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2. Kinetic experiments have been carried out in a batch reactor in the laboratory at two different temperatures, 25°C and 65°C. It was found that 58.1% of conversion was achieved within 1 minute @ 65°C and 60% conversion was achieved in 10 minutes at 25°C. Calculate the values of EA and ko for the forward and backward rate constants knowing that equilibrium conversions are 89% and 98% at 65°C and 25°C, respectively.

2. Kinetic experiments have been carried out in a batch reactor in the laboratory at two different temperatures, 25°C and 65°C. It was found that 58.1% of conversion was achieved within 1 minute @ 65°C and 60% conversion was achieved in 10 minutes at 25°C. Calculate the values of EA and ko for the forward and backward rate constants knowing that equilibrium conversions are 89% and 98% at 65°C and 25°C, respectively.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Integrate the Performance Equation. For a reversible first-order reaction, the performance equation for a batch reactor is dX 1=CM XACM RCA-K₂Cgk₁ 1-X₁/XM 1x, dX, = TA According to Eq. 3.54 this integrates to give Xx--In (1-XA) (1) Calculate the Forward Rate Constant. From the batch run at 65°C, noting from Example 9.2 that X 0.89, we find with Eq. (i) or k, (1 min) 0.89 (1- k₁-0.942 min¹ gol beSimilarly, for the batch run at 25°C we find 0.581 0.89 K₁2=0.0909 min¹ 218 Chapter 9 Temperature and Pressure Effects Assuming an Arrhenius temperature dependency, the ratio of forward rate con- stants at these two temperatures gives k₁, 0.942 k = 20.0909 KR from which the activation energy of the forward reaction can be evaluated, giving E, 48 900 J/mol AR: k, 34 x 10 exp Note that there are two activation energies for this reaction, one for the forward reaction, another for the reverse. (ii) Now for the complete rate constant for the forward reaction. From either the numerator or the denominator of Eq. (iv) first evaluate kethen use it as shown below: Equilibrium: (iii) -48 900 Noting that K = , thus k₂= k₂ we can find the value of k₂. Summary. For the reversible first-order reaction of Example 9.2 we have K= -exp 17.2- K = exp where K is given by Eq. (iii) of Example 9.2, -CM-TA-FR = k₁C₂-K₂CR 75 300 RT 48 900 RT -24.7 Rate constants: k, -exp 17.2-48 900 R7 -exp[41.9 (iv) 123 800 RT min-1 min-1
Transcribed Image Text:Integrate the Performance Equation. For a reversible first-order reaction, the performance equation for a batch reactor is dX 1=CM XACM RCA-K₂Cgk₁ 1-X₁/XM 1x, dX, = TA According to Eq. 3.54 this integrates to give Xx--In (1-XA) (1) Calculate the Forward Rate Constant. From the batch run at 65°C, noting from Example 9.2 that X 0.89, we find with Eq. (i) or k, (1 min) 0.89 (1- k₁-0.942 min¹ gol beSimilarly, for the batch run at 25°C we find 0.581 0.89 K₁2=0.0909 min¹ 218 Chapter 9 Temperature and Pressure Effects Assuming an Arrhenius temperature dependency, the ratio of forward rate con- stants at these two temperatures gives k₁, 0.942 k = 20.0909 KR from which the activation energy of the forward reaction can be evaluated, giving E, 48 900 J/mol AR: k, 34 x 10 exp Note that there are two activation energies for this reaction, one for the forward reaction, another for the reverse. (ii) Now for the complete rate constant for the forward reaction. From either the numerator or the denominator of Eq. (iv) first evaluate kethen use it as shown below: Equilibrium: (iii) -48 900 Noting that K = , thus k₂= k₂ we can find the value of k₂. Summary. For the reversible first-order reaction of Example 9.2 we have K= -exp 17.2- K = exp where K is given by Eq. (iii) of Example 9.2, -CM-TA-FR = k₁C₂-K₂CR 75 300 RT 48 900 RT -24.7 Rate constants: k, -exp 17.2-48 900 R7 -exp[41.9 (iv) 123 800 RT min-1 min-1
2. Kinetic experiments have been carried out in a batch reactor in the laboratory at two different temperatures, 25°C and 65°C. It was found that 58.1% of conversion was achieved within 1 minute @ 65°C and 60% conversion was achieved in 10 minutes at 25°C. Calculate the values of EA and ko for the forward and backward rate constants knowing that equilibrium conversions are 89% and 98% at 65°C and 25°C, respectively. Clues for Solution: 1. You will need to use the Arrhenius expressions for the forward and backward reactions at the two temperatures. 2. Use the equation for-ra for the rate of disappearance of A using forward and backward reactions. Interim answer: EA for the forward reaction is approx. 48 kJ mol¹¹.
Transcribed Image Text:2. Kinetic experiments have been carried out in a batch reactor in the laboratory at two different temperatures, 25°C and 65°C. It was found that 58.1% of conversion was achieved within 1 minute @ 65°C and 60% conversion was achieved in 10 minutes at 25°C. Calculate the values of EA and ko for the forward and backward rate constants knowing that equilibrium conversions are 89% and 98% at 65°C and 25°C, respectively. Clues for Solution: 1. You will need to use the Arrhenius expressions for the forward and backward reactions at the two temperatures. 2. Use the equation for-ra for the rate of disappearance of A using forward and backward reactions. Interim answer: EA for the forward reaction is approx. 48 kJ mol¹¹.
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