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Two enzymes are immobilized on the same flat, nonporous surface. For enzyme A, the substrate is S₁. For enzyme B, the substrate is S₂. The product of the first reaction is S₂ S₁ EA S₂ Ев P

Two enzymes are immobilized on the same flat, nonporous surface. For enzyme A, the substrate is S₁. For enzyme B, the substrate is S₂. The product of the first reaction is S₂ S₁ EA S₂ Ев P

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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**Immobilized Enzyme Reactions** Two enzymes are immobilized on the same flat, nonporous surface. For enzyme A, the substrate is \( S_1 \). For enzyme B, the substrate is \( S_2 \). The product of the first reaction is \( S_2 \). **Diagram Explanation:** The diagram illustrates a sequential enzymatic reaction: 1. **Step 1**: - Substrate \( S_1 \) is converted to \( S_2 \) by enzyme \( E_A \). - This is represented by an arrow pointing from \( S_1 \) to \( S_2 \) labeled \( E_A \). 2. **Step 2**: - Substrate \( S_2 \) is then converted to product \( P \) by enzyme \( E_B \). - An arrow labeled \( E_B \) points from \( S_2 \) to \( P \). This sequence demonstrates how multiple enzymes can work on a surface to transform substrates through a series of reactions, leading to the desired product \( P \).
Transcribed Image Text:**Immobilized Enzyme Reactions** Two enzymes are immobilized on the same flat, nonporous surface. For enzyme A, the substrate is \( S_1 \). For enzyme B, the substrate is \( S_2 \). The product of the first reaction is \( S_2 \). **Diagram Explanation:** The diagram illustrates a sequential enzymatic reaction: 1. **Step 1**: - Substrate \( S_1 \) is converted to \( S_2 \) by enzyme \( E_A \). - This is represented by an arrow pointing from \( S_1 \) to \( S_2 \) labeled \( E_A \). 2. **Step 2**: - Substrate \( S_2 \) is then converted to product \( P \) by enzyme \( E_B \). - An arrow labeled \( E_B \) points from \( S_2 \) to \( P \). This sequence demonstrates how multiple enzymes can work on a surface to transform substrates through a series of reactions, leading to the desired product \( P \).
### Reaction Rate Analysis and Calculation #### Part (a) **Figure Description:** Figure 1 shows the reaction rate on a surface as a function of the local concentration of substrate \( S_1 \). The graph plots the reaction rate \( V \) (or flux \( J \)) in units of \( 10^6 \, \text{mg/cm}^2\cdot \text{s} \) against the concentration of \( S_1 \) in \( \text{mg/L} \). - **Graph Details:** - The x-axis represents the concentration of \( S_1 \) ranging from 0 to 120 mg/L. - The y-axis represents the reaction rate, ranging from 0 to 4 \( \times 10^6 \, \text{mg/cm}^2\cdot \text{s} \). - The curve increases steeply at low \( S_1 \) concentrations and then levels off, indicating a saturation effect at higher concentrations. **Problem Statement:** - Bulk concentration of \( S_1 \): 100 mg/L - Mass-transfer coefficient: \( 4 \times 10^{-5} \, \text{cm/s} \) **Tasks:** 1. Calculate the rate of consumption of \( S_1 \) for a 1 cm\(^2\) surface. 2. Determine the surface concentration of \( S_1 \). --- #### Part (b) **Rate of the Second Reaction:** The rate equation for the second reaction is given by: \[ -\frac{d[S_2]}{dt} = \frac{d[P]}{dt} = \frac{V_m'' S_{2 \, \text{surface}}}{K_m + S_{2 \, \text{surface}}} \] **Given Data:** - \( K_m = 5 \, \text{mg/L} \) (or \( 5 \times 10^{-3} \, \text{mg/cm}^3 \)) - \( V_m'' = 4 \times 10^{-6} \, \text{mg/cm}^2 \cdot \text{s} \) - Bulk concentration of \( S_2 \): 5 mg/L - Surface to bulk mass-transfer condition: Same mass-transfer coefficient as \( S_1 \) **Tasks:** 1. Calculate
Transcribed Image Text:### Reaction Rate Analysis and Calculation #### Part (a) **Figure Description:** Figure 1 shows the reaction rate on a surface as a function of the local concentration of substrate \( S_1 \). The graph plots the reaction rate \( V \) (or flux \( J \)) in units of \( 10^6 \, \text{mg/cm}^2\cdot \text{s} \) against the concentration of \( S_1 \) in \( \text{mg/L} \). - **Graph Details:** - The x-axis represents the concentration of \( S_1 \) ranging from 0 to 120 mg/L. - The y-axis represents the reaction rate, ranging from 0 to 4 \( \times 10^6 \, \text{mg/cm}^2\cdot \text{s} \). - The curve increases steeply at low \( S_1 \) concentrations and then levels off, indicating a saturation effect at higher concentrations. **Problem Statement:** - Bulk concentration of \( S_1 \): 100 mg/L - Mass-transfer coefficient: \( 4 \times 10^{-5} \, \text{cm/s} \) **Tasks:** 1. Calculate the rate of consumption of \( S_1 \) for a 1 cm\(^2\) surface. 2. Determine the surface concentration of \( S_1 \). --- #### Part (b) **Rate of the Second Reaction:** The rate equation for the second reaction is given by: \[ -\frac{d[S_2]}{dt} = \frac{d[P]}{dt} = \frac{V_m'' S_{2 \, \text{surface}}}{K_m + S_{2 \, \text{surface}}} \] **Given Data:** - \( K_m = 5 \, \text{mg/L} \) (or \( 5 \times 10^{-3} \, \text{mg/cm}^3 \)) - \( V_m'' = 4 \times 10^{-6} \, \text{mg/cm}^2 \cdot \text{s} \) - Bulk concentration of \( S_2 \): 5 mg/L - Surface to bulk mass-transfer condition: Same mass-transfer coefficient as \( S_1 \) **Tasks:** 1. Calculate
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