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If the initial concentration of reactant in certain reaction is doubled, the half life period of the reaction is also doubled. The order of reaction is (a) zero (b) first (c) second (d) \(1.5\)

Short Answer

Expert verified
The order of the reaction is zero (option a).

Step by step solution

01

Understanding the Half-Life Concept

The half-life of a reaction is the time taken for the concentration of a reactant to reduce to half its initial value. The way the half-life changes with initial concentration helps in determining the order of the reaction.
02

Analyzing the Order-Dependent Half-Life Behavior

For a zero-order reaction, the half-life is given by \( t_{1/2} = \frac{[A]_0}{2k} \), which is directly proportional to the initial concentration \([A]_0\). For a first-order reaction, the half-life \( t_{1/2} = \frac{0.693}{k} \), which is independent of the initial concentration. For a second-order reaction, the half-life \( t_{1/2} = \frac{1}{k[A]_0} \), which is inversely proportional to the initial concentration. We will need to decide which behavior matches the situation described.
03

Matching the Reaction Order to Given Conditions

In the given problem, the half-life period doubles when the initial concentration of the reactant is doubled. This suggests that the half-life is directly proportional to the initial concentration, typical of a zero-order reaction where doubling \([A]_0\) results in doubling \( t_{1/2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zero-Order Reactions
Zero-order reactions are fascinating because their rate is independent of the concentration of the reactant. In a zero-order reaction, the rate remains constant over time. This means the rate at which the reactant is used up doesn't change as the concentration goes down. Instead, it relies on the rate constant (k).

Mathematically, the rate of a zero-order reaction is given by:
  • Rate = k
In terms of concentration over time for a zero-order reaction, you can use the equation:
  • \[ [A] = [A]_0 - kt \]
Here,
  • \([A]\) is the concentration at time \(t\)
  • \([A]_0\) is the initial concentration
  • \(k\) is the rate constant

One key feature of zero-order reactions is how their half-life, the time it takes for half of the reactant to vanish, depends on concentration. The half-life formula in this case is:
  • \[ t_{1/2} = \frac{[A]_0}{2k} \]
This makes the half-life directly proportional to the initial concentration, meaning if you double the concentration, the half-life also doubles. This characteristic helped us determine the reaction order in the original problem.
Half-Life
Half-life is a significant concept in chemical kinetics as it tells how long it takes for a reactant's concentration to reduce to half of its original value. Different orders of reactions affect how half-lives change based on reactant concentration.

For zero-order reactions, the half-life depends on the initial concentration. It's calculated using:
  • \[ t_{1/2} = \frac{[A]_0}{2k} \]

In first-order reactions, the half-life doesn't change with the concentration of reactants. It remains constant throughout the reaction. The formula for a first-order half-life is:
  • \[ t_{1/2} = \frac{0.693}{k} \]
This is because the rate of a first-order reaction depends exponentially on concentration.

For second-order reactions, the half-life is inversely proportional to the initial concentration:
  • \[ t_{1/2} = \frac{1}{k[A]_0} \]
In the practice problem, understanding how half-life escalates with the initial concentration helped conclude that it was a zero-order reaction, as it fit the doubling pattern.
Concentration Effects
Concentration effects refer to how alterations in reactant concentration affect the rate of reaction and its kinetics. Different reaction orders exhibit distinct concentration dependencies.

In a zero-order reaction, concentration changes do not affect the reaction rate, which remains fixed. This effect can illustrate operation limits of catalysts or surface reactions where saturation has been reached.

For first-order reactions, the rate is proportional to the concentration of one reactant. This means if you double the concentration, the rate doubles as well.
Second-order reactions have more complex behaviors, being influenced by the concentration of one reactant or two reactants in a bimolecular reaction. Rate here is proportional to the square of the reactant's concentration or the product of concentrations in bimolecular cases.

If a reaction shows that the rate, and in turn the half-life, changes in direct proportion to the concentration, like in the original exercise, it is a distinct marker for it being a zero-order reaction.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors affecting them. It helps us understand the speed at which chemical reactions occur and the steps involved.

Kinetics involves rate laws, which show the relationship between reaction rate and concentrations. The order of reaction is key in kinetics. It tells us how concentration changes impact the rate.

In determining reaction order, chemists often look at the formula derived from the rate law, which usually follows this pattern:
  • \[ ext{Rate} = k[A]^x[B]^y \]
Where:
  • \(k\) is the rate constant
  • \([A]\) and \([B]\) are reactant concentrations
  • \(x\) and \(y\) are the reaction orders for the reactants
In the broader context of kinetics, factors such as temperature, catalysts, and reaction conditions can also majorly impact rates.

Understanding these principles of kinetics allows us to predict how changes like concentration doubling will alter the reaction's dynamics as observed in the provided problem.

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Most popular questions from this chapter

The experimental data for the reaction \(2 \mathrm{~A}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) is \(10.2\) Table \begin{tabular}{llll} \hline Exp. & [A] & [B_ ] & Rate \(\left(\mathrm{Ms}^{-1}\right)\) \\ \hline 1. & \(0.50 \mathrm{M}\) & \(0.50 \mathrm{M}\) & \(1.6 \times 10^{-4}\) \\ \(2 .\) & \(0.50 \mathrm{M}\) & \(1.00 \mathrm{M}\) & \(3.2 \times 10^{-4}\) \\ \(3 .\) & \(1.00 \mathrm{M}\) & \(1.00 \mathrm{M}\) & \(3.2 \times 10^{-4}\) \\ \hline \end{tabular} the rate equation for the above data is (a) rate \(=\mathrm{k}\left[\mathrm{B}_{2}\right]\) (b) rate \(=k\left[\mathrm{~B}_{2}\right]^{2}\) (c) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}\) (d) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)

The following data pertains to the reaction between \(\mathrm{A}\) and \(\underline{B}\) \begin{tabular}{llll} \multicolumn{4}{c} { Table \(10.5\)} \\ \hline S. No. & {\([\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}\)} & Rate mol \(\mathrm{L}^{-1} \mathrm{~S}^{-1}\) \\ \hline 1. & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ 2\. & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ 3\. & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3

Graph between \(\log _{10} \mathrm{~K}\) and \((1 / t)\) is linear with slope \(S\). Hence \(\mathrm{E}\) is (a) \(\mathrm{R} \times \mathrm{S}\) (b) \(\mathrm{S} / \mathrm{R}\) (c) \(\mathrm{R} / \mathrm{S}\) (d) \(2.303 \mathrm{RS}\)

Consider the following statements: (1) rate of a process is directly proportional to its free energy change (2) the order of an elementary reaction step can be determined by examining the stoichiometry (3) the first-order reaction describe exponential time course. Of the statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct

\(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}\), rate of reaction \(+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\) is equal to (a) \(-\frac{3}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (b) \(-\frac{2}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (c) \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) \((\mathrm{d})+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)

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