(a)
Interpretation: The suitable operating conditions have to be recommended.
Concept Introduction:
Second order reaction: When the rate of a reaction is proportional to the concentration of two reacting molecules, then such a reaction is known as second order reaction
General representation of a second order reaction is as follows:
Rate law for second order reaction is given below:
(b)
Interpretation: The wrong happenings on choosing the given condition has to be found.
Concept Introduction:
Activity: It is the measure of effective concentration of a chemical species under non-ideal conditions.
(c)
Interpretation: The gas velocity for the given data has to be recommended.
Concept Introduction:
Gas velocity:
The velocity of a gaseous molecule is known as its gas velocity. The average velocity of all gas will be zero in a given sample. This is because gaseous molecules move in all random directions. So the velocity in one particular direction will be cancelled by the velocity of another gas molecule which is acting in the exactly opposite direction.
(d)
Interpretation: The corresponding conversion for the given data with respect to the chosen temperature has to be found.
Concept Introduction:
STTR: Spinning Tube-in-Tube Reactor: It is an area-based reactor system that has applications in petrochemical, biodiesel and pharmaceutical fields.
(e)
Interpretation: The appearance of temperature-time trajectory for the given CSTR has to be found.
Concept Introduction:
CSTR means Continuous stirred-tank reactor. It serves as a common model for a chemical reactor in chemical engineering. It is a mathematical model that works for all types of fluids, liquids, gases and slurries. Prefect mixing of substances can be achieved from this reactor by its action of continuous stirring.
Temperature-time trajectory: In the word temperature-time trajectory, the word trajectory literally means the path. So temperature-time trajectory means the curve that is described by the temperature and time co-ordinates.
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Chapter 10 Solutions
Elements of Chemical Reaction Engineering (5th Edition) (Prentice Hall International Series in the Physical and Chemical Engineering Sciences)
- A semi-truck tire is inflated to 110 psig with nitrogen. What will be the initial gas discharge ratein lbm/s due to a 1/16-inch diameter hole? Assume at temperature of 80℉ and an ambientpressure of 1 atm.arrow_forward# 4 The reaction, AB, is to be carried out isothermally in a continuous flow reactor. The entering volumetric flow rate, vo is 10 L/h and is constant (v=vo). Calculate both the CSTR and PFR volumes necessary to reduce the entering concentration of species A from CAD to CA = 0.01 CAO when the entering molar flow rate of species A is 5 mol/h. (a) This reaction is a second order reaction. The reaction rate constant, k is given as 300 L/mol.h. (b) This reaction is a zeroth order reaction. The reaction rate constant, k is given as 0.05 mol/h.L.arrow_forward#3 Using the initial rates method and the given experimental data below to determine the rate law and the value of the rate constant for the reaction, as shown below. All trials are performed at the same temperature. 2NO + Cl2 → 2NCOCI Trial [NO] (mol/L) [Cl₂] (mol/L) Initial rates (mol/L.s) 1 0.10 0.10 0.00300 2 0.10 0.15 0.00450 3 0.15 0.10 0.00675arrow_forward
- #2 The reaction rate constant at temperature, T₁, is 15 mol/L-s while at the reaction rate constant changed to 7 mol/L-s when temperature changed to T2 at 398 K. What is T₁? Given the activation energy is 600 kJ/mol. Assume at this temperature interval, pre-exponential factor and activation energy are constant.arrow_forward#1 Chloral is consumed at a rate of 10 mol/L-s when reacting with chlorobenzene to form DDT and water in the reaction given below. Determine: i) the rate of disappearance of chlorobenzene. ii) the rate of formation of DDT. CCI CHO (Chloral) + 2C6H5Cl (Chlorobenzene) → (C6H4Cl)2CHCCI 3 (DDT) + H2Oarrow_forward#5 The irreversible liquid phase second order reaction, 2A → B, is carried out in a CSTR. The entering concentration of A, CAD is 2 mol/L, and the exit concentration of A, CA is 0.1 mol/L. The volumetric flow rate, vo, is at 3 L/s and is constant (v=vo). The reaction rate constant, k is 0.03 L/mol's. What is the corresponding reactor volume?arrow_forward
- Problem 9.11 An 80 mm long line MN has its end M 15 mm in front of the V.P. The distance between the ends projector is 50 mm. The front view is parallel to and 20 mm above reference line. Draw the projections of the line and determine its inclination with the V.P. Also, locate the traces. Interpretation Front view of a line is parallel to xy, therefore, 1. The line is parallel to the H.P. 2. The top view of the line has true length. 3. The front view has projected length equal to the distance be- tween the projectors. Construction Refer to Fig. 9.11. 1. Draw a reference line xy. Mark point m' 20 mm above xy and point m 15 mm below xy. 2. Draw a 50 mm long line m'n' parallel to xy. 3. Draw an arc with centre m and radius 80 mm to meet projec- tor from point n' at point n. Join mn to represent the top view. Determine its inclination with xy as the inclination of line MN with the V.P. Here = 51°. 4. Traces Extend line mn to meet xy at point v. Project point v to meet m'n' produced at…arrow_forwardoh 30 20 D и D P 60 60 80arrow_forward⑤ b Δε m ab C 40arrow_forward
- Problem 10.16 An isosceles triangle of base 40 mm and altitude 54 mm has its base in the V.P. The surface of the plane is inclined at 50° to the V.P. and perpendicular to the H.P. Draw its projections. Construction Refer to Fig. 10.17. An isosceles triangle has its base in the V.P., so con- sider that initially the triangle ABC is placed in the V.P. with base AB perpendicular to the H.P. 1. First stage Draw a triangle a'b'c' keeping a'b' perpendicular to xy to represent the front view. Project the corners to xy and obtain ac as the top view. 2. Second stage Reproduce the top view of first stage keeping ab on xy and ac inclined at 50° to xy. Obtain new points a', b' and c' in the front view by joining the points of intersection of the vertical projectors from a, b and c of the second stage with the corresponding horizontal locus lines from a', b' and c' of the first stage. Join a'b'c' to represent the final front view. Here, the front view is an equilateral triangle of side 40 mm. X 54…arrow_forward%9..+ ۱:۱۹ X خطأ عذرا ، الرقم الذي أدخلته خاطئ. يرجى إدخال رقم بطاقة الشحن الصالحة والمحاولة مرة أخرى. رصيد هاتفك قم بمسح الرمز = رقم بطاقة التعبئة 7794839909080 رمز مكون من 13 او 14 رقماً طريقة إعادة التعبئة قم باعادة تعبئة الرصيد إعادة تعبئة الإنترنت إعادة تعبئة الرصيد O >arrow_forwardProblem 10.14 A hexagonal plane of side 30 mm has a corner in the V.P. The surface of the plane is inclined at 45° to the V.P. and perpendicular to the H.P. Draw its projections. Assume that the diagonal through the corner in the V.P. is parallel to the H.P. d' a 2 b b.f C' c.e b 'C' H.P. (a) V.P E HEX 30 e' O' d' a a' b' C' b' X y a b,f c,e d b,f (b) c,earrow_forward
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