Errors? Please direct to Jay Joshi 2. Vinyl chloride can be synthesized by reaction of acetylene with hydrochloric h ydrochloric acid over a mercuric chloride catalyst at 500 K and 5.0 atm total pressure. An undesirable side reaction is the subsequent reaction of vinyl chloride with HCl. These reactions are illustrated below.
The equilibrium constants at 500 K are 6.6 x 103 and 0.88 for reaction 1 and 2, respectively. Assume ideal behavior behavio r. Solution a) Find the equilibrium composition at 5.0 atm and 500 K for the case when acetylene and HCl are present initially as an equimolar mixture. What is the equilibrium conversion of acetylene?
First, write out the ICE table for the two reactions. Keep in mind, the feed is equimolar:
The respective equilibrium constants can also be broken up into K y and K P, using the ideal gas assumption to ignore K φ:
= = −− = = You can then substitute in ξ1 and ξ2 for the vapor mole fractions in both equations. Plugging in the given equilibrium constants at reaction con ditions as well transforms the equations:
6.6×10 =1 15 ∗ 0.ξ5 ξ1 ξξ0.ξ 0.1 5ξξξ ξξ 0.88 = 5 ∗ ξ ξ0.0. 5 ξ ξ
Errors? Please direct to Jay Joshi Now, we can construct our “ICE table” to relate the equilibrium constants to composition: !"#$#%&
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/ 01ξ/-2ξ1-2ξ2
/01ξ/-2ξ1-2ξ2
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5 ξ/
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ξ1
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ξ2
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So we can relate our individual equilibrium constants to composition directly, since we made the ideal gas assumption, and are operating at an assumed standard state pressure of 1 atm:
= () = 12 ++2 ++ 2 = 0.075 = = 12 + 2 + 2 = 0.078 = = 12 2 2 = 0.167
This is a system of three equations and three unknowns. The ξ’s can be solved to find: ξ1 = 0.062 ξ2 = 0.065 ξ3 = 0.138 Now, the composition can be evaluated: Toluene: Benzene: Ortho-xylene: Meta-xylene: Para-xylene:
= =1 +2 +2 = 20.062 = +0.12065∗0.+0.0622 ∗ 0. 0 65 2∗ 0. 1 38 = . 1 38 = . = = . = = . = = .
A shape-selective catalyst can be utilized to increase para-xylene selectivity, since paraxylene has a smaller kinetic diameter than the other isomers. The catalyst could allow mostly para-xylene to diffuse out of it to shift the product distribution.
Errors? Please direct to Jay Joshi Solving the system of equations gives ξ1 and ξ2, which are can then be directly plugged in to the equations found in the ICE table above to find the outlet composition:
To find equilibrium conversion of acetylene, we u se the equation:
= 1 ° = 1 0.0.0125 = .
b) Redo part (a) with a large excess of inert gas. Assume the inert gas constitute 90 vol. % of the initial gas mixture. We now have to rewrite the ICE table. To account for the addition of inerts, add 9 moles to the previous 1 mole basis to obtain 90 vol% of inerts:
Relating the information on the table above to the equilibrium constants, our system of equations now becomes:
ξξ0.10ξ5ξ ξξ 6.6×10 =1 15 ∗ 0.ξ5 10ξ 0.88 = 5 ∗ ξ ξ0. 5ξ ξ ξ
Solving for ξ1 and ξ2, we can evaluate the compositions of the final mixture:
Errors? Please direct to Jay Joshi
Analogously to part (a), the equilibrium conversion is found using the new final acetylene amount:
, = 1 0.00.0145 = .
Errors? Please direct to Jay Joshi 3. A first-order homogenous reaction of A going to 3B is carried out in a constant pressure batch reactor. It is found that starting with pure A, the volume after 12 min is increased b y 70 percent at a pressure of 1.8 atm. If the same reaction is to be carried out in a constant volume reactor and the initial pressure is 1.8 atm, calculate the time required to bring the pressure to 2.5 atm. Solution
First, we need to solve for the reaction rate constant . Start with the batch reactor design equation. Remember, we are told the reaction is first-order, so we can write:
= = ° =
Integrate the differential equation to obtain the solution form for a first order reaction:
We do not know
°
. However, we can find it through the molar expansion factor . To do this,
start by using the ideal gas law to define the ratio of initial and final total moles:
° = °
The vessel is constant pressure, so P does not change. We are also given that there is a 70% increase in volume after 12 minutes. This information let us simplify our ratio:
° = ° = 0.7° °+ °= 1.7 ° = 1 +
We relate this ratio of total moles to conversion using the following equation:
The expansion factor can be found using the given information
Write conversion in terms of
°
= 1 3|11| = 2
. Then, plug in the appropriate numbers and solve:
1.7 = 1 +2∗ 1 ° ° = 0.65
Errors? Please direct to Jay Joshi
Now, we can plug this into our solution of the batch reactor balance to solve for the reaction rate constant :
°
0.65 = ∗12 = 0.0359 −
Now, we can start to consider the actual variable pressure case in the problem statement. However, from the design equation will not be the same as before. We can still find it using the same method though. So, we write out the ideal gas law again:
° = ° ° = ° = 21..58 = 1.389 ° 1.389 = 1+2 ∗1 ° ° = 0.8055 0.8055 = 0.0359− ≈
The expansion factor is unchanged. Solving for
:
Now, plug this into the batch reactor design equation we derived before to solve for time:
Errors? Please direct to Jay Joshi 4. At room temperature, sucrose is hydrolyzed by the catal ytic action of the enzyme sucrase as follows: sucrose → products Starting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentration C E0 = 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements): C A, millimol/liter t , hr
0.84 0.68 1
2
0.53 3
0.38 0.27 4
0.16
5
6
0.09
0.04
0.018
7
8
9
0.006 0.0025 10
Determine whether these data can be reasonably fitted by a kinetic equation of the MichaelisMenten type, or
where C M = Michaelis constant. If the fit is reasonable, evaluate the constants k 3 and CM. Solve by the integral method.
Solution
This problem is an example of using linear regression to fit data. The Michaelis-Menten equation can be put into a linear form, where it can then be compared to the collected data to see if the data fits the model. With this in mind, we can attempt to linearize the given rate equ ation. To simplify this, we make some substitutions:
Substituting:
= = 1 = = + = 1 +
11
Errors? Please direct to Jay Joshi Now, we have an ODE, which we can separate and integrate. This is the integral method of fitting rate equations: HINT: This integral appears difficult, but try separating the fraction, and then integrating it in two, simpler parts
1+ = + = + =
Transforming to put into a more useful form (linear equation):
“Dependent variable”
+ = = “Slope”
“y-intercept
“Independent Variable”
We can now plot our data, and see if it fits our derived model. A sketch of the graph is shown below:
Errors? Please direct to Jay Joshi As written on the figure, fitting the data to our mo del reveals our unknown parameters:
So our final rate equation becomes:
== . . − = . +.
To verify if our fit is appropriate, we can comp are the given data to that generated from our linearized Michaelis-Menten equation. The data that was used to generate the plot above is shown. Because we successfully fit our data to the Michaelis-Menten model, we have shown that it can be reasonably fitted :
Errors? Please direct to Jay Joshi 5. An ampoule of radioactive Kr-89 (half life = 76 minutes) is set aside for a day. What does this do to the activity of the ampoule? Note that radioactive decay is a first-order process Solution
To begin, “activity” is a thermodynamic quantity, which is defined from the chemical potential of a species relative to a selected standard state. Iin this context, it can be assumed to be the effective concentration of Kr-89. We are given that this is a first order process. From the design equation, we know that the general equation for this reaction becomes:
=
We are also given information about the half-life of the ampoule. At t = 76 min, we know that CA = 0.5CA0. This allows us to solve the reaction rate constant :
0.5 = |2| = 76 = 0.00912 − = 0.5 = [0.00912 − 1 × 24ℎ × 6ℎ 0 ] = 1.98 ×10− 2 ×10−
So to gauge the relative change in activity, we can pick another time, and see how the concentration changes. For the half-life, we found
. Let’s find what this ratio
becomes after one day has elapsed:
So, the activity has dropped to about
of its original value
Errors? Please direct to Jay Joshi 6. Find the first-order rate constant for the disappearance of A in the gas reaction 2A → R if, on holding the pressure constant, the volume of the reaction mixture, starting with 80% A, decreases by 20% in 3 min.
Solution
Assume: Ideal gas This problem is very similar to #3 on this assignment. To begin, we write the batch reactor design equation. We are told implicitly through the information regardin g the reaction rate constant that the reaction must be first order:
Which yields the solution:
To find k, we must first find
°
= = ° =
. We can do this by first evaluating what the change in the
total number of moles are. This can be done by through the ideal gas assumption:
° = °
Pressure and temperature are held constant. We are also told that the volume decreases by 20% at a given time, so:
° = ° = ° 0.° 2 °= 0.8 Additionally, we write the relationship relating the change in total moles to the conversion of our reactant and the molar expansion factor :
° = 1 + = 1 °
Conversion of A can be written in terms of moles of A:
Plugging in:
Errors? Please direct to Jay Joshi
° = 1 + 1 ° + 1 ° = ° + 1 ° = + 1 ° 1 =
Now, rearrange this equation so that it can be used to plug in for
°
:
Plug this into our derived solution to the design equation:
We can almost solve for the reaction rate constant through the design equation, but we still need to solve for the molar expansion factor . Refer to the equation from Fogler:
= The problem statement gives us that the initial mole fraction of A is 80%. So, we can evaluate εA:
= 0.8 1 = 0.812 1 = 0.4 0. 8 = 3 1 0.4+1 0.4 = . −
--Where "r" and "a" are the stoichiometric coefficients of species R and A, respectively
Now, we solve for the reaction rate constant directly through our modified design equation:
