TUTORIAL 3 SOLUTIONS
Lecturer: Miss Anis Atikah Ahmad
Tel: +604 976 3245
Email: anisatikah@unimap.edu.my
QUESTIONS
1.
Write the rate laws for the following reactions assuming each
reaction follows an elementary rate laws.
a) C2H6 → C2H4 + H2
b) (CH3)3COOC(CH3)3 ⇌ C2H6 + 2CH3COCH3
2.
Write the rate law for the reaction:
2A + B → C
if the reaction
a) is second order in B and overall third order,
b) is zero order in A and first order in B
c) is zero order in both A and B
d) is first order in A and overall zero order
3.
The formation of ortho-nitroaniline is formed from the reaction
of ortho-nitrochlorobenzene (ONCB) and aqueous ammonia.
NO 2
NO2
Cl
NH 2
+ 2NH3
+ NH4Cl
The liquid-phase reaction is first order in both ONCB and
ammonia with k= 0.0017 m3/kmol-min at 188°C with E = 11,273
cal/mol. The initial entering concentration of ONCB and
ammonia are 1.8 kmol/m3 and 6.6 kmol/m3 respectively.
a)
b)
c)
d)
e)
f)
g)
Write the rate law for the rate of disappearance of ONCB in terms of
concentration.
Set up stoichiometric table for this reaction for a flow system.
Explain how part (a) and (b) would be different for a batch system.
Write –rA solely as a function of conversion.
What is the initial rate of reaction (X=0) at 188°C and at 25°C?
What is the rate of reaction when X = 0.9 at 188°C and at 25°C?
What would be the corresponding CSTR volume at 25°C to achieve 90%
conversion at 188°C for a feed rate of 2 dm3/min
QUESTION (1)
Write the rate laws for the following reactions
assuming each reaction follows an
elementary rate laws.
a) C2H6 → C2H4 + H2
b) (CH3)3COOC(CH3)3 ⇌ C2H6 + 2CH3COCH3
(a) C2H6 → C2H4 + H2
A →
B + C
 rA  kCA
(b) (CH3)3COOC(CH3)3 ⇌ C2H6 + 2CH3COCH3
A
⇌
B

CBCC2 
 rA  k C A 

KC 

+
2C
QUESTION (2)
(a)  rA  kCACB
2
(b)  rA  kCB
(c)  rA  k
(d)  rA  kCACB1
Write the rate law for the reaction:
2A + B → C
if the reaction
a) is second order in B and overall third order
b) is zero order in A and first order in B
c) is zero order in both A and B
d) is first order in A and overall zero order
QUESTION (3)(A)
The formation of ortho-nitroaniline is formed from the reaction
of ortho-nitrochlorobenzene (ONCB) and aqueous ammonia.
NO 2
NO2
Cl
NH 2
+ 2NH3
+ NH4Cl
The liquid-phase reaction is first order in both ONCB and
ammonia with k= 0.0017 m3/kmol-min at 188°C with E = 11,273
cal/mol. The initial entering concentration of ONCB and
ammonia are 1.8 kmol/m3 and 6.6 kmol/m3 respectively.
(a)
Write the rate law for the rate of disappearance of ONCB in terms of
concentration.
Let A = ONCB, B = NH3, C = Nitroaniline , D = NH4Cl
 rA  kCACB
QUESTION (3)(B)
A + 2B → C + D
b)
Set up stoichiometric table for this reaction for a flow system
Species
A
Entering
FA0
Change
 FA0 X
Leaving
FA0 1  X 
FB  FB 0  2 FA0 X
B
FB 0
 2 FA0 X
F

 FA0  B 0  2 X 
 FA0

 FA0  B  2 X 
C
0
FA0 X
FC  FA0 X
D
0
FA0 X
FD  FA0 X
QUESTION (3)(C)
c)
Explain how part (a) and (b) would be different for a batch
system
For batch system,
NA
CA 
V
N A  NB 
 rA  k


V V 
 kNA N B /V 2
QUESTION (3)(D)
d)
Write –rA solely as a function of conversion.
 rA  kCACB
FA


F
1

X
A
0
CA 



For liquid phase rxn, υ = υ0
FA0 1  X 
CA 
0
 C A0 1  X 
 rA  kC 1  X  B  2 X 
2
A0
FB FA0  B  2 X 
CB 



FA0  B  2 X 
CB 
0
 C A0  B  2 X 
QUESTION (3)(D)
 rA  kCA2 0 1  X  B  2 X 
Substituting the concentration of A & B;
C A0  1.8kmol / m3
C B 0  6.6kmol / m 3
CB0
B 
C A0
 6.6

 rA  k 1.8 1  X 
 2X 
 1.8

2
 3.24k 1  X 3.67  2 X 
QUESTION (3)(E)
e)
What is the initial rate of reaction (X=0) at 188°C and at 25°C
 rA  3.24k 1  X 3.67  2 X  ---(1)
i) At T= 188°C, k =0.017m3/kmol-min
Substituting X=0 and k =0.017m3/kmol-min into (1);


 rA  3.24 0.0017m / kmol  min 1  03.67  2  0
3
 0.0202kmol / m 3  min
QUESTION (3)(E)
e)
What is the initial rate of reaction (X=0) at 188°C and at 25°C
 rA  3.24k 1  X 3.67  2 X  ---(1)
ii) At T= 25°C (298.15 K), k =? m3/kmol-min
Find k at T =25°C first
k at initial T is
k T0   Ae  E / RT0
k at any temperature is
k T   Ae E / RT
Taking the ratio;
k T   k T0 e
E 1 1 
  
R  T0 T 
11273  1
1 
 0.0017m3kmol  min exp 



 1.987  461.15 298.15 
 2.039 x10 6 m3kmol  min
QUESTION (3)(E)
e)
What is the initial rate of reaction (X=0) at 188°C and at 25°C
 rA  3.24k 1  X 3.67  2 X  ---(1)
ii) Now we know that, at T= 25°C (298.15 K),
k = 2.039 x 10-6 m3/kmol-min
Therefore, we can calculate –rA at 25° by susbtituting
k= 2.039 x 10-6 m3/kmol-min, and X = 0 in eq (1).


 rA  3.24 2.039 x106 1  03.67  2  0
 2.42 x10 5 kmol / m3  min
QUESTION (3)(F)
f)
What is the rate of reaction when X = 0.9 at 188°C and at 25°C?
 rA  3.24k 1  X 3.67  2 X  ---(1)
(i) At T= 188°C, k =0.0017m3/kmol-min
Substituting X=0.9 and k =0.017m3/kmol-min into (1);
 rA  3.240.00171  0.93.67  20.9
 1.03x10 3 kmol / m3  min
QUESTION (3)(F)
f)
What is the rate of reaction when X = 0.9 at 188°C and at 25°C?
 rA  3.24k 1  X 3.67  2 X  ---(1)
(ii) From part (e) when T= 25°C, k = 2.039 x 10-6 m3/kmol-min
Substituting X=0.9 and k = k = 2.039 x 10-6 m3/kmol-min into (1);


 rA  3.24 2.039 x106 1  0.93.67  20.9
 1.235 x10 6 kmol / m3  min
QUESTION (3)(G)
g.
What would be the corresponding CSTR volume at 25°C to achieve
90% conversion at 188°C for a feed rate of 2 dm3/min?
VCSTR
FA0 X

 rA
C A0 0 X

 rA
Substituting the value of CA0, υ0 and –rA(at 288°C & X=0.9);
VCSTR


1.8kmol / m3 2dm3 / min 0.9

1.03x10 3 kmol / m3  min
 3,145.6m3