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In a zero-order reaction, \(47.5 \%\) of the reactant remains at the end of \(2.5\) hours. The amount of reactant consumed in one hour is (a) \(11.0 \%\) (b) \(33.0 \%\) (c) \(42.0 \%\) (d) \(21.0 \%\)

Short Answer

Expert verified
(d) 21.0%

Step by step solution

01

Understand Zero-Order Reaction

In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactant. The rate equation is given by \( [R] = [R]_0 - kt \), where \([R]\) is the concentration of the reactant at time \( t \), \([R]_0\) is the initial concentration, and \( k \) is the rate constant.
02

Calculate Initial and Remaining Concentration

Since 47.5% of the reactant remains after 2.5 hours, initially, it was 100%. Thus, 47.5% remains, and hence, 52.5% of the reactant is consumed in 2.5 hours.
03

Determine Rate of Reaction

To find the rate \( k \), we use the proportion of the reactant consumed over 2.5 hours. Since 52.5% is consumed, the rate \( k \) is \( \frac{52.5}{2.5} \) % per hour, which simplifies to 21% per hour. This indicates the amount consumed each hour.
04

Find Amount Consumed in One Hour

With the rate of 21% per hour, the reactant consumed in one hour is 21%.

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Key Concepts

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

Rate Constant
The rate constant, often denoted as "k" in chemical kinetics, is a crucial parameter that defines the speed of a reaction. It quantifies how quickly reactants turn into products, specifically for a given reaction under certain conditions.

In a zero-order reaction, the rate constant has a unique interpretation. Unlike reactions where the rate depends on reactant concentration, a zero-order reaction's rate constant signifies a constant rate. This means the transformation rate remains the same regardless of how much reactant is present. The formula, \[ [R] = [R]_0 - kt \] illustrates this by showing us how the concentration \([R]\) changes over time \(t\), using the initial concentration \([R]_0\)\ and the rate constant \(k\).
  • The rate constant \(k\) gives us the specific number of consumed reactants per time unit.
  • In the provided problem, it was found that \(k\) is equal to 21% per hour.
This stability in rate makes zero-order reactions predictable for calculating how much reactant remains at any point in time.
Reaction Rate
Understanding the reaction rate is essential in chemical kinetics. It describes how quickly a chemical reaction proceeds. For zero-order reactions, the reaction rate is consistent, meaning that the reactants become products at a uniform rate.
This rate is determined by the rate constant \(k\).When we refer to reaction rate, we often mean the rate of consumption of reactants or formation of products.

In the exercise, 52.5% of reactants were consumed over 2.5 hours. This consistency in rate shows that 21% of the reactant was transformed each hour, which makes it easy to predict future concentrations.
  • Unlike other reaction orders, for zero-order, the rate does not reduce as reactants deplete.
  • Even when only 47.5% of the reactant remains after 2.5 hours, the rate stays at 21% per hour.
This predictable behavior simplifies calculations and helps in controlling processes where zero-order reactions apply.
Chemical Kinetics
Chemical kinetics explores the mechanics of how chemical reactions occur and progress over time. It investigates the range of factors that influence reaction rates, including temperature, reactant concentration, and the presence of catalysts.

In zero-order reactions, kinetics tell us that the rate is not affected by changes in reactant concentration. This peculiarity requires a specific approach when analyzing data and predicting outcomes. For instance, zero-order processes might occur due to surface saturation effects in catalyzed reactions, where further increases in concentration do not enhance the rate.
  • It gives a deeper understanding of how reactions proceed at a molecular level, explaining variables like rate constant and reaction rate.
  • Kinetics allows us to connect theoretical expressions like \([R] = [R]_0 - kt\) with real-world observations, such as the percentage of reactants consumed over time.
The area of chemical kinetics helps scientists and engineers accurately model reactions, ensuring that they understand how to adjust conditions for desired outcomes effectively, even in complex scenarios.

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Most popular questions from this chapter

\(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}\), 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) \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) \((\mathrm{d})+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)

The rate constant of a first-order reaction is \(6 \times 10^{-3}\) \(\mathrm{s}^{-1}\). If the initial concentration is \(0.10 \mathrm{M}\), the initial rate of reaction is (a) \(6 \times 10^{-3} \mathrm{Ms}^{-1}\) (b) \(6 \times 10^{-1} \mathrm{Ms}^{-1}\) (c) \(6 \times 10^{-6} \mathrm{Ms}^{-1}\) (d) \(6 \times 10^{-8} \mathrm{Ms}^{-1}\)

A follows parallel path Ist order reactions giving B and C as shown: If initial concentration of \(\mathrm{A}\) is \(0.25 \mathrm{M}\), calculate the concentration of \(\mathrm{C}\) after 5 hour of reaction. Given, \(\lambda_{1}=1.5 \times 10^{-5} \mathrm{~s}^{-1}, \lambda_{2}=5 \times 10^{-6} \mathrm{~s}^{-1}\) (a) \(7.55 \times 10^{-3} \mathrm{M}\) (b) \(1.89 \times 10^{-2} \mathrm{M}\) (c) \(5.53 \times 10^{-3} \mathrm{M}\) (d) \(3.51 \times 10^{-3} \mathrm{M}\)

For the reaction, \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) The following mechanism has been proposed \(\mathrm{NO}+\rightleftharpoons \mathrm{Cl} \quad \mathrm{NOCl}_{2}\) (fast) \(\mathrm{NOCl}_{2}+\mathrm{NO} \longrightarrow 2 \mathrm{NOCl}\) (slow) (a) Rate \(=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]^{2}\) (b) Rate \(=\mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]\) (c) Rate \(=\mathrm{k}[\mathrm{NOCl}]^{2}\) (d) Rate \(=\mathrm{k}\left[\mathrm{NOCl}_{2}\right][\mathrm{NO}]\)

For a chemical reaction which can never be a fractional number. (a) order (b) half-life (c) molecularity (d) rate constant

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