Answered: The irreversible liquid-phase reaction…

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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Question

The irreversible liquid-phase reaction A à 2B + C is carried out isothermally at 127 ^oC. in a well-mixed constant volume batch reactor.

The reaction is half order in A (( ). During lab experiments, it was shown that a production rate of B (r_B) of 2.3 mol L^-1 min^-1 was obtained as the concentration of A was 5.0 mol L^-1. Also an activation energy of 94 kJ mol^-1 was obtained for this reaction.

Find an analytical equation that describes the concentration of A (C_A) as a function of the initial concentration of A (C_A0) and time (t)

Expert Solution

Step 1

The reactor consisting of a large vessel with an impeller and a system for heating/cooling is a batch reactor. The reactants are fed into the reactor initially and the products are taken out after the reaction is completed.

The constant volume batch reactor typically refers to the constant volume of the reaction mixture. The term actually means the constant-density reaction system.

Step 2

The reaction taking place is the irreversible liquid phase reaction which is a single-phase reaction. For a single-phase reaction, $A \to 2 B + C$ , the rate of reaction of all the substances is given by

$\frac{- r_{A}}{1} = \frac{r_{B}}{2} = \frac{r_{C}}{1}$

For a constant volume reactor, $\in_{A} = 0$

Also, the concentration of A at any instant

$C_{A} = \frac{C_{A 0} (1 - x_{A})}{1 + \in_{A} x_{A}} P u t t i n g \in_{A} = 0 C_{A} = C_{A 0} (1 - x_{A})$

where C_A0is the initial concentration of A and x_A is the extent of reaction.

It is given that reaction is half-order with respect to A.

$- r_{A} = {kC}_{A}^{\frac{1}{2}} - \frac{d C_{A}}{d t} = {kC}_{A}^{\frac{1}{2}} \int_{C_{A}}^{C_{A 0}} \frac{d C_{A}}{{C_{A}}^{\frac{1}{2}}} = - k \int_{0}^{t} d t - \frac{1}{2} [{C_{A 0}}^{\frac{1}{2}} - {C_{A}}^{\frac{1}{2}}] = - k (t - 0) \frac{1}{2} [{C_{A 0}}^{\frac{1}{2}} - {C_{A}}^{\frac{1}{2}}] = k (t) [{C_{A 0}}^{\frac{1}{2}} - {C_{A}}^{\frac{1}{2}}] = 2 \times k \times t {C_{A}}^{\frac{1}{2}} = {C_{A 0}}^{\frac{1}{2}} - 2 \times k \times t$

Also, $\frac{- r_{A}}{1} = \frac{r_{B}}{2}$

It is given that

$r_{B} = 2.3 m o l . L^{- 1} . m i n^{- 1} a t C_{A} = 5 m o l / L T h e r e f o r e, - r_{A} = \frac{r_{B}}{2} = \frac{2.3}{2} = 1.15 m o l . L^{- 1} . m i n^{- 1} - r_{A} = {kC}_{A}^{\frac{1}{2}} 1.15 = {k (5.0)}^{\frac{1}{2}} k = 0.5143 m o l^{\frac{3}{2}} L^{\frac{- 3}{2}} m i n^{- 1}$

The equation that describes the concentration of A (C_A) as a function of the initial concentration of A (C_A0) and time (t)

${C_{A}}^{\frac{1}{2}} = {C_{A 0}}^{\frac{1}{2}} - 2 \times 0.5143 \times t {C_{A}}^{\frac{1}{2}} = {C_{A 0}}^{\frac{1}{2}} - 1.0286 \times t$

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