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Chapter 14: Problem 82

Consider this elementary step: $$ X+2 Y \longrightarrow X Y_{2} $$ (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{XY}_{2}\) is \(3.8 \times 10^{-3} \mathrm{M} / \mathrm{s}\) and the initial concentrations of \(X\) and \(Y\) are \(0.26 M\) and \(0.88 M\) what is the rate constant of the reaction?

Short Answer

Expert verified
(a) Rate law: \(r = k [X]^1 [Y]^2\). (b) Rate constant: \(1.887 \times 10^{-2} \mathrm{M}^{-2} \mathrm{s}^{-1}\).

Step by step solution

01

Identify the Rate Law

For the elementary reaction \(X + 2Y \longrightarrow XY_2\), the rate law can be directly derived from the stoichiometry of the reactants. The rate \(r\) is proportional to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient. Thus, the rate law is given by \( r = k [X]^1 [Y]^2 \), where \(k\) is the rate constant.
02

Set Up the Equation for Initial Rate

Use the given initial rate of formation for \(XY_2\) which is \(3.8 \times 10^{-3} \mathrm{M/s}\). Substitute the known concentrations into the rate law: \( 3.8 \times 10^{-3} = k [0.26]^1 [0.88]^2 \).
03

Calculate the Rate Constant

Calculate \( [0.26]^1 = 0.26 \) and \( [0.88]^2 = 0.7744 \), then plug them into the equation to solve for \(k\): \[ 3.8 \times 10^{-3} = k \times 0.26 \times 0.7744 \]. Now, solve for \(k\): \[ k = \frac{3.8 \times 10^{-3}}{0.26 \times 0.7744} \].
04

Solve for the Value of k

Perform the division to find \(k\): \[ k = \frac{3.8 \times 10^{-3}}{0.201344} \approx 1.887 \times 10^{-2} \mathrm{M}^{-2} \mathrm{s}^{-1} \].

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Key Concepts

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

Rate Constant
In a chemical reaction, the rate constant, often denoted by "k," is a proportionality factor in the rate law that links the reaction rate to the concentrations of the reactants. It is unique for each reaction and dependent on temperature. Understanding the rate constant is essential, as it provides insight into the speed of a reaction under specific conditions.
The rate constant demonstrates how quickly reactants are converted to products. In the rate law expression, given as: \[ \text{Rate} = k [X]^m [Y]^n \]"k" quantifies how the rate is affected by the concentration of the reactants.
  • The units of "k" vary depending on the overall order of the reaction.
  • For the given reaction \(X + 2Y \to XY_2\), "k" is determined to be \(1.887 \times 10^{-2} \text{ M}^{-2} \text{s}^{-1}\), indicating a second-order dependence on both X and Y.
  • The value of "k" can give insights into the activation energy required and the efficiency of the reaction under specified conditions.
Concentration
Concentration refers to the abundance of a constituent divided by the total volume of a mixture. In chemical reactions, it usually represents the amount of a substance present in a given volume of solution, expressed in molarity (M, moles per liter). Understanding concentration is crucial, as it is one of the key factors influencing the rate of reaction.
  • In the rate law \( \text{Rate} = k [X]^1 [Y]^2 \), concentrations of X and Y affect the initial rate.
  • A higher concentration of reactants can increase the reaction rate. For instance, with initial concentrations of 0.26 M for X and 0.88 M for Y, the reaction proceeds at a specific rate.
  • Changes in concentration lead to shifts in equilibrium and rate, emphasizing its importance in reaction dynamics.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships or ratios between reactants and products in a chemical reaction. It is essential for writing balanced equations and determining amounts of subjects needed or produced. For the given elementary reaction \(X + 2Y \rightarrow XY_2\), stoichiometry tells us that one mole of X reacts with two moles of Y to produce one mole of \(XY_2\).
  • The stoichiometric coefficients in the reaction (1 for X and 2 for Y) become the powers in the rate law expression, \( \text{Rate} = k [X]^1[Y]^2 \).
  • These coefficients indicate how changes in concentrations affect the rate of the reaction.
  • Balancing stoichiometry is vital for accurately predicting reaction outcomes and yields.
Initial Rate
The initial rate of a reaction refers to the change in concentration of a reactant or product per unit time at the start of the reaction. It provides critical information about how fast the reaction begins.
For the reaction \(X + 2Y \rightarrow XY_2\), the initial rate of formation of \(XY_2\) is given as \(3.8 \times 10^{-3} \text{ M/s}\). This serves as a useful parameter in determining the rate constant and understanding first interactions among reactants.
  • The initial rate is used to derive the rate law, by considering initial concentrations and comparing to the rate of reaction observed.
  • It helps in constructing the reaction mechanism hypotheses and understanding early-stage dynamics.
  • Initial rates are foundational in kinetic studies, illustrating how reactant concentrations impact the beginning of the reaction.

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Most popular questions from this chapter

Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the rate drops off abruptly at a certain temperature. Account for this behavior.

What are the units of the rate of a reaction?

Can you suggest two reactions that are very slow (take days or longer to complete) and two reactions that are very fast (are over in minutes or seconds)?

For each of these pairs of reaction conditions, indicate which has the faster rate of formation of hydrogen gas: (a) sodium or potassium with water, (b) magnesium or iron with \(1.0 \mathrm{M} \mathrm{HCl}\), (c) magnesium rod or magnesium powder with \(1.0 \mathrm{M} \mathrm{HCl}\) (d) magnesium with \(0.10 \mathrm{M} \mathrm{HCl}\) or magnesium with \(1.0 M \mathrm{HCl}\)

Polyethylene is used in many items such as water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules (the basic unit is called a monomer) together (see p. 377 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min} .\) (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\) or \(438 \mathrm{~min},\) at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?
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