Chapter 1: Problem 7
Calculate the time to reduce the number of moles of \(\mathrm{A}\) to \(1 \%\) of its initial value in a constant-volume batch reactor for the reaction and data in Example \(1-1\).
Short Answer
Expert verified
The time it takes for the number of moles of A to reduce to 1% of its initial value in the constant-volume batch reactor is approximately 19.69 hours.
Step by step solution
01
Write the rate equation
The given reaction is a first-order reaction, and we are told that it follows elementary kinetics. Therefore, the rate equation for this reaction is given by:
\( -\dfrac{d[A]}{dt} = k[A] \)
02
Integration strategy
In this step, we will separate variables and integrate both sides of the rate equation to obtain an expression relating the number of moles of A and time t.
Divide both sides by [A] and multiply both sides by dt to get:
\(-\dfrac{d[A]}{[A]} = k dt \)
Now, integrate both sides:
\(\int_{[A]_0}^{[A]} -\dfrac{1}{[A]} d[A] = \int_0^t k dt \)
Where [A]_0 is the initial concentration of A.
03
Solve the integral
After integrating both sides, we get:
\(-\ln{([A]/[A]_0)} = kt + C \)
Here, C is the constant of integration. As the boundary condition, when t=0, [A] = [A]_0. Thus, C=0. The equation simplifies to:
\( \ln{([A]_0/[A])} = kt \)
04
Determine the time for the number of moles of A to reduce to 1% of its initial value
We need to find the time, t, when the number of moles of A is 1% of its initial value, i.e., [A] = 0.01[A]_0.
Substitute [A] with 0.01[A]_0 in the previous equation:
\( \ln{(0.01)} = kt \)
Now we need to solve for t. The reaction rate constant, k, is given in Example 1-1 as \( k = 0.234 \, h^{-1} \).
05
Solve for t
Plug in the value of k and solve the equation for t:
\( t = \dfrac{\ln{(0.01)}}{0.234} \)
\( t \approx -\dfrac{4.605}{0.234} = 19.685 h \)
So, it takes approximately 19.69 hours for the number of moles of A to reduce to 1% of its initial value in the constant-volume batch reactor.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order reaction
In chemical reaction engineering, a first-order reaction is one wherein the rate of reaction is directly proportional to the concentration of one reactant. This type of reaction is common in many chemical processes and is characterized by a simple, linear relationship between reactant concentration and reaction rate.
For a reaction where A converts to products, the rate can be expressed as:
For a reaction where A converts to products, the rate can be expressed as:
- The rate of decrease in concentration of reactant A is proportional to the concentration of A itself.
- The mathematical form for a first-order reaction is \(-\frac{d[A]}{dt} = k[A]\), where \([A]\) is the concentration of A, and \(k\) is the rate constant.
- Because the reaction rate depends only on the concentration of A, changes in the amount of other substances don't impact the rate of the reaction.
Rate equation
The rate equation, also known as the rate law, defines how the rate of a chemical reaction depends on the concentration of its reactants. For first-order reactions, this equation expresses the relationship between the rate of reaction and the concentration of a single reactant.
Some key components include:
In many industrial settings, determining this rate equation allows for better control over manufacturing processes that rely on chemical reactions.
Some key components include:
- The form \(-\frac{d[A]}{dt} = k[A]\) describes how the concentration of A decreases over time.
- The rate constant \(k\) has dimensions that depend on the order of the reaction. For a first-order reaction, its units are \(\text{time}^{-1}\), such as \(\text{sec}^{-1}\) or \(\text{h}^{-1}\).
- The negative sign in the rate equation indicates the decrease of the concentration of the reactant with time.
In many industrial settings, determining this rate equation allows for better control over manufacturing processes that rely on chemical reactions.
Integration strategy
The integration strategy in solving rate equations involves mathematical integration to transform the differential equation into a format that links reactant concentration with time. This is a key step in determining how long a reaction will take to reach a certain level of completion.
For a first-order reaction, this strategy entails:
For a first-order reaction, this strategy entails:
- Separating variables by taking \(-\frac{d[A]}{[A]} = k \, dt\), which allows us to integrate both sides independently.
- Integrating \(\int_{[A]_0}^{[A]} -\frac{1}{[A]} \, d[A]\) gives the left side of the equation, and \(\int_0^t k \, dt\) solves for the right side.
- The result is the expression \(\ln{([A]_0/[A])} = kt\), which can be used to solve for time when given rate constant \(k\) and concentrations.
Constant-volume batch reactor
A constant-volume batch reactor is a type of reactor where no material enters or leaves during the reaction. The volume of the reactor remains unchanged throughout the process. This type of reactor is commonly used in laboratory studies and small-scale industrial applications for a variety of chemical reactions.
Key features of constant-volume batch reactors include:
Such reactors serve as a vital tool in developing reaction kinetic models, scaling up processes, and optimizing reaction conditions for the chemical industry.
Key features of constant-volume batch reactors include:
- No flow of reactants or products, which means that the system can be studied as a closed system.
- Concentrations of reactants change as the reaction progresses due to the consumption of reactants without additional feedstock or removal of products.
- Because the reactor volume is constant, the concentration changes depend solely on the reaction rate.
Such reactors serve as a vital tool in developing reaction kinetic models, scaling up processes, and optimizing reaction conditions for the chemical industry.