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In a hypothetical reaction given below $$ 2 \mathrm{XY}_{2}(\mathrm{aq})+2 \mathrm{Z}^{-}(\mathrm{aq}) \rightarrow $$ (Excess) $$ 2 \mathrm{XY}_{2}^{-}(\mathrm{aq})+\mathrm{Z}_{2}(\mathrm{aq}) $$ \(\mathrm{XY}_{2}\) oxidizes \(\mathrm{Z}\) - ion in aqueous solution to \(\mathrm{Z}_{2}\) and gets reduced to \(\mathrm{XY}_{2}-\) The order of the reaction with respect to \(\mathrm{XY}_{2}\) as concentration of \(Z\) - is essentially constant. Rate \(=\mathrm{k}\left[\mathrm{XY}_{2}\right]^{\mathrm{m}}\) Given below the time and concentration of \(\mathrm{XY}_{2}\) taken (s) Time \(\left(\mathrm{XY}_{2}\right) \mathrm{M}\) \(0.00\) \(4.75 \times 10^{-4}\) \(1.00\) \(4.30 \times 10^{-4}\) \(2.00\) \(3.83 \times 10^{-4}\) The order with respect to \(\mathrm{XY}_{2}\) is a. 0 b. 1 c. 2 d. 3

Short Answer

Expert verified
The order of the reaction with respect to \( \mathrm{XY}_2 \) is 1.

Step by step solution

01

Understand the Rate Law

The rate law of a reaction gives the relationship between the rate of the reaction and the concentration of reactants. For this reaction, the rate is given by \( R = k [\mathrm{XY}_2]^{m} \), where \( m \) is the order of the reaction with respect to \( \mathrm{XY}_2 \). The task is to find \( m \).
02

Initial Concentrations and Rate Calculation

From the problem, determine the initial concentration \( [\mathrm{XY}_2]_0 = 4.75 \times 10^{-4} \) M. If the initial rate is proportional to \( [\mathrm{XY}_2]^{m} \), calculate for different times to see how the concentration changes affect the rate.
03

Use the Method of Initial Rates

Using the concentration data at different times, calculate the change in concentration over time to infer the order. We calculate the initial rate using two concentrations: \( (4.75 - 4.30) \times 10^{-4} \) M over 1 second, and similarly for the subsequent interval. Compare successive rates to the changes in concentration to determine \( m \).
04

Calculate and Simplify

The drop in \( [\mathrm{XY}_2] \) over the first second is \( (4.75 - 4.30) \times 10^{-4} = 0.45 \times 10^{-4} \) M/s. This approximates to \( 0.45 \) M/s. A similar calculation for the second second gives another rate. If these rates are directly proportional to the initial concentration, the reaction is first order.
05

Determine the Order

By comparing the relative changes in rate and concentration, if they are equal (i.e., a constant factor), then \( m = 1 \), indicating a first-order reaction with respect to \( \mathrm{XY}_2 \). The rate decrease is consistent with first-order kinetics.

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Key Concepts

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

Rate Law
Understanding the rate law of a chemical reaction is crucial for identifying how various factors affect the speed of the reaction. The rate law is expressed as an equation:- It provides the relation between the rate of reaction and the concentration of reactants.- For a reaction such as the one given, it is expressed as \( R = k [\mathrm{XY}_2]^{m} \), where \( R \) is the rate, \( k \) is the rate constant, \( [\mathrm{XY}_2] \) is the concentration of reactant \( \mathrm{XY}_2 \), and \( m \) is the order of the reaction with respect to \( \mathrm{XY}_2 \).To determine \( m \), which reveals how the reaction rate changes with varying \( \mathrm{XY}_2 \) concentrations, you can analyze the method of initial rates.
Concentration
Concentration plays a pivotal role in the rate law and overall chemical reaction. In this problem, we focus on how the concentration of \( \mathrm{XY}_2 \) affects the reaction rate:- Initial concentration refers to the amount of \( \mathrm{XY}_2 \) available at the start of the reaction.- Measurements are typically given in moles per liter (Molarity, M), such as \( 4.75 \times 10^{-4} \) M initially in this problem.As the reaction progresses, these concentrations change, providing informative data about the reaction's behavior.
Method of Initial Rates
The method of initial rates is a direct and reliable technique to ascertain the reaction order:- It involves measuring the initial rate of reaction at varying initial concentrations of a reactant, keeping other conditions constant.- By observing how the initial rate varies with the initial concentration, one can deduce the order of reaction with respect to a particular reactant.For this problem:- Consider the change in concentration of \( \mathrm{XY}_2 \) over time intervals.- Calculate how these changes correlate with time to indicate how quickly the reaction proceeds at different concentrations.- If the rate decreases proportionally with concentration, it supports first-order kinetics.
First-Order Kinetics
First-order kinetics describes reactions where the rate is directly proportional to the concentration of one reactant:- The mathematical expression is \( R = k [\mathrm{XY}_2]^1 \).- This means the reaction rate decreases at a constant rate as the reactant is consumed.Using the data:- Evaluate how the rate changes as the concentration of \( \mathrm{XY}_2 \) decreases.- The decrease in reaction rate that directly mirrors the decrease in \( [\mathrm{XY}_2] \) confirms the first-order nature of the reaction.This understanding allows chemists to predict how the reaction progresses over time and adjust conditions to control the reaction speed.

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Most popular questions from this chapter

Observe the reaction given below \(\mathrm{A}+2 \mathrm{~B} \rightarrow 3 \mathrm{C}\) \(\begin{array}{lll}\mathrm{g} & \mathrm{g} & \mathrm{g}\end{array}\) If the rate of this reaction \(-\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) is \(2 \times 10^{-3} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) than the value of \(\frac{\mathrm{dB}}{\mathrm{dt}}\) and \(\frac{\mathrm{dC}}{\mathrm{dt}}\) will be respectively a. \(1 \times 10^{-3}, 2 / 3 \times 10^{-3}\) b. \(4 \times 10^{-3}, 6 \times 10^{-3}\) c. \(6 \times 10^{-3}, 4 \times 10^{-3}\) d. \(2 / 3 \times 10^{-3}, 1 \times 10^{-3}\)

The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH}\) (aq) \(\rightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=\mathrm{k}[\mathrm{RCl}] .\) The rate of the reaction will be a. Doubled on doubling the concentration of sodium hydroxide b. Halved on reducing the concentration of alkyl halide to one half c. Decreased on increasing the temperature of reaction d. Unaffected by increasing the temperature of the reaction.

The calculation of the Arrhenius factor is based on the a. Idea that the reactant species must come together, leading to the formation of the transition state which then transforms into the products b. Idea that, for a reaction to take place, the reactant species must come together c. Calculation of the order of thereaction d. Calculation of the molecularity of the reaction

The reaction of hydrogen and iodine monochloride is given as: \(\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{ICl}(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g})\) This reaction is of first order with respect to \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{ICl}(\mathrm{g})\), following mechanisms were proposed: Mechanism (1): \(\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{ICl}(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g})\) Mechanism (2): \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{HI}_{2}(\mathrm{~g}) ;\) slow \(\mathrm{HI}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~g}) ;\) fast Which of the above mechanism(s) can be consistent with the given information about the reaction? a. 2 only b. Both 1 and 2 c. Neither 1 nor 2 d. 1 only

In a first order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times\) \(10^{4} \mathrm{sec}\). The rate constant of reaction in \(\mathrm{sec}^{-1}\) is a. \(2 \times 10^{4}\) b. \(3.45 \times 10^{-5}\) c. \(1.386 \times 10^{-4}\) d. \(2 \times 10^{-4}\)

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