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If the rate of a gaseous reaction is independent of partial pressure of reactant, the order of reaction is (a) 0 (b) 1 (c) 2 (d) 3

Short Answer

Expert verified
The order of reaction is zero (\textbf{0}), option (a).

Step by step solution

01

Understanding Reaction Order

Reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. When the rate is independent of the concentration of reactants, the reaction is said to be zero order because raising the concentration to the power of zero gives 1, indicating no change in rate with concentration changes.
02

Identifying the Correct Option

Since the rate is independent of the partial pressure or concentration of the reactant, this means the concentration term in the rate equation is raised to the power of 0, which is characteristic of a zero-order reaction.
03

Determining the Reaction Order

With the given information, for the reaction to be independent of the concentration of the reactants (partial pressure for gases), the order must be zero. This means that changing the concentration does not affect the rate of the reaction, which is the definition of a zero-order reaction.

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Key Concepts

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

Zero Order Reaction
In the world of chemical reactions, understanding how the rate of a reaction changes with variations in reactant concentration is crucial. In a zero order reaction, the reaction rate remains constant regardless of the changes in concentration of the reactants.

Imagine you're baking cookies, and no matter how many ingredients you add or remove, the number of cookies made in an hour doesn't change. That's similar to a zero order reaction; altering the amount of reactant does not speed up or slow down the reaction at all. For a reaction of zero order, the rate law expression simplifies to \( rate = k \), where \( k \) is the rate constant.

From the perspective of analysis and calculations, zero order kinetics can seem counterintuitive because we might expect that more reactants would naturally lead to a faster reaction. However, due to specific conditions like catalytic surfaces or saturated reactant sites, increasing concentration doesn't impact the rate, which is precisely what students need to learn when faced with such reactions.
Rate Law Equation
When talking about the pace at which a chemical reaction occurs, the rate law equation is an indispensable tool. It's like a recipe that tells you how the amount of ingredients (reactants) affects the speed of cooking (reaction rate). The rate law links the reaction rate to the concentrations of reactants, often in a power relationship, and is usually expressed as \( rate = k[A]^m[B]^n \), where \( [A] \) and \( [B] \) are the concentrations of reactants, and \( m \) and \( n \) represent the reaction orders with respect to each reactant.

The value of \( k \) is the rate constant, a direct indicator of how quickly a reaction proceeds. It varies with temperature but is independent of reactant concentrations. The total reaction order is the sum of \( m+n \). For students grappling with rate laws, remembering that this is an empirical equation—determined experimentally rather than theoretically—is key. And it's crucial to note that reaction orders aren't always integers; they can be fractions, zero, or even negative, reflecting the complex nature of chemical processes.
Chemical Kinetics
Chemical kinetics is akin to the study of motion, but instead of cars and planes, we're looking at atoms and molecules in a chemical reaction. It delves into the rates of chemical processes and the factors that influence them. Factors like temperature, concentration, surface area, and catalysts can all change how quickly a reaction occurs.

One fundamental principle students should grasp is that chemical kinetics gives insight into the reaction mechanism—the step-by-step path from reactants to products. The kinetics study helps chemists to understand whether a reaction occurs in a single step or multiple steps, each with its own speed.

Furthermore, kinetics plays a pivotal role in designing chemical processes, producing medicines, and even in environmental modeling to understand pollutant behavior. Mastering the basics of chemical kinetics equips students with the tools to predict and control the outcomes of chemical reactions, an essential skill in scientific research and many industrial applications.

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Most popular questions from this chapter

For a reaction \(2 \mathrm{~A}+\mathrm{B}+3 \mathrm{C} \rightarrow \mathrm{D}+3 \mathrm{E}\), the following date is obtained: $$ \begin{array}{ccccc} \hline \text { Reaction } & \multicolumn{2}{c} {\text { Concentration in }} & \text { Initial rate of } \\ & \multicolumn{2}{c} {\text { mole per litre }} & \text { formation of } \\ & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D}\left(\text { torr } \mathbf{s}^{-1}\right) \\ \hline 1 & 0.01 & 0.01 & 0.01 & 2.5 \times 10^{-4} \\ 2 & 0.02 & 0.01 & 0.01 & 1.0 \times 10^{-3} \\ 3 & 0.01 & 0.02 & 0.01 & 2.5 \times 10^{-4} \\ 4 & 0.01 & 0.02 & 0.02 & 5.0 \times 10^{-4} \\ \hline \end{array} $$ The order with respect to \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are, respectively, (a) \(0,1,2\) (b) \(2,0,1\) (c) \(1,0,2\) (d) \(2,1,1\)

As the initial concentration increases from \(0.75\) to \(1.55 \mathrm{M}\) in a reaction, \(t_{1 / 2}\) decreases from 60 to \(29 \mathrm{~s}\). The order of the reaction is (a) zero (b) first (c) second (d) third

Iodide ion is oxidized to hypoiodite ion, \(\mathrm{IO}^{-}\), by hypochlorite ion, \(\mathrm{ClO}^{-}\), in basic solution as: $$ \begin{array}{ccccc} & \mathbf{I}^{-} & \mathbf{C l O}^{-} & \mathbf{O H}^{-} & \left(\mathrm{mol} \mathbf{L}^{-1} \mathbf{s}^{-1}\right) \\ \hline 1 & 0.010 & 0.020 & 0.010 & 12.2 \times 10^{-2} \\ 2 & 0.020 & 0.010 & 0.010 & 12.2 \times 10^{-2} \\ 3 & 0.010 & 0.010 & 0.010 & 6.1 \times 10^{-2} \\ 4 & 0.010 & 0.010 & 0.020 & 3.0 \times 10^{-2} \\ \hline \end{array} $$ The correct rate law for the reaction is (a) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]^{0}\) (b) \(r=K\left[\mathrm{I}^{-}\right]^{2}\left[\mathrm{ClO}^{-}\right]^{2}\left[\mathrm{OH}^{-}\right]^{0}\) (c) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]\) (d) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]^{-1}\)

A zero-order reaction \(\mathrm{A} \rightarrow \mathrm{B}\). At the end of \(1 \mathrm{~h}, \mathrm{~A}\) is \(75 \%\) reacted. How much of it will be left unreacted at the end of \(2 \mathrm{~h}\). (a) \(12.5 \%\) (b) \(6.25 \%\) (c) \(3.12 \%\) (d) \(0 \%\)

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}\), decomposes according to the following equation: \(\mathrm{CH}_{3}-\mathrm{N} \equiv \mathrm{N}-\mathrm{CH}_{3}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g})\) The initial concentration of azomethane was \(1.50 \times 10^{-2}\) M. After 10 min, the concentration decreased to \(1.29 \times 10^{-2} \mathrm{M}\). The average rate of reaction during this time interval is (a) \(3.5 \times 10^{-6} \mathrm{Ms}^{-1}\) (b) \(2.1 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3.5 \times 10^{-6} \mathrm{M} \mathrm{h}^{-1}\) (d) \(2.1 \times 10^{-3} \mathrm{Mmin}^{-1}\)

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