Problem 47
Question
Consider the chemical reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} $$ If \(x(t)\) denotes the concentration of \(\mathrm{AB}\) at time \(t\), then $$ \frac{d x}{d t}=k(a-x)(b-x) $$ where \(k\) is a positive constant and \(a\) and \(b\) denote the concentrations of \(A\) and \(B\), respectively, at time \(0 .\) Assume that \(k=3, a=7\), and \(b=4\). For what values of \(x\) is \(d x / d t=0 ?\) Interpret the meaning of \(d x / d t=0\).
Step-by-Step Solution
Verified Answer
The values of \( x \) for which \( \frac{d x}{d t} = 0 \) are 7 and 4, indicating the reaction stops as one reactant is fully consumed.
1Step 1: Identify the Zero Condition
The condition for which the rate of reaction is zero is when the derivative \( \frac{d x}{d t} = 0 \). We need to solve the equation \( k(a-x)(b-x) = 0 \) to find the values of \( x \).
2Step 2: Substitute Known Values
Substitute the given values into the equation: \( 3(7-x)(4-x) = 0 \). Since the constant \( 3 eq 0 \), this implies \( (7-x)(4-x) = 0 \).
3Step 3: Solve for x
Using the fact that the product of the terms is zero, set each factor equal to zero: \( 7-x = 0 \) and \( 4-x = 0 \). Solving these gives \( x = 7 \) and \( x = 4 \).
4Step 4: Interpret Zero Values
When \( x = 7 \) (or \( x = 4 \)), the concentration of one reactant is zero, meaning all of one reactant could be converted to the product without any further reaction. Thus, \( \frac{d x}{d t} = 0 \) indicates the reaction has stopped because at least one reactant is completely consumed.
Key Concepts
Chemical KineticsConcentrationRate of Change
Chemical Kinetics
Chemical kinetics is the study of the speed or rate at which chemical reactions occur. It helps us understand how concentrations of substances change over time in a reaction. When we know the rate of reaction, we can predict how quickly products will form or how soon reactants will be consumed. This is crucial in industrial processes, environmental science, and many other fields where reaction times are critical.
In the context of the given problem, the reaction between substances A and B to form AB is analyzed using kinetics. The equation \( \frac{d x}{d t} = k(a-x)(b-x) \) represents the rate of change of the concentration of AB over time \( t \). Here, \( k \) is a constant that gives us an indication of the reaction speed. Knowing the rate helps us determine whether adjustments are needed to change how fast or slow a reaction progresses.
In the context of the given problem, the reaction between substances A and B to form AB is analyzed using kinetics. The equation \( \frac{d x}{d t} = k(a-x)(b-x) \) represents the rate of change of the concentration of AB over time \( t \). Here, \( k \) is a constant that gives us an indication of the reaction speed. Knowing the rate helps us determine whether adjustments are needed to change how fast or slow a reaction progresses.
- Reaction rates can be affected by various factors, including temperature, pressure, concentration, and the presence of catalysts.
Concentration
Concentration is the amount of a substance present in a mixture or solution. It is usually expressed in terms of moles per liter (mol/L). Understanding concentration is essential in chemical reactions, as it directly influences the reaction's rate.
In the exercise, \( a \) and \( b \) are the initial concentrations of reactants A and B, respectively. The concentration of the product AB at any time \( t \) is denoted by \( x(t) \). As the reaction proceeds, concentrations change, impacting how the reaction progresses.
In the exercise, \( a \) and \( b \) are the initial concentrations of reactants A and B, respectively. The concentration of the product AB at any time \( t \) is denoted by \( x(t) \). As the reaction proceeds, concentrations change, impacting how the reaction progresses.
- Higher concentrations of reactants typically increase the rate of reaction.
- The reaction slows as reactants are consumed and products are formed.
Rate of Change
The rate of change refers to how quickly the concentration of a substance changes over time. It's represented mathematically by derivatives in the differential equation.
For any reaction, \( \frac{d x}{d t} \) indicates the rate at which product AB is being formed. When \( \frac{d x}{d t} = 0 \), it means there is no change in the concentration of the product over time, implying the reaction has reached completion or equilibrium.
For any reaction, \( \frac{d x}{d t} \) indicates the rate at which product AB is being formed. When \( \frac{d x}{d t} = 0 \), it means there is no change in the concentration of the product over time, implying the reaction has reached completion or equilibrium.
- A positive \( \frac{d x}{d t} \) indicates that the product is being formed.
- A negative \( \frac{d x}{d t} \) would suggest decomposition, which is not applicable in this context.
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