Chemical Engineering (GATE & PSUs)
Postal Correspondence GATE GA TE & Public Sectors S ectors
Chemical Reaction Engineering
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Chemical Engineering (GATE & PSUs) GATE 2015 Cut-off Marks BRANCH
GENERAL SC/ST/PD
Chemical Engineering
27.52
OBC(Non-Creamy)
18.34
Total Appeared
24.77
15874
GATE GATE Syllabus Syllabus:: Chemical Engineerin Engineering g Chemical Reaction Engineering: Theories of reaction rates; kinetics of homogeneous reactions, reactions, interpretation interpretation of kinetic data, single and multiple reactions reactions in ideal reactors, reactors, non-ideal reactors; reactors; residence residence time distribution, single parameter model; non-isothermal reactors; kinetics of heterogeneous catalytic reactions; diffusion effects in catalysis.
CONTENT 1.
INT RODUCTION RODUCTION AND BASIC BASIC CONCEP CONCEPT T ……………………………………. 03-9
2 . K I NE N E T I CS CS O F R E A C T I ON ON
……………………………………………………….
1 0 -1 -1 7
3.
DESIGN DESIGN EQUATION EQUATION FOR FOR IDEAL IDEAL REACTO REACTORS RS ……………………………… 18-31
4.
INT ERPR ERPRETATION ETATION OF BATCH BATCH REACTOR REACTOR DATA DATA …… …………………………. 32-44
5.
DESIGN FOR FOR SINGLE REAC REACTI TI ON……… ON……………………………………………… 45-66
6.
D ES ES I GN GN F OR OR M U LT LT I P L E R EA EA CT CT I O N S
7.
H E T ER ER OG OG EN EN EO E O U S R EA EA CT CT I O N
………………………………………………
8 3 -8 -8 7
8.
T E M PE PE RA RA T U RE RE AN AN D PR PR ES ES SU SU R E E FF FF EC EC T S ………………………………
8 8 -9 -9 7
9.
BASIC BASIC ASPE ASPECTS CTS OF NON IDEAL IDEAL FLOW FLOW… …………………………………….
98-108
1 0 . Q U ES ES T I O N BA BA SE SE D O N GA GA T E PA PA PE PE R
……………………………………… 6 7 -8 -8 2
……………………………………. . 1 0 9 -1 -1 2 0
1 1 . M O DE DE L T E ST ST P A PE PE RR-I W I T H S OL OL U T I O N …………………………………
1 2 1 -1 -1 3 3
12. PRACTICE PRACTICE SET-1, 2 , 3……… 3……………………………………………………………
134 -171
13 . Int erview Pract ice Bank ……… ……………… ……………… ……………… ……………… ………………… ……………… ……… 17 2-182
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Chemical Engineering (GATE & PSUs)
CHAPTER-1 IN TRODUCTI ON AND BASIC CONCEPT
1. Every industrial chemical process is designed to produce economically a designed product from a variety of starting material through a succession of treatment steps.
Figure: Show typical chemical process:
2. Raw material undergoes a number of physical steps to put them in the form in which they can be reacted chemically. Then material passes through the reactor. The products of reaction must then undergo further treatment. 3. In chemical reaction engineering subject we are concerned with chemical treatment step of a process. CRE
combines the study of chemical kinetics (reaction rates and reaction mechanisms) with the reactors in which the reaction occurs.
A chemical reaction takes place by decomposition, combination or change in configuration of molecules.
4. We define rate of reaction in terms of component. If the rate of change in number of moles of this component due to reaction is r i
1 d Ni V dt
d Ni dt
then
Moles of i formed (Volume of fluid) (Time) 3
i.e., the number of moles of component i formed per unit time per unit volume (mol/dm s). If species i is a 3
3
reactant, the numerical value of ri is a negative number (e.g. ri = -3mol/dm s or – ri = 3mol/dm s) If species i is a product, then ri= will be a positive number.
r i
1 dN i V dt
For constant volume system
ri
d N i
d C dC i ( i ) dt dt V dt
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Chemical Engineering (GATE & PSUs) Ci is the concentration of species i in reaction mixture. Q.1. For a homogeneous reaction system, where Ci is the concentration of i at time t, Ni is the no. of moles of i at time t, V is the reaction volume at time t and t is the reaction time The rate of reaction for species i is defined as (a.)
dC i
dC i dt
(b.)
dt
(c.)
1 dN i V dt
(d.)
1 V
dN i dt
Ans. (c) 5. (i) Chemical reaction may be classified depending upon no. of phases involved as homogeneous and heterogeneous reaction. Homogeneous reaction is one which takes place in only one phase. A heterogeneous reaction is one which involves presence of more than one phases. (ii) Reaction may be classified according to catalytic and non catalytic reaction. (iii) Reaction may be classified on the basis of molecularity of reaction. (iv) The reaction may be classified based upon the heating effect as exothermic and endothermic reaction. (v) Reaction may be classified as reversible and irreversible reaction. (vi) Reaction may be classified based upon the order of reaction.
6. Relative rates of reaction:
For the reaction:
aA bB cC dD The relation between the rate of participants A, B, C, D in chemical reaction with their stoichiometric coefficient is
r A a
rB b
r C c
r D d
Also, if we compare
r A
a c
r A a
r C c
r c a
Rate of disappearance of A =
c
(Rate of formation of C)
Q.2. For an elementary reaction k
3C A 2B (a.) Rate of appearance of C is equal to rate of disappearance of A (b.) Rate of disappearance of A is equal to rate of disappearance of B. (c.) Rate of appearance of C is equal to rate of disappearance of B. (d.) Rate of appearance of C is 3 times rate of disappearance of A. Solution:
r A 1
r B 2
r C Relative Rates
3 of reaction
r A = rate of disappearance of A Postal Course ( GATE & PSUs) © 2015 ENGINEERS INSTITUTE OF INDIA® . 28-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016. Ph. 011-26514888.
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Chemical Engineering (GATE & PSUs) r B = rate of disappearance of B rC = Rate of appearance of C
rC 3(r A ) Option (D) is correct
7. Many variable may affect the rate of a chemical reaction. In homogeneous system the temperature, pressure and composition affect the rate of reaction. In heterogeneous system factors that affect the rate of reaction are rate of mass transfer i.e., heat transfer nature of catalyst if present. Heterogeneous reactions involve more than one phase.
Rate of reaction is expressed in measure other than volume such as reaction surface area or catalyst weight. r i'
Number of moles of i formed per unit time per unit mass of catalyst (mol/s.g catalyst0
8. Rate law
An algebraic equation that relates the reaction rate and concentration
A B PRODUCTS
-rA
k A (T ) f n ( CA, C B ) or
r A k A C AC B β k A Rate constant/ specific reaction rate Order of reaction (n) = + For the reaction where all the stoichiometric coefficients are 1 e.g. A + B C + D
k A
kB kC kD k
Imp. Rate Laws Are determined by experimental observation.
Q.3. Rate expression of reaction
r A 0.02C A2 mol / cm3 min If concentration is to be expressed in mol/l and time in hours, the value of rate constant would be -1
-1
(a.) 0.12hr (mol/L) -1
-1
(c.) 12hr (mol/L)
-1
-1
(b.) 0.012hr (mol/L) -1
-1
(d.) 1.2hr (mol/L)
Solution: 1L = 10cm3 1hr = 60min
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Chemical Engineering (GATE & PSUs) mol
r A 0.02C A2
2
1 1 hr 10 60
r A 0.02C A2 600 r A 12CA2 mol
mol L hr
L1hr 1
Clearly k 12 unity of k
r A C A2
molL1hr 1 mol 2 L2
unit o f k mol 1 L hr 1 k
12hr 1 (mol / L)1
Option C is correct
Q.4. The rate of reaction for a gas phase reaction is expressed as
dp A dt
2.56 P A2atm / hr
The Units of rate constant are (a.) hr 1 ( atm)1
Solution: k
(b.) hr 1 ( atm)3
dp A / dt p A2
atm / hr 2
atm
1 (c.) hr atm
(d.) hr 1 ( mol / L) 1
atm1hr 1 TYPES OF REACTIONS
9. (i) A homogeneous reaction is one which takes place in only one phase; i.e., all the reacting material and products and catalyst (if any) will be present in a similar phases. Example: NO( g )
1 2
O2 ( g ) NO2 (g )
A hetrogeneous reaction is one which involves the presence of more than one phase.
Example:
SO 2 ( g )
1 2
2 5 O2 ( g ) SO3 (g )
V O (s )
(ii) Catalytic reaction are those reaction which involves use of catalyst to enhance the rate of reaction.
Example:
C2 H 4 Ethylene
Ni H 2 C 2 H6 Heat Ethane
Non catalytic reactions are those reaction which does not involve use of catalyst. Example:
NO
1 2
O2
NO2
(iii) The reaction may be classified based upon the molecularity of a reaction i.e., based upon the number of molecules that takes part in the reaction as unimolecular, bimolecular and trimolecular reaction.
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Chemical Engineering (GATE & PSUs) (a) Unimolecular Reaction:
H 2C — CH 2 Decomposition | | 2C2 H 4 H 2C — CH 2
Example:
(b) Bimolecular Reaction: Decomposition 2HI( g ) I2 ( g ) H 2 ( g )
(c) Trimolecular Reaction:
2 NO O2
Oxidation 2 NO2
(iv) The reaction may be classified as exothermic and endothermic reaction. Exothermic reaction is one in which heat is evolved. CO 2 H 2
Example:
Cu CH 3 OH Heat
Endothermic reaction is one in which heat is absorbed 2 3 C2 H 5OH C2 H 4 Dehydration
Al O
H 2O Heat
ethylalcohol
(v) The reaction may be classified as reversible and irreversible reaction. Reversible reaction are those in which forward and reverse reaction takes place simultaneously. H
CH COOC H C2 H 5OH CH3COOH 3 2 5
H2 O
Irreversible reaction are those which can proceed only in one direction. C6 H 6
HNO3
C6 H 5 NO 2
H 2O
(vi) The reaction may be classified as first order, second order and third order reaction.
First order reaction: Example: N 2 O5 NO2 + Decomposition
1 2
O2
Second order reaction: (Saponification of ester) Example:
CH 3COOC 2H 5
NaOH
CH 3COONa
C 2H 5OH
Third order reaction: Example:
2 NO H 2
N 2 O H 2O
Single and Multiple Reaction: (i) When a single stoichiometric equation and single rate e quation are chosen to represent the progress of the reaction, we have a single reaction. When more than one stoichiometric equation is chosen to observed changes then more than one kinetic expression is needed to follow the changes in the concentration of all t he reaction component and we have multiple reaction.
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Chemical Engineering (GATE & PSUs) (ii) Multiple reaction may be classified as A B S
Series Reaction:
Parallel Reaction: B
and
A S
A B
C S Competitive Side by side Type Complicated Reaction: AB C CB D Here reaction proceeds in parallel with respect to B but in series with respect to A, C and D.
Elementary and Non Elementary Reaction: Consider a single reaction with stoichiometric equation A B C According to rate law, Rate of disappearance of A is given by
r A K C A C B Such reaction in which the rate equation corresponds to a stoichiometric equation are called elementary reaction. When there is no direct corresponding rate and stoichiometry, then the reaction is called non-elementary reaction. The best example of a non-elementary reaction is that between hydrogen and bromine. H2
Br2 2 HBr
Which has rate expression 1
r HBr
K1[H 2 ] [Br2 ] 2 [HBr] K2 + [Br2 ]
10. Molecularity and Order of Reaction The molecularity of an elementary reaction is the number of molecules involved in the reaction and this has been found to have the values of one, two or occasionally three. Molecularity refers only to an elementary reaction. Consider a reaction a A b B c C d D
P
then rate expression
r A K CAa CbB CcC CdD
a+b+c+d=n
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Chemical Engineering (GATE & PSUs) where a, b, …….. d are not necessarily related to the stoichiometric coefficient. We call the power to which the concentration are raised the order of reaction. Hence the reaction is ath order with respect to A bth order with respect to B nth order overall
Since the order refers to the empirically found rate expression, it may not to be an integer and can have a fractional value. But, the molecularity of a reaction must be an integer because it refers to mechanism of reaction and applies only to elementary reaction.
11. Rate constant (K) The rate constant of a chemical reaction is a measure of the rate of reaction when all the reactants are at unit concentration. So, for any reaction say A B
r A K C A If we assume that the concentration are equal to unity then
r A K Where K is rate constant. The rate constant K depends on the temperature. It increasees with temperature. K is also called specific rate constant. The units of K depends upon the units of time and concentration.
r A K CAn
Suppose :
Then we know units of ( r A ) mol/l. sec and
CA
mol lit.
mol
K
1 n sec. 1 mol lit. (sec) n lit. mol
lit.
K (time) 1 (concentration)1 n
Where n is the order of reaction.
Q.5. For a gas reaction at T(K), the rate is given by
dp A dt
k ' p A2
atm / hr
If the rate equation is expressed as
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Chemical Engineering (GATE & PSUs) r A
1 dN A V
dt
kC A2 (mol / L hr )
The rate constant k is given by (a.) k = k’rT
(c.) k = k’ (RT) 2
(b.) k = k’
(d.) k = k’/RT
Solution:
dp A dt
d dt
cA
pA RT
(c A RT ) k ' c 2A ( RT ) 2
RT
k ' p A2
dc A
dc A dt
dt
k ' c A2 ( RT )2
k ' c A2 ( RT )
r A k ' cA2 ( RT ) 2
kc A k
k ' c2A ( RT )
k '( RT )
Option (A) is correct
FORWARD
B C 12. For reversible reactions: A
BACKWARD
Equilibrium constant (k C) =
k A k A
k A Rate constant of forward reaction k -A Rate constant of backward reaction
During the course of a chemical reaction, the rate of a reaction decreases as the reaction proceeds
Q.6. The irreversible reaction is the special case of reversible reaction if Concentration of reactant at equilibrium conditi on is ___________ Solution: Zero
Conceptual Numerical Answer Practice Set
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Chemical Engineering (GATE & PSUs) CHAPTER-2 KI NET ICS OF REACTI ON
1. In homogeneous system, the possible factors that affect the rate of reaction are temperature, composition and pressure. Hence for homogeneous system we can write rate of reaction as a function of concentration and temperature.
rA
f [Temperature dependent term, Concentration dependent term]
mol l. s
2. According to rate law rate of reaction is directly proportional to the reactant concentration. Rate law is also called rate equation, kinetic expression.
3. Consider a irreversible reaction a A b B .......
rR
s S ..........
According to rate law we can write
rA K f (CA , CB ) K CAa CbB .......... Where K is rate constant a and b are the power of concentration of reactant A, B ….. on which rate of reaction depends. 4. Kinetics Model for Non-Elementary Reactions: To explain the kinetics of non elementary reaction we assume that a sequence of elementary reaction is actually occurring but we can not observe the intermediate formed because they are only present in very minute quantity. Hence we observe only initial reactant and final products that appears to be a single reaction. For example:
A2
D 2 2 AD
If the kinetic of reaction shows that reaction is non-elementary then we can postulate a series of elementary steps. A2
A* D 2
2 A*
AD D*
AD A* D*
where * sign shows unobserved intermediate.
Types of Intermediate: (i) Free radicals: Free atoms that contain one or more unpaired electrons are called free radicals. Postal Course ( GATE & PSUs) © 2015 ENGINEERS INSTITUTE OF INDIA® . 28-B/7, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016. Ph. 011-26514888.
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Chemical Engineering (GATE & PSUs) Example: CH3 , C2 H5 , I , H , CCl3 They are highly reactive and unstable. (ii) Ions and Polar Substance: Electrically charged atoms, molecules or fragments of molecules are called ions. Example: N3 , Na , OH , I
(iii) Molecules: Consider a consecutive reaction A B C If the intermediate B is highly reactive its mean life time will be very small and its concentration in the reacting mixture can become too small to measure. In this condition B may not be observed and can be considered to be a reactive intermediate.
(iv) Transition Complex: Unstable forms of molecule or unstable association of molecules which can either decompose to give product or by further collision return to molecules in the normal state. Such unstable forms are called transition complex. Reaction scheme involving these four kinds of intermediate can be of two types: ( a) Non chain reaction: Reactants (Intermediates)* (intermediates) *
Products
( b) Chain Reaction: Reactant (Intermediate) * (Intermediate) * Reactant (Intermediate) *
[Initiation]
Product
(Intermediate) * Product
[Propagation] [Termination]
Testing Kinetics Models: Now we will see that how to test correspondence between experiment and a proposed mechanism. In matching the predicted rate expression with the experimentally found, we rely on the following rule: (i) If any component: Takes part in more than one reaction then its net rate of change is equal to sum of the rates of change of components i in each of the elementary reaction in which it is participated. ri , net
r i = All elementary reaction participated.
(ii) As the intermediates are highly reactive they have very short life time and they are present in very small quantities, hence their net rate of formation is taken as zero.
*
r A
0
This is called the steady state approximation.
5. Temperature Dependent Term of a Rate Equation
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Chemical Engineering (GATE & PSUs) For many reactions and particularly elementary reacti ons, the rate expression can be written as a product of a temperature dependent term and a concentration dependent term or ri
ri
K
f1 (temperature) f 2 (concentration) f 2 (concentration)
For such reactions the temperature dependent term, the reaction rate constant has been found in practically all cases. (a) Temperature dependency from Arrhenius law: From Arrhenius law rate constant K is given by: K K o e
E RT
Where K o is called the frequency or pre-exponential factor. E is called activation energy of the reaction. R = gas constant T = absolute temperature, K At the same concentration, but at two different temperature, Arrhenius law indicate that
ln
r 2 r 1
ln
K2 K1
E 1
1 R T1 T2
t 1 E 1 1 t 2 R T1 T2
ln
Unit of Ko = unit of K Ko is not a function of temperature Ko (also known as frequency factor) does not affect the temperature sensitivity of reaction
K (T 2 ) H R 1 1 K T1 R T1 T2
Also ln
x
Here K(T2) and K (T1) are equilibrium constant at T1 & T2 respectively & HRX is the heat of reaction Also
d dT
(ln K )
E RT 2
If temperature is low change in K w.r.t T is high and vice versa. Q.1. From Arrhenius law, a plot of ln k versus 1/T gives a straight line with a slope of (-E/R). The units of E/R are (a.) K/cal Solution:
E R
(b.) cal/K cal mol 1
cal mol 1 K 1
(c.) cal
(d.) K
K
Option (d) is correct
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Chemical Engineering (GATE & PSUs) k
B an increase in temperature from 295K to 305K causes the rate of Q.2. From the chemical reaction A reaction to double. The energy of activation for this reaction must be (a.) 12389Kcal/mol
(b.) 12.38Kcal/mol
(c.) 1.2Kcal/mol
(d.) 1238Kcal/mol
Solution:
1 r1 R T T2 2r E 1 1 ln 1 r 1 1.987 295 305 ln
r 2
R 1.987cal mol 1 K 1
E 1
r2
2r 1
On solving E = 12389.47 cal/mol Or E = 12.38Kcal/mol
Option (b) is correct
Figure: Sketch showing temperature dependency of the reaction rate. Activation Energy and Temperature Dependency: The temperature dependency of reaction is determined by the activation energy and temperature level of the reaction as shown in figure. Conclusion from figure are as follows: (i) From the Arrhenius law, the value of the frequency factor K o does not affect the temperature sensitivity. (ii) Any given reaction is much more temperature sensitive at a low temperature than at a high temperature. (iii) Reactions with high activation energy are very temperature sensitive, reaction with low activation energy are relatively temperature insensitive. (iv) From Arrhenius law a plot of ln K vs
1 T
given a straight line with large slope for large E and small slope
for small E.
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Chemical Engineering (GATE & PSUs) Activation Energy:
The excess energy of the reactants required to dissociate into products.
In Exothermic reaction, reactants have more energy than products
H Reaction Energy of product - Energy of Reactant H Reaction ve (for exothermic reactions)
(b) Temperature Dependency from the Collision Theory: Collision theory is based upon the concept that before molecule react, there must be collision between the reactant molecules and only those collision i n which the colliding molecules have a certain minimum amount of energy are effective. Further this concept leads to the rate expression based upon the frequency of molecular collisions and the fraction collisions that are effective. From collision theory rate constant K is given by 1
KT e 2
E RT
k2 E 1 1 1 T 2 ln k1 R T1 T2 2 T1
ln
(c) Temperature Dependency from Transition State Theory: The fundamental postulate of the theory is that ( i) The reacting molecules must form unstable intermediate called activated complex before being converted (i.e., there is formation of activated complex) to product. (ii) There exists an equilibrium between the activated complex and reactants at all the time. According to transition state theory rate constant is given by:
K T e
E RT
or
k2 E 1 1 T 2 ln k1 R T1 T2 T1
ln
This equation describe the temperature dependency from Transition State Theory.
6. Difference between Collision Theory and Transition State Theory:
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Chemical Engineering (GATE & PSUs) Collision Theory
Transition State Theory
1. Collision theory does not agree more closely with
1. Transition state theory agrees more closely with
experiments.
experiments.
2. Collision theory is based on Kinetic theory of 2. Transition state theory is based on statistical gases.
mechanics.
3. Collision theory view that decomposition of 3. Transition state theory view that formation of activated complex is very rapid.
activated complex is very rapid.
4. Collision theory view that formation of activated
4. Transition state theory view that decomposition
complex is slow and rate controlling.
of activated complex is slow and rate controlling.
1
5. K T 2 e
E
5. K T e
RT
E RT
SOLVED EXAMPLE 1. The decomposition of N 2 O5 is postulated to occur by the following mechanism.
N 2 O5
K1 K2
NO2
NO*3
3 NO*3 NO* O 2
K
NO* NO3 *
K 2 NO2 4
using steady state approximation derive an expression for rate of decomposition of N 2 O 5 . Solution: We know that Rate of decomposition is given by
d [N 2O5 ] dt
K1[N 2O5 ] K 2 [NO2 ] [NO*3 ]
…(a)
By steady state hypothesis. d [NO*] dt d [NO*3 ]
K 3 [NO*3 ] K 4 [NO*] [NO*3 ] 0
…(i)
K1[N 2O5 ] K 2 [NO2 ] [NO*3 ] K3 [NO*3 ] K 4 [NO*] [NO*3 ]
…(ii)
dt From equation (i)
[NO*]
K 3 [NO*3 ] * 3
K 4 [NO ]
K3
…(iii)
K4
Put equation (iii) in equation (ii)
K1[N 2 O5 ] K 2 [NO2 ] [NO*3 ] K 3 [NO*3 ] K3 [NO3* ]
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Chemical Engineering (GATE & PSUs)
[NO*3 ]
K1[N 2 O5 ]
…(iv)
K 2 [NO 2 ] 2K 3
Put equation (iv) in equation (a) We get,
d [N 2 O5 ] dt
K1[N 2 O 5 ] K1[N 2O 5 ] K 2 [NO 2 ] K 2 [NO 2 ] 2 K 3 K 2 [NO 2 ] K1[N 2O 5 ] 1 K 2 [NO 2 ] 2 K 3
2. At 500 K the rate of a bimolecular reaction is ten times the rate at 400 K. Find the activati on energy for this reaction. (a) From Arrhenius law
(b) From Collision theory
Solution: ( a) We have from Arrhenius law
ln(r ) r 2
ln
r 1
E RT
ln Ko
1 1 R T2 T1 E
r 2 E 1 1 R T1 T2 r 1
ln
We have,
r2
10 r 1
T1
400 K
T2
500 K
R = 1.987 cal/mol K
10 r 1 E 1 1 r 1 1.987 400 500
ln
E = 9150 cal/mol K ( b) From Collision theory we know that 1
KT e 2
E RT
1
r
ln r
K o
E1
T2 e
E RT
1
ln T 2 ln K o
RT
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Chemical Engineering (GATE & PSUs) E 1
ln r 1 ln T1 ln K o R T1 1
2
E 1
ln r 2 ln T2 ln K o R T2 1
2
r 2 E 1 T2 1 ln ln 1 r 1 R T1 T2 T12 1
2
T1 = 400 K, r2
T2 = 500 K
10 r 1 1 2 (500) 1.987 400 500 ln 1 (400) 2
10 r 1
ln
r 1
E
1
E = 8707 cal/mol
1
Ans.
3. On doubling the concentration reactant rate of reaction triples. Find the reaction order. Solution:
r A K CAn
Let
Suppose at concentration CA1 rate of reaction is r A1 At concentration C A2 rate of reaction is If
2CA
C A2
1
r A K C nA 2
2
we have
1
rA 3 (r A 2
K C nA
1
)
r A K C nA r A K C nA 2
2
1
1
r A C nA r A C nA 2
2
1
1
3(r A1 )
r A
1
(2 C A1 ) n (C A1 ) n
3 2n
n 1.58 1.6
ln 3 = n ln 2
4. A certain reaction has a rate given by
r A 0.005 C A2
mol/cm3 min
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Chemical Engineering (GATE & PSUs) If the concentration is expressed in mole/lit and time in hours. What would be the value and units of rate constants.
r A 0.005 CA2 , mol/cm3 min
Solution:
cm3 2 mol 2 mol r A 0.005 CA 3 , 3 mol . min cm cm min K = 0.005
cm3 mol. min
3 We know that 1 liter = 1000 cm
1 cm 3
1 l 1000
60 min = 1h 1 min =
1 60
h
1 lit 1000 cm3 1 h mol. min 60 min
(0.005 cm3 ) K=
lit mol. h
K = 3 10 4 Value of K = 3 10
4
Units of K =
lit mol h
5. The pyrolysis of ethane proceeds with an activation energy of about 75000 cal. How much faster is the decomposition at 650ºC than at 500ºC ? C 2 H 6 P
Solution:
E = 75000 cal/mol Let K1 be the rate constant at T1 and K 2 be the rate constant at T2 T1
500º C 773K
T2
650º C 923K
K2 E 1 1 K1 R T1 T2
ln
R = 1.987 cal/mol K
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Chemical Engineering (GATE & PSUs) K 2 75000 1 1 K1 1.987 773 923
ln
K 2 7.951 K1
ln
K2 K1 K2
2838 2838 K1
At 650ºC, the decomposition is faster by a factor 2838 than at 500ºC.
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