Derive the relationship between rate of reaction, rate of disappearance of \(X, Y\) and rate of formation of \(X_{2} Y_{2}\) for the reaction : $$ 2 X+3 Y \longrightarrow X_{2} Y_{3} $$

Short Answer

Expert verified
The relationship between the rates is \( -\frac{1}{2}\frac{d[X]}{dt} = -\frac{1}{3}\frac{d[Y]}{dt} = \frac{d[X_2Y_3]}{dt} \.\)

Step by step solution

01

Expressing the Rate of Reaction in Terms of Reactive Disappearance and Product Formation

The rate of a reaction can be expressed in terms of the rate of disappearance of the reactants or the rate of appearance of the products. For the given reaction, if the rate of disappearance of reactant X is represented by \( -\frac{d[X]}{dt} \) and for Y as \( -\frac{d[Y]}{dt} \) (negative sign indicates disappearance), and rate of formation of product \( X_2Y_3 \) is \( \frac{d[X_2Y_3]}{dt} \) then we need to relate these rates to the stoichiometric coefficients in the balanced chemical equation.
02

Incorporating Stoichiometry of the Reaction

According to the stoichiometry of the reaction, \(2 \) moles of \(X\) and \(3\) moles of \(Y\) are required to form \(1\) mole of \(X_2Y_3\). This means that for every mole of \(X_2Y_3\) formed, \(2\) moles of \(X\) are consumed and \(3\) moles of \(Y\) are consumed. Therefore, we can relate the rates using their stoichiometric coefficients:
03

Writing the Rate Expression

The rate of the reaction in terms of the disappearance of reagents and appearance of products is given by: \[\text{Rate of reaction} = -\frac{1}{2}\frac{d[X]}{dt} = -\frac{1}{3}\frac{d[Y]}{dt} = \frac{1}{1}\frac{d[X_2Y_3]}{dt}\] This means that the rate of disappearance of \(X\) is twice as fast as the rate of formation of \(X_2Y_3\), and the rate of disappearance of \(Y\) is three times as fast as the rate of formation of \(X_2Y_3\).
04

Final Relationship Between the Rates

The final relationship between the rates can be expressed as: \[ -\frac{1}{2}\frac{d[X]}{dt} = -\frac{1}{3}\frac{d[Y]}{dt} = \frac{d[X_2Y_3]}{dt} \.\] These are the required relationships for the rate of the reaction in terms of the rate of disappearance of \(X\) and \(Y\), and the rate of formation of \(X_2Y_3\).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is a field of chemistry that deals with understanding the speed or rate at which a chemical reaction occurs. It involves studying how different variables such as concentration, temperature, and catalysts impact the reaction rate. A reaction rate can be defined as the change in concentration of a reactant or product per unit time.

For example, if we consider the reaction provided, \(2 X + 3 Y \longrightarrow X_{2}Y_{3}\), chemical kinetics would allow us to determine how quickly the reactants are getting consumed and how fast the product is being formed over time. By studying these rates, chemists can optimize conditions to increase the efficiency of reactions for industrial applications, understand reaction mechanisms, and predict the behavior of chemical systems.

Understanding the kinetics of a chemical reaction is fundamental in fields such as pharmaceuticals, environmental science, and energy conversion, as it can vastly affect the outcome and practicality of chemical processes.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between the amounts of reactants and products in a chemical reaction. It is based on the principle that during a chemical reaction, atoms are neither created nor destroyed—they simply rearrange. The stoichiometry of a reaction gives us the ratios of the amount of each substance involved.

In the exercise's reaction, \(2 X + 3 Y \longrightarrow X_{2}Y_{3}\), stoichiometry tells us that two moles of \(X\) will react with three moles of \(Y\) to produce one mole of \(X_{2}Y_{3}\). When determining the rate of reaction, this stoichiometric relationship is crucial as it allows us to express the rate at which reactants are consumed and products are formed in terms of each other. This balanced reaction equation is a tool for predicting how much of each substance is needed or produced, which is vital in chemical manufacturing and research settings.
Differential Rate Expressions
Differential rate expressions relate the instantaneous rate of reaction to the concentrations of reactants and products. They are derived from the stoichiometry of the balanced chemical equation and are represented mathematically as derivatives with respect to time.

For instance, the rate of disappearance of \(X\) in the provided reaction is written as \(-\frac{d[X]}{dt}\), whereas the rate of appearance of product \(X_{2}Y_{3}\) is described by \(\frac{d[X_{2}Y_{3]}{dt}\). The negative sign in the rate of disappearance denotes that the reactant's concentration is decreasing over time.

The differential rate expression becomes a powerful tool when combined with the stoichiometry of the reaction. It allows us to establish a direct quantitative connection between the rate at which reactants are used up and the rate at which products are formed. In this reaction, the rate at which \(X\) disappears is twice the rate of formation of \(X_{2}Y_{3}\), which directly relates to the stoichiometric coefficient of \(X\) being twice as large as that of \(X_{2}Y_{3}\).

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Most popular questions from this chapter

The reaction; \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\), shows an increase in concentration of \(\mathrm{NO}_{2}\) by \(20 \times 10^{-3}\) mol litre \(^{-1}\) in 5 second. Calculate (a) rate of appearance of \(\mathrm{NO}_{2}\), (b) rate of reaction and (c) rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) takes place according to I order as $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ Calculate : (a) The rate constant, if instantaneous rate is \(1.4 \times 10^{-6}\) mol litre \(^{-1}\) \(\mathrm{sec}^{-1}\) when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.04 \mathrm{M}\) (b) The rate of reaction when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(1.20 \mathrm{M}\) (c) The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when the rate of reaction will be \(2.45 \times 10^{-5} \mathrm{~mol}\) litre \(^{-1} \mathrm{sec}^{-\mathrm{i}}\).

The rate constant for a first order reaction was found to be \(0.082\) \(\min ^{-1}\). If initial concentration of reactant is \(0.15\) M.how long would it take, (a) to reduce the concentration of \(A\) to \(0.03 M\). (b) to reduce the concentration of \(A\) by \(0.03 M\)

The rate for the decomposition of \(\mathrm{NH}_{3}\) on platinum surface is zero order. What are the rate of production of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) if \(K=2.5 \times 10^{-4}\) mol litre \(^{-1} \mathrm{~s}^{-1}\).

A reaction is second order with respect to a reaction. How is the rate of reaction affected if the concentration of the reactant is : (a) doubled, (b) reduced to \(1 / 2 ?\)

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