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Chapter 3: Problem 1

Using Equation (3.6), derive a material balance for a reactant in a closed system.

Short Answer

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Question: Derive the material balance for a reactant in a closed system using Equation (3.6). Answer: The material balance for a reactant in a closed system using Equation (3.6) is derived as follows: \begin{align*} \frac{d[A]}{dt} + r\nu_A = k[A] \end{align*} Where, d[A]/dt is the rate of change of the reactant's concentration with respect to time, r is the chemical reaction rate, ν_A is the stoichiometric coefficient of the reactant, and k is the reaction rate constant.

Step by step solution

01

Understand Equation (3.6)

Equation (3.6) is the general equation for calculating the rate of change of a reactant in a closed system. It's defined as follows: \begin{align*} \text{Rate of accumulation of reactant} = \text{Rate of input} - \text{Rate of output} + \text{Rate of generation} - \text{Rate of consumption} \end{align*} We'll be using this equation to derive the material balance for the reactant.
02

Identify rate of input and rate of output in a closed system

In a closed system, there is no input or output of reactant. The reactant is either consumed or generated within the system. So, in this case, the rate of input and rate of output are both zero. Now, the equation becomes: \begin{align*} \text{Rate of accumulation of reactant} = \text{Rate of generation} - \text{Rate of consumption} \end{align*}
03

Analyze the reactant's rate of generation and rate of consumption

Consider a reactant A that participates in a chemical reaction. The rate of generation of A is given by the product of the reaction rate constant (k) and the concentration of A ([A]). The rate of consumption of A is given by the product of the chemical reaction rate (-r) and the stoichiometric coefficient of A (ν_A). Now, the equation becomes: \begin{align*} \text{Rate of accumulation of A} = k[A] - (-r\nu_A) \end{align*}
04

Derive the material balance for the reactant

We're now ready to derive the material balance for reactant A. Let's substitute the rate of accumulation of A with the rate of change of its concentration with respect to time: \begin{align*} \frac{d[A]}{dt} = k[A] - (-r\nu_A) \end{align*} Now, rearrange the equation to get the material balance for reactant A: \begin{align*} \frac{d[A]}{dt} + r\nu_A = k[A] \end{align*} This is the material balance for a reactant in a closed system using Equation (3.6).

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