professor Alan Lane's chemical reaction engineering class at the University of Alabama Tuscaloosa
|Elementary Rate Laws (p. 82)||top|
A reaction follows an elementary rate law if and only if the (iff) stoichiometric coefficients are the same as the individual reaction order of each species. For the reaction in the previous example (), the rate law would be:
if 2NO+O2 2NO2 then -rNO = kNO (CNO)2 CO2 if elementary!
See the example below for more examples of rate laws.
|Non-Elementary Rate Laws||top|
Example: If the rate law for the non-elementary reaction
is found to be
then the reaction is said to be 2nd order in A, 1st order in B, and 3rd order
|Reversible Reactions (p. 88)||top|
The net rate of formation of any species is equal to its rate of formation in the forward reaction plus its rate of formation in the reverse reaction:
ratenet = rateforward + ratereverse
At equilibrium, ratenet0
and the rate law must reduce to an equation that is thermodynamically consistent
with the equilibrium constant for the reaction.
Example: Consider the exothermic, heterogeneous reaction
At low temperature, the rate law for the disappearance of A is
At high temperature, the exothermic reaction is significantly reversible:
What is the corresponding rate law? Let's see.
If the rate of formation of A for the forward reaction (A + BC) is
then we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (CA + B) is
Does this rate law satisfy our requirement that, at equilibrium, it must reduce to an equation that is thermodynamically consistent with KP? Let's see.
From Appendix C we know that for a reaction at equilibrium:
At equilibrium, rnet 0,
Solving for KP gives:
The conditions are satisfied.
We shall set up Stoichiometric Tables using A as our basis of calculation
in the following reaction. We will use the stoichiometric tables to express
the concentration as a function of conversion. We will combine Ci
= f(X) with the appropriate rate law to obtain -rA = f(X).
|Batch System Stoichiometric Table (p.101)||top|
Concentration -- Batch System:
Constant Volume Batch:
Note: if the reaction occurs in the liquid phase
if a gas phase reaction occurs in a rigid (e.g., steel) batch reactor
Constant Volume Batch
|Flow System Stoichiometric Table (p. 107)||top|
|Species||Symbol||Reactor Feed||Change||Reactor Effluent|
Concentration -- Flow System:
Liquid Phase Flow System:
Flow Liquid Phase
If the rate of reaction were -rA = kCACB then
we would have
This gives us -rA = f(X). Consequently, we can use the methods discussed in Chapter 2 to size a large number of reactors, either alone or in series.
Again, these equations give us information about -rA = f(X), which we can use to size reactors.
Flow Gas Phase
Calculating the equilibrium conversion for gas phase reaction
Consider the following elementary reaction with KC and = 20 dm3/mol and CA0 = 0.2 mol/dm3. Pure A fed. Calculate the equilibrium conversion, Xe, for both a batch reactor and a flow reactor.
Species Initial Change Remaining A NA0 -NA0X NA = NA0(1-X) B 0 +NA0X/2 NB = NA0X/2 NT0 = NA0 NT = NA0 - NA0X/2
Constant Volume V = V0
Species Fed Change Remaining A FA0 -FA0X FA = F A0(1-X) B 0 +FA0X/2 FB = F A0X/2 FT0 = F A0 FT = F A0 - F A0X/2
Old Exam Questions
* All chapter references are for the 4th Edition of the text Elements of Chemical Reaction Engineering .