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Chapter 18: Problem 62

Rate of a reaction \(A+B \rightarrow\) products, is given below as a function of different initial concentrations of \(A\) and \(B\) : [1982-4 Marks] 119 \(\begin{array}{lll}{[A](\operatorname{mol} / l)} & {[B](\mathrm{mol} / l)} & \text { Initial rate }(\mathrm{mol} / l / \mathrm{min}) \\ 0.01 & 0.01 & 0.005 \\ 0.02 & 0.01 & 0.010 \\ 0.01 & 0.02 & 0.005\end{array}\) 1

Short Answer

Expert verified
The rate law is \( \text{Rate} = k[A] \).

Step by step solution

01

Write the rate law equation

The rate law for this reaction can be expressed as \( \text{Rate} = k[A]^m[B]^n \), where \( k \) is the rate constant, and \( m \) and \( n \) are the orders of the reaction with respect to reactants \( A \) and \( B \) respectively.
02

Use the experimental data

Using the data provided, set up the equations based on the initial rate and concentrations: 1. \( 0.005 = k (0.01)^m (0.01)^n \) 2. \( 0.010 = k (0.02)^m (0.01)^n \) 3. \( 0.005 = k (0.01)^m (0.02)^n \)
03

Determine the order with respect to A

Divide equation 2 by equation 1: \[ \frac{0.010}{0.005} = \frac{k (0.02)^m (0.01)^n}{k (0.01)^m (0.01)^n} \2 = \left( \frac{0.02}{0.01} \right)^m \m = 1\] Thus, the reaction is first order with respect to \( A \).
04

Determine the order with respect to B

Divide equation 3 by equation 1: \[ \frac{0.005}{0.005} = \frac{k (0.01)^m (0.02)^n}{k (0.01)^m (0.01)^n} \1 = \left( \frac{0.02}{0.01} \right)^n = 0\] Thus, the reaction is zero order with respect to \( B \).
05

Combine the results to write the rate law

Based on the findings from Steps 3 and 4, the rate law is \( \text{Rate} = k[A]^1[B]^0 \). This simplifies to \( \text{Rate} = k[A] \).

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Key Concepts

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

Rate Law
A rate law in chemical kinetics describes how the concentration of reactants affects the rate of a chemical reaction. It is expressed with an equation that includes the rate constant and the concentrations of the reactants raised to some power. This general form can be written as \( \text{Rate} = k[A]^m[B]^n \), where \( k \) is the rate constant, and \( [A] \) and \( [B] \) represent the concentrations of the reactants. The exponents \( m \) and \( n \) depict the reaction order with respect to each reactant.

In simple terms, the rate law helps to predict how changing the amount of each reactant will affect the speed at which the reaction occurs. It is derived from experimental data rather than the stoichiometric coefficients in the balanced equation, which many students often confuse. Understanding the rate law allows scientists to control reaction rates, vital in industrial applications, pharmaceuticals, and other fields.
Reaction Order
Reaction order is a crucial concept in understanding the dynamics of chemical reactions. It indicates the dependency of the reaction rate on the concentration of one or more reactants. The order with respect to a reactant is defined by the exponent in the rate law equation. For instance, in \( \text{Rate} = k[A]^1[B]^0 \), the reaction is first order in \( A \) and zero order in \( B \).

This tells us that the rate of reaction increases linearly with an increase in \( A \), while changes in \( B \) have no effect on the rate. Overall, the reaction order is significant as it provides insight into the mechanism of the reaction and often influences the strategy for controlling the rate of product formation. To determine the reaction order, one must rely on experimental data since it cannot be determined directly from the balanced chemical equation.
Rate Constant
The rate constant, represented by \( k \), is a proportionality factor in the rate law equation. It provides valuable information about the speed of the reaction. Unlike the reaction rate or the concentration, the rate constant remains unaffected by changes in concentrations. It is influenced by temperature and the presence of catalysts.

The units of \( k \) can vary depending on the overall order of the reaction and can help verify the correctness of derived rate laws. For the rate law \( \text{Rate} = k[A] \), the units of \( k \) would be \( \text{L mol}^{-1} \text{min}^{-1} \) for a first-order reaction. Understanding the rate constant is vital as it provides comprehensive insights into the kinetics of a reaction and helps predict the behavior of the reaction under different conditions.

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Most popular questions from this chapter

A catalyst: (a) increases the average kinetic energy of reacting molecules (b) decreases the activation energy (c) alters the reaction mechanism (d) increases the frequency of collisions of reacting species

Under the same reaction conditions, initial concentration of \(1.386 \mathrm{~mol}\) \(\mathrm{dm}^{-3}\) of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio \(\left(k_{1} / k_{0}\right)\) of the rate constant for first order \(\left(k_{1}\right)\) and zero order \(\left(k_{0}\right)\) of the reaction is \(-\) [2008](a) \(0.5 \mathrm{~mol}^{1} \mathrm{dm}^{3}\) (b) \(1.0 \mathrm{~mol} \mathrm{dm}^{3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2.0 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(C^{14}\) concentration is \(C_{1}\) in nearby areas and \(C_{2}\) in areas far away. If the age of the fossil is determined to be \(T_{1}\) and \(T_{2}\) at the respective places then (a) The age of the fossil will increase at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (b) The age of the fossil will decrease at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (c) The age of fossil will be determined to be same (d) \(\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}\)

An experiment requires minimum beta activity product at the rate of 346 beta particles per minute. The half life period of \({ }_{42}^{99} \mathrm{Mo}\), which is a beta emitter is \(66.6\) hours. Find the minimum amount of \({ }_{42}^{99} \mathrm{Mo}\) required to carry out the experiment in \(6.909\) hours.

Complete and balance the following reactions. \({ }_{92} \mathrm{Th}^{234} \longrightarrow \ldots \ldots \ldots . .+7{ }_{2} \mathrm{He}^{4}+6_{-1}^{(i)} \beta^{0}\) (ii) \({ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \longrightarrow \ldots \ldots \ldots .+{ }_{52} \mathrm{Te}^{137}+{ }_{40} \mathrm{Zr}^{92}\) (iii) \({ }_{34} \mathrm{Se}^{86} \longrightarrow 2_{-1} \mathrm{e}^{0}+\ldots \ldots \ldots \ldots .\)
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