Problem 48
Question
Consider the chemical reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} $$ If \(x\) denotes the concentration of \(\mathrm{AB}\) at time \(t\) and \(a, b\) are the initial amounts of \(A\) and \(B\) present, then the reaction rate \(R(x)\) is given by $$ R(x)=k(a-x)(b-x) $$ where \(k, a\), and \(b\) are positive constants. Differentiate \(R(x)\).
Step-by-Step Solution
Verified Answer
The derivative of the reaction rate is: \( R'(x) = k(-a-b+2x) \).
1Step 1: Identify the Function to Differentiate
The reaction rate function is given by \( R(x) = k(a-x)(b-x) \). We need to differentiate this function with respect to \( x \).
2Step 2: Use the Product Rule
To differentiate \( R(x) = k(a-x)(b-x) \), we must use the product rule since it is the product of two functions, \( (a-x) \) and \( (b-x) \). The product rule states that the derivative of two functions multiplied together is: \((u imes v)' = u' imes v + u imes v'\). Let \( u = a-x \) and \( v = b-x \).
3Step 3: Differentiate Each Component
First, find the derivatives of \( u \) and \( v \) with respect to \( x \).
\( \frac{du}{dx} = -1 \) since \( u = a-x \).
\( \frac{dv}{dx} = -1 \) since \( v = b-x \).
\( \frac{du}{dx} = -1 \) since \( u = a-x \).
\( \frac{dv}{dx} = -1 \) since \( v = b-x \).
4Step 4: Apply the Product Rule
Apply the product rule as determined in Step 2:
\( R'(x) = k \left[ (a-x) \left( -1 \right) + (b-x) \left( -1 \right) \right] \).
\( R'(x) = k \left[ (a-x) \left( -1 \right) + (b-x) \left( -1 \right) \right] \).
5Step 5: Simplify the Derivative
Simplify the expression obtained after applying the product rule:
\( R'(x) = k \left[ -(a-x) - (b-x) \right] \).
Distribute the negative signs:
\( R'(x) = k \left[ -a+x-b+x \right] \).
Combine like terms:
\( R'(x) = k(-a-b+2x) \).
\( R'(x) = k \left[ -(a-x) - (b-x) \right] \).
Distribute the negative signs:
\( R'(x) = k \left[ -a+x-b+x \right] \).
Combine like terms:
\( R'(x) = k(-a-b+2x) \).
Key Concepts
Reaction Rate in Chemical ReactionsUnderstanding the Product RuleDifferentiation and Derivatives
Reaction Rate in Chemical Reactions
The reaction rate in a chemical process refers to how quickly reactants are converted into products. In this case, we're talking about a reaction where chemical species A and B react to form compound AB. Understanding how the reaction rate changes with concentration is important in predicting how the reaction will proceed over time.
The equation for the reaction rate is given as \( R(x) = k(a-x)(b-x) \). Here:
The equation for the reaction rate is given as \( R(x) = k(a-x)(b-x) \). Here:
- \( k \) is the rate constant, which influences the speed of the reaction.
- \( a \) is the initial concentration of reactant A.
- \( b \) is the initial concentration of reactant B.
- \( x \) represents the concentration of the product AB at time \( t \).
Understanding the Product Rule
The product rule is a useful tool in calculus differentiation, especially when dealing with functions that are products of two or more functions. It provides a way to find the derivative of such functions. The rule states:
If you have a function \( h(x) = u(x) \times v(x) \), then the derivative \( h'(x) \) is given by:
In the context of the reaction rate equation, \( R(x) = k(a-x)(b-x) \), identifying \( u \) as \( a-x \) and \( v \) as \( b-x \) allows us to apply the product rule effectively in order to differentiate \( R(x) \).
This involves calculating the derivatives \( \frac{du}{dx} = -1 \) and \( \frac{dv}{dx} = -1 \), and then applying the product rule to find \( R'(x) \). This methodology is essential in achieving the accurate derivative of the function.
If you have a function \( h(x) = u(x) \times v(x) \), then the derivative \( h'(x) \) is given by:
- \((u \cdot v)' = u' \cdot v + u \cdot v' \),
In the context of the reaction rate equation, \( R(x) = k(a-x)(b-x) \), identifying \( u \) as \( a-x \) and \( v \) as \( b-x \) allows us to apply the product rule effectively in order to differentiate \( R(x) \).
This involves calculating the derivatives \( \frac{du}{dx} = -1 \) and \( \frac{dv}{dx} = -1 \), and then applying the product rule to find \( R'(x) \). This methodology is essential in achieving the accurate derivative of the function.
Differentiation and Derivatives
Differentiation is a fundamental concept in calculus used to find the derivative of a function. The derivative itself represents the rate at which one quantity changes with respect to another. In the case of the reaction rate function, differentiation helps in determining how fast or slow the reaction rate \( R(x) \) is changing as the product AB forms.
To solve the given exercise, obtaining the derivative \( R'(x) \) plays a crucial role in understanding how the reaction rate is influenced by changes in concentration of reactants A and B. Calculus provides the tools needed, and by applying the product rule, we were able to simplify and find:
To solve the given exercise, obtaining the derivative \( R'(x) \) plays a crucial role in understanding how the reaction rate is influenced by changes in concentration of reactants A and B. Calculus provides the tools needed, and by applying the product rule, we were able to simplify and find:
- \( R'(x) = k(-a-b+2x) \)
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