Problem 36
Question
Which of the following is correct for a first order reaction? \(\left(k=\right.\) rate constant \(t_{1 / 2}=\) half-life \()\) (a) \(t_{1 / 2}=0.693 \times k\) (b) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=1 / 0.693\) (c) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=0.693\) (d) \(6.93 \times k \times t_{1 / 2}=1\)
Step-by-Step Solution
Verified Answer
Option (c) is correct: \( k \cdot t_{1/2} = 0.693.\)
1Step 1: Understanding First-order Reactions
For a first-order reaction, the rate of reaction depends on the concentration of one reactant raised to the first power. It is characterized by the differential rate law \( \text{Rate} = k[A] \), where \( k \) is the rate constant and \([A]\) is the concentration of reactant \(A\).
2Step 2: Determine the Half-life Expression
For first-order reactions, the half-life \( t_{1/2} \) is the time taken for the concentration of the reactant to reduce to half of its original value. It is given by the formula \( t_{1/2} = \frac{0.693}{k} \).
3Step 3: Evaluate Each Option Against the Correct Formula
Compare each given option with the expression derived in Step 2.- Option (a) states \( t_{1/2} = 0.693 \times k \), which is incorrect according to the formula.- Option (b) states \( k \cdot t_{1/2} = \frac{1}{0.693} \), which is incorrect.- Option (c) states \( k \cdot t_{1/2} = 0.693 \), which matches the correct formula rearranged as \( k = \frac{0.693}{t_{1/2}} \), this is the correct option.- Option (d) makes an incorrect assertion about the relationship between \( k \), \( t_{1/2} \), and a constant.
Key Concepts
Rate Constant in First-order ReactionsHalf-life Expression for First-order ReactionsDifferential Rate Law
Rate Constant in First-order Reactions
In the context of first-order reactions, the rate constant, denoted by the symbol \( k \), is a crucial parameter. It quantitatively describes the speed at which the reaction proceeds. For a reaction to be classified as first-order, its rate is directly proportional to the concentration of one reactant, usually expressed as \( \text{Rate} = k[A] \), where \( [A] \) is the concentration of the reactant.
The rate constant \( k \) remains constant throughout the reaction, assuming temperature does not change. This means it is independent of the concentration of the reactants. Typically, the units of the rate constant for a first-order reaction are \( ext{s}^{-1} \). Understanding \( k \) helps chemists determine how quickly a reactant is transformed into products, making it a valuable tool in reaction kinetics.
The rate constant \( k \) remains constant throughout the reaction, assuming temperature does not change. This means it is independent of the concentration of the reactants. Typically, the units of the rate constant for a first-order reaction are \( ext{s}^{-1} \). Understanding \( k \) helps chemists determine how quickly a reactant is transformed into products, making it a valuable tool in reaction kinetics.
Half-life Expression for First-order Reactions
In kinetics, the concept of half-life, denoted by \( t_{1/2} \), is significant, especially when analyzing first-order reactions. The half-life is the time required for the concentration of the reactant to decrease to half of its initial value. A unique feature of first-order reactions is that their half-life remains constant throughout the reaction, irrespective of the initial concentration. This is because the rate depends on the concentration raised to the first power.
The relationship between the half-life and the rate constant for a first-order reaction is given by the formula:
The relationship between the half-life and the rate constant for a first-order reaction is given by the formula:
- \( t_{1/2} = \frac{0.693}{k} \)
Differential Rate Law
The differential rate law in chemistry is a mathematical expression that describes how the rate of a reaction depends on the concentration of its reactants. For first-order reactions, the differential rate law is:
Understanding the differential rate law is essential because it provides a direct relationship between concentration and rate, unlike integrated rate laws which relate concentrations to time. This law is vital for predicting how changes in concentration affect the rate, thus allowing chemists to control reaction speeds in industrial and laboratory settings.
- \( \text{Rate} = k[A] \)
Understanding the differential rate law is essential because it provides a direct relationship between concentration and rate, unlike integrated rate laws which relate concentrations to time. This law is vital for predicting how changes in concentration affect the rate, thus allowing chemists to control reaction speeds in industrial and laboratory settings.
Other exercises in this chapter
Problem 34
According to the collision theory of reaction rates, an increase of the temperature at which the reaction oc curs will inturn increase the rate of the reaction.
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Which of the following relation is correct for a first order reaction? \((k=\) rate constant; \(\mathrm{r}=\) rate of reaction; \(\mathrm{C}=\) conc, of reactan
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If the rate law of a reaction \(\mathrm{nA} \longrightarrow \mathrm{B}\) is expressed as Rate \(=-\frac{1}{n} \frac{d[A]}{d t}=+\frac{d[B]}{d t}=k[A]^{x}\) The
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