Problem 199
Question
Units of rate constant for the first and zero-order reactions in terms of molarity \(\mathrm{M}\), units are respectively [2002] (a) \(\mathrm{s}^{-1}, \mathrm{Ms}^{-1}\) (b) \(\mathrm{s}^{-1}, \mathrm{M}\) (c) \(\mathrm{M} \mathrm{s}^{-1}, \mathrm{~s}^{-1}\) (d) \(\mathrm{M}, \mathrm{s}^{-1}\)
Step-by-Step Solution
Verified Answer
The correct answer is (a) \(\mathrm{s}^{-1}, \mathrm{M} \mathrm{s}^{-1}\).
1Step 1: Understand the Reaction Order
Identify the order of reactions given in the problem. The first-order reaction depends linearly on the concentration of one reactant, while the zero-order reaction is independent of the concentration of the reactants.
2Step 2: Formula for Rate Constant Units
For a first-order reaction, the rate equation is \(r = k[A]\). Solving for \(k\), we have \(k = \frac{r}{[A]}\). For rate \(r\) with units \(\mathrm{M} \mathrm{s}^{-1}\) and concentration \([A]\) with units \(\mathrm{M}\), the units for \(k\) are \(\mathrm{s}^{-1}\). For a zero-order reaction, the rate equation is \(r = k\). Thus, \(k = r\) with units \(\mathrm{M} \mathrm{s}^{-1}\).
3Step 3: Match Units to Answer Options
The units for the first-order reaction are \( \mathrm{s}^{-1} \), and the units for the zero-order reaction are \( \mathrm{M} \mathrm{s}^{-1} \). Compare these units with the provided options. Option (a) indicates the correct units: \(\mathrm{s}^{-1}\) for the first-order and \(\mathrm{M} \mathrm{s}^{-1}\) for the zero-order reactions.
Key Concepts
Reaction OrderFirst-Order ReactionZero-Order ReactionRate Equation
Reaction Order
The reaction order of a chemical reaction offers insights into how the rate of the reaction is affected by the concentration of its reactants. It is essentially a summary of how these concentrations impact the reaction speed. A reaction can be zero-order, first-order, second-order, or even fractional order. Each order reflects the exponent on the concentration of the reactant in the rate equation for the reaction.
In a zero-order reaction, the rate is independent of the concentration of the reactants. This means that changing the concentration does not affect the rate of the reaction. On the other hand, a first-order reaction is linearly dependent on the concentration of one reactant. This linear relationship implies that if you double the concentration, you would also double the rate of the reaction. Understanding the reaction order is crucial to predicting how a change in concentration will affect the rate of a reaction.
In a zero-order reaction, the rate is independent of the concentration of the reactants. This means that changing the concentration does not affect the rate of the reaction. On the other hand, a first-order reaction is linearly dependent on the concentration of one reactant. This linear relationship implies that if you double the concentration, you would also double the rate of the reaction. Understanding the reaction order is crucial to predicting how a change in concentration will affect the rate of a reaction.
First-Order Reaction
A first-order reaction is characterized by its direct proportionality to the concentration of a single reactant. The rate equation for a typical first-order reaction is represented as \(r = k[A]\), where \(r\) is the reaction rate, \(k\) is the rate constant, and \([A]\) is the concentration of reactant A.
Since the rate is directly dependent on the concentration of one reactant, changes in this concentration significantly impact the reaction's speed. When solving for the rate constant \(k\), the relationship can be expressed as:
Since the rate is directly dependent on the concentration of one reactant, changes in this concentration significantly impact the reaction's speed. When solving for the rate constant \(k\), the relationship can be expressed as:
- \(k = \frac{r}{[A]}\)
- This shows that the units for \(k\) will be \(\mathrm{s}^{-1}\) when \(r\) is in the units of \(\mathrm{M} \mathrm{s}^{-1}\) (molarity per second) and \([A]\) in \(\mathrm{M}\) (molarity).
Zero-Order Reaction
In zero-order reactions, the rate of reaction does not depend on the concentration of the reactants. The rate equation for such a reaction is given by \(r = k\), implying that the rate is equal to the rate constant. This means the reaction proceeds at a constant rate, regardless of varying concentrations of the reactant involved.
When examining the units of the rate constant \(k\) for zero-order reactions, it becomes evident that:
When examining the units of the rate constant \(k\) for zero-order reactions, it becomes evident that:
- \(k\) has the units of \(\mathrm{M} \mathrm{s}^{-1}\).
- This is because the dimensions of the reaction rate \(r\) itself are \(\mathrm{M} \mathrm{s}^{-1}\), and since \(r = k\), the rate constant directly adopts this unit.
Rate Equation
The rate equation provides a crucial insight into the mechanics of a reaction by detailing how the concentration of reactants affects the reaction rate. It is a mathematical expression that relates the reaction rate to the concentration of reactants and includes the rate constant and reaction order.
The general form of a rate equation can be written as:
The general form of a rate equation can be written as:
- \(r = k[A]^n\)
- \(r\) is the reaction rate,
- \(k\) is the rate constant,
- \([A]^n\) represents the concentration of reactant A raised to the power of its reaction order \(n\).
Other exercises in this chapter
Problem 194
Two first order reactions \(\mathrm{A}\) and \(\mathrm{B}\) have the same frequency factor and activation energy of \(\mathrm{A}\) exceeds that of \(\mathrm{B}\
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For a reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C}\), rate is given by \(+\mathrm{d}[\mathrm{C}] / \mathrm{dt}=k[\mathrm{~A}][\mathrm{B}]\), h
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Rate constant of the first-order reaction when initial concentration \(\mathrm{C}_{\mathrm{o}}\) and concentration \(\mathrm{C}_{\mathrm{t}}\) at time \(\mathrm
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