Question

A spherical particle is dissolving in a liquid. The rate of dissolution is first order in the solvent concentration, \(C\). Assuming that the solvent is in excess, show that the following conversion time relationships hold. $$\begin{array}{cl} \hline \begin{array}{c} \text {Rate} \text { -Limiting} \\ \text {Regime} \end{array} & \text {Conversion Time Relationship} \\ \hline \text { Surface reaction } & 1-(1-X)^{1 / 3}=\frac{\alpha t}{D_{i}} \\ \text { Mass transfer } & \frac{D_{i}}{2 D^{*}} 11-(1-X)^{2 n} 1=\frac{\alpha t}{D_{i}} \\ \text { Mixed } & 11-(1-X)^{1 / 3} 1+\frac{D_{i}}{2 D^{*}}\left[1-(1-X)^{2 / 3}\right]=\frac{\alpha t}{D_{i}} \\ \hline \end{array}$$

Short answer

In summary, we showed that the conversion time relationships provided for surface reaction, mass transfer, and mixed regimes hold for a spherical particle dissolving in a liquid, considering the dissolution is first-order with respect to solvent concentration. We derived the relationships by integrating respective dissolution rates for each regime with respect to time and solving the integrals. The conversion time relationships are as follows: Surface reaction: \(1-(1-X)^{1 / 3}=\frac{\alpha t}{D_{i}}\) Mass transfer: \(\frac{D_{i}}{2 D^{*}} 11-(1-X)^{2 n} 1=\frac{\alpha t}{D_{i}}\) Mixed regime: \(11-(1-X)^{1 / 3} 1+\frac{D_{i}}{2 D^{*}}\left[1-(1-X)^{2 / 3}\right]=\frac{\alpha t}{D_{i}}\)

Step-by-step solution

Understand the different regimes of dissolution

The given problem considers three regimes of dissolution: 1. Surface reaction: In this regime, the rate-determining step is the reaction occurring on the surface of the particle. 2. Mass transfer: In this regime, the rate-determining step is the slow diffusion of the solvent molecules through a boundary layer surrounding the particle. 3. Mixed: In this regime, both surface reaction and mass transfer have significant contributions to the rate-determining step. In all of these regimes, we assume the dissolution is first-order with respect to the concentration of the solvent (C).

Demonstrate the Surface Reaction Regime

The dissolution rate for the surface reaction regime can be defined as: \(R = k_s (1 - X)^\frac{1}{3}\) where \(R\) is the dissolution rate, \(k_s\) is the surface reaction rate constant, and \(X\) is the conversion fraction. Integrate with respect to time \(t\) to find the conversion time relationship: \(\int_{0}^{t} R\,dt = \int_{0}^{X} k_s (1 - x)^\frac{1}{3}\,dx\) Solve the integral: \(k_s t = -3D_i[1 - (1-X)^\frac{1}{3}]\) Rearrange the above equation for the conversion time relationship: \(1-(1-X)^\frac{1}{3}=\frac{\alpha t}{D_{i}}\)

Demonstrate the Mass Transfer Regime

The dissolution rate for the mass transfer regime can be defined as: \(R = k_D (1 - X)^n\) where \(R\) is the dissolution rate, \(k_D\) is the mass transfer rate constant, and \(n\) is a constant. Integrate with respect to time \(t\) to find the conversion time relationship: \(\int_{0}^{t} R\,dt = \int_{0}^{X} k_D (1 - x)^n\,dx\) Solve the integral and rearrange: \(\frac{D_{i}}{2 D^{*}} 11-(1-X)^{2 n} 1=\frac{\alpha t}{D_{i}}\)

Demonstrate the Mixed Regime

The dissolution rate for the mixed regime can be defined as the sum of the surface reaction and mass transfer contributions: \(R = k_s (1 - X)^\frac{1}{3} + k_D (1 - X)^n\) Integrate with respect to time \(t\) to find the conversion time relationship: \(\int_{0}^{t} R\,dt = \int_{0}^{X} k_s (1 - x)^\frac{1}{3} + k_D (1 - x)^n\,dx\) Solve the integral and rearrange: \(11-(1-X)^{1 / 3} 1+\frac{D_{i}}{2 D^{*}}\left[1-(1-X)^{2 / 3}\right]=\frac{\alpha t}{D_{i}}\) In conclusion, we have demonstrated that the conversion time relationships provided are valid for each regime of dissolution, considering that the dissolution is first-order with respect to the solvent concentration.

Key concepts

Reaction Kinetics

Reaction kinetics is a fascinating area of chemical engineering that explores how chemical reactions occur and proceed over time. At its core, it examines the speed or rate at which reactants transform into products. This rate is influenced by several factors, including the nature of the chemical substances involved, temperature, and the concentration of reactants.

In the context of dissolution of a spherical particle, reaction kinetics helps us determine which step controls or limits the rate of the overall process. There are often different rate-determining steps that can dictate the kinetics of a reaction. These include surface reactions, where reactions occur at the interface of the particle, and mass transfer, where movement of molecules is limited by diffusion through a boundary layer around the particle.

Understanding reaction kinetics allows engineers to manipulate conditions to optimize reaction speeds. This is crucial in industrial applications where efficiency can lead to cost savings and better product yields.

By integrating the rate equations with respect to time, we can derive important relationships like conversion time, which tells us how long it will take for a certain percentage of the reactants to convert to products under given conditions.

Mass Transfer

Mass transfer refers to the movement of mass from one location, usually meaning stream, phase, fraction or component, to another. It's a vital concept in chemical and process engineering where it often interacts with reaction kinetics.

In the dissolution exercise, the mass transfer regime deals with how solvent molecules diffuse through a boundary layer surrounding the dissolving particle. This layer acts as a barrier, controlling how quickly molecules can move and therefore how quickly the particle dissolves.

When mass transfer is the rate-limiting step, the overall rate is limited not by chemical reactions occurring, but by how quickly molecules can travel through this boundary layer. The mass transfer rate constant, denoted as \(k_D\), quantifies this process. During design and operation of various processes, maintaining an optimal rate of mass transfer can ensure adequate reaction times and improved process performance.

In practical terms, engineers can engage strategies like increasing mixing or reducing particle size to enhance mass transfer, speeding up the dissolution in systems where mass transfer is the bottleneck.

First-Order Reactions

First-order reactions are a simplification used to describe the rate at which reactions proceed. In a first-order reaction, the rate is directly proportional to the concentration of one reactant. This simplification is particularly useful for reactions that involve a single reactant or where one reactant is in large excess.

In the dissolution problem where a particle dissolves in a solvent, the reaction is assumed to be first-order with respect to the solvent concentration \(C\). This means that the rate of dissolution changes linearly with the concentration of the solvent.

This concept is important because it allows for straightforward mathematical modeling of the reaction kinetics. The application of a first-order rate law simplifies the integration needed to analyze the reaction over time, thereby aiding in the derivation of conversion time relationships for different dissolution scenarios.

For students understanding chemical reaction engineering, grasping first-order kinetics provides a foundational tool for tackling more complex reaction systems later in their studies.