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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 \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)

Short Answer

Expert verified
The rate equation is (a) rate \( = k[\text{B}_2]\).

Step by step solution

01

Understand the Experimental Data

We have a reaction \(2\, \text{A} + \text{B}_2 \rightarrow 2\, \text{AB}\), and an experimental data table with concentrations and rates provided. We need to find the rate equation from the given options.
02

Establish the General Form of the Rate Law

The general rate law can be expressed as \(\text{Rate} = k[\text{A}]^m[\text{B}_2]^n\), where \(m\) and \(n\) are the reaction orders we need to determine.
03

Compare Experimental Data 1 and 2

In trials 1 and 2, the concentration of \([A]\) remains constant at 0.50 M, whereas \([B_2]\) doubles from 0.50 M to 1.00 M. The rate also doubles from \(1.6 \times 10^{-4} \text{ M/s}\) to \(3.2 \times 10^{-4} \text{ M/s}\). This suggests a first-order reaction with respect to \([B_2]\) (\(n = 1\)).
04

Compare Experimental Data 2 and 3

In trials 2 and 3, the concentration of \([B_2]\) remains constant at 1.00 M, while \([A]\) doubles from 0.50 M to 1.00 M. However, the rate does not change and remains \(3.2 \times 10^{-4} \text{ M/s}\). This indicates that \([A]\) does not affect the rate, or \(m = 0\).
05

Formulate the Rate Law

Given the findings from Steps 3 and 4, the rate law is \(\text{Rate} = k[\text{B}_2]^1\). Thus, the correct rate equation is option (a): rate \( = k[\text{B}_2]\).

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Key Concepts

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

Rate Law
In the field of chemical kinetics, understanding the Rate Law is crucial. The Rate Law is an equation that expresses the rate of a chemical reaction as a function of the concentration of its reactants. In a general format, the Rate Law can be written as:
  • \[\text{Rate} = k[A]^m[B]^n \]
Here, \(k\) is the rate constant, \([A]\) and \([B]\) are the concentrations of reactants, and \(m\) and \(n\) represent the orders of the reaction with respect to each reactant.
The rate constant \(k\) is a proportionality constant that remains unchanged with concentration changes but varies with temperature and other factors. By analyzing experimental data, the Rate Law helps in determining how the concentration of each reactant affects the overall rate of the reaction.
Reaction Order
The concept of Reaction Order is key to understanding how reactants influence the speed of a chemical reaction. Simply put, Reaction Order is the power to which the concentration of a reactant is raised in the Rate Law. This can be done for each reactant individually, leading to terms known as 'partial orders.'
Overall, the reaction order is the sum of these partial orders. For instance, if a reaction has a Rate Law of
  • \[\text{Rate} = k[A]^1[B]^2 \]
Then, the reaction is first-order with respect to \([A]\) and second-order with respect to \([B]\). Therefore, the overall reaction order is \(1 + 2 = 3\), making it a third-order reaction.
The order of reaction dictates how altering the concentrations of reactants affects the reaction rate. First order reactions change rate linearly with concentration change, while second order and higher can see quadratic or more complex changes.
Experimental Data Analysis
Understanding how to analyze Experimental Data is essential in determining the Rate Law and Reaction Order. In experiments, data is usually collected by varying the concentrations of reactants and measuring the corresponding reaction rates. From this collected data, we can deduce the values of \(m\) and \(n\), which tell us the relationship each reactant has with the reaction rate.
  • Start by observing one reactant while keeping others constant. See how the rate changes as you alter its concentration. This helps find its reaction order.
  • Then, repeat the process as you evaluate other reactants.
By carefully analyzing changes, it is possible to formulate the correct Rate Law. For example, if doubling the concentration of a reactant causes the rate to double, the reaction is first order in that reactant. This was observed with \([B_2]\) in the given experiment, confirming its first-order nature. Such detailed analysis allows chemists to predict how systems behave with varying concentrations and is invaluable for controlling reactions in industrial applications.

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Most popular questions from this chapter

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 \(=\mathrm{k}\left[\mathrm{NOCl}_{2}\right][\mathrm{NO}]\)

Identify the correct statements: (a) The order of an elementary reaction is equal to its molecularity (b) The order of a reaction can be zero (c) For second order reaction, order of reaction \(=2 \times\) molecularity. (d) The order of inversion of cane sugar is 2 .

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) \(2.303 \mathrm{RS}\)

The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)

A graph plotted between concentration of reactant, consumed at any time \((\mathrm{x})\) and time ' \(\mathrm{t}\) ' is found to be a straight line passing through the origin. The reaction is of (a) first-order (b) zero-order (c) third-order (d) second-order

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