Problem 50
Question
Which of the following statement is correct? (a) A plot of \(\log k\) vs \(1 / t\) is linear (b) A plot of \(\log [\mathrm{X}]\) vs time is linear for a first-order reaction, \(\mathrm{X} \longrightarrow \mathrm{P}\) (c) A plot of log P vs \(1 / t\) is linear at constant volume (d) A plot of \(\mathrm{P}\) vs \(1 / \mathrm{V}\) is linear at constant pressure
Step-by-Step Solution
Verified Answer
Statement (b) is correct; it's linear for a first-order reaction.
1Step 1: Read the Statements Carefully
We are given four statements about various plots involving logarithmic and inverse transformations. Each statement suggests a specific relationship between two variables. The task is to determine which of these plots are linear based on known chemical kinetics and mathematical relationships.
2Step 2: Analyze Statement (a)
Statement (a) suggests that a plot of \( \log k \) versus \( \frac{1}{t} \) is linear. \( k \) typically represents a rate constant in chemical kinetics, and \( t \) is time. However, there are no known equations relating \( \log k \) and \( \frac{1}{t} \) linearly for common reaction orders.
3Step 3: Analyze Statement (b)
Statement (b) suggests that a plot of \( \log [\text{X}] \) versus time is linear for a first-order reaction. For a first-order reaction, \( \text{X} \rightarrow \text{P} \), the concentration of \( X \) changes according to the equation \( \log [\text{X}] = -kt + \log [\text{X}]_0 \), which indeed describes a linear plot of \( \log [\text{X}] \) versus time.
4Step 4: Analyze Statement (c)
Statement (c) suggests that a plot of \( \log P \) versus \( \frac{1}{t} \) is linear at constant volume. This statement does not correspond to any known gas laws or kinetic relationships that result in a linear plot under these conditions.
5Step 5: Analyze Statement (d)
Statement (d) suggests that a plot of \( P \) versus \( \frac{1}{V} \) is linear at constant pressure. Given \( P = nRT/V \) from the ideal gas law, if the volume changes while \( nRT \) remains constant, \( P \) will not vary linearly with \( \frac{1}{V} \) under constant pressure but rather will maintain a constant value.
6Step 6: Determine the Correct Statement
Based on our analysis, Statement (b) is correct. A plot of \( \log [\text{X}] \) versus time is linear for a first-order reaction, \( \text{X} \rightarrow \text{P} \). This is verified by the first-order reaction rate law, which predicts a linear form.
Key Concepts
First-Order ReactionRate ConstantReaction Rate Laws
First-Order Reaction
A first-order reaction is one where the rate of reaction depends on the concentration of a single reactant raised to the power of one. This means that if you double the concentration of the reactant, the reaction rate will also double. First-order reactions are quite common and often involve processes like radioactive decay and the decomposition of some chemical compounds.
To recognize a first-order reaction mathematically, we note that the rate of reaction is proportional to the concentration of the reactant, expressed as: - Rate = k[X], where: - Rate is the reaction rate, - k is the rate constant, - [X] is the concentration of reactant X.
This unique relationship affects how certain variables relate graphically. For instance, plotting the natural logarithm of the concentration of the reactant versus time yields a straight line, with the slope being the negative of the rate constant.
To recognize a first-order reaction mathematically, we note that the rate of reaction is proportional to the concentration of the reactant, expressed as: - Rate = k[X], where: - Rate is the reaction rate, - k is the rate constant, - [X] is the concentration of reactant X.
This unique relationship affects how certain variables relate graphically. For instance, plotting the natural logarithm of the concentration of the reactant versus time yields a straight line, with the slope being the negative of the rate constant.
Rate Constant
The rate constant ( k ) is a proportional factor in the rate law of a chemical reaction, revealing how fast or slow a reaction occurs. It is an essential parameter in the study of reaction rates and is unique for every reaction at a given temperature.
The value of k is influenced by: - Temperature: As temperature increases, the rate constant typically increases. This is because the frequency of reactant collisions increases. - Activation energy: Reactions with lower activation energy have higher rate constants, as they require less energy to proceed.
The dimensions of the rate constant vary depending on the overall order of the reaction. For a first-order reaction, k has the dimension of time inverse (e.g., s⁻¹). Understanding the rate constant allows chemists to predict how changing conditions will impact reaction speed and to compare the dynamics of different reactions.
The value of k is influenced by: - Temperature: As temperature increases, the rate constant typically increases. This is because the frequency of reactant collisions increases. - Activation energy: Reactions with lower activation energy have higher rate constants, as they require less energy to proceed.
The dimensions of the rate constant vary depending on the overall order of the reaction. For a first-order reaction, k has the dimension of time inverse (e.g., s⁻¹). Understanding the rate constant allows chemists to predict how changing conditions will impact reaction speed and to compare the dynamics of different reactions.
Reaction Rate Laws
Reaction rate laws are mathematical expressions that describe the relationship between the reaction rate and the concentrations of reactants. These expressions are crucial because they help predict how rates change under different concentrations and conditions. A rate law is usually in the form:
- Rate = k[A]^m[B]^n, where:
- Rate is the reaction rate,
- k is the rate constant,
- [A] and [B] are concentrations of reactants,
- m and n are the reaction orders with respect to each reactant.
The overall order of a reaction is the sum of the exponents m and n, which reveals how many molecules are involved in the rate-determining step. The orders m and n are often determined experimentally, as they don't always correspond to the stoichiometric coefficients in the balanced equation.
By studying the rate laws, chemists can engineer conditions for optimal reaction speed and yield, playing a pivotal role in various industrial and laboratory processes.
The overall order of a reaction is the sum of the exponents m and n, which reveals how many molecules are involved in the rate-determining step. The orders m and n are often determined experimentally, as they don't always correspond to the stoichiometric coefficients in the balanced equation.
By studying the rate laws, chemists can engineer conditions for optimal reaction speed and yield, playing a pivotal role in various industrial and laboratory processes.
Other exercises in this chapter
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