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Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2},(\mathrm{~b})\) rate \(=k\), (c) rate \(=k\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{Br}_{2}\right]^{1 / 2}\) (d) rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

Short Answer

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(a) 2, (b) 0, (c) 2.5, (d) 3

Step by step solution

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01

Understand Reaction Order

The reaction order indicates how the rate is affected by the concentration of reactants. It is the sum of the powers of the concentration terms in the rate law.
02

Analyze Rate Law for Part (a)

For rate law rate \( = k[\mathrm{NO}_{2}]^{2} \), only one reactant \( [\mathrm{NO}_{2}] \) appears with a power of 2. This means the reaction order is simply \( 2 \).
03

Analyze Rate Law for Part (b)

For rate law rate \( = k \), there are no concentration terms. This indicates a zero-order reaction, as the rate does not depend on the concentration of any reactant.
04

Analyze Rate Law for Part (c)

In the rate law rate \( = k [\mathrm{H}_{2}]^{2} [\mathrm{Br}_{2}]^{1/2} \), the order with respect to \( [\mathrm{H}_{2}] \) is 2, and with respect to \( [\mathrm{Br}_{2}] \) is \( 1/2 \). The overall reaction order is \( 2 + 1/2 = 2.5 \).
05

Analyze Rate Law for Part (d)

For rate law rate \( = k [\mathrm{NO}]^{2} [\mathrm{O}_{2}] \), the order with respect to \( [\mathrm{NO}] \) is 2 and with respect to \( [\mathrm{O}_{2}] \) is 1. The overall reaction order is \( 2 + 1 = 3 \).

Key Concepts

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

Rate Law
A rate law is an equation that shows how the rate of a chemical reaction depends on the concentration of its reactants. It typically takes the form: \[ \text{rate} = k[A]^x[B]^y... \]where:
  • rate is the speed at which the reaction occurs.
  • k is the rate constant, a value that changes with temperature and different reactions.
  • [A] and [B] are the concentrations of the reactants.
  • x and y are the reaction orders with respect to each reactant.
In any given rate law, increasing the concentration of a reactant with a positive reaction order increases the rate of reaction. Each exponent in the rate law indicates how much the concentration of that reactant affects the rate.
Concentration of Reactants
The concentration of reactants is a crucial factor in chemical kinetics and determining how quickly a reaction proceeds. Concentration is typically measured in molarity (M), which represents the number of moles of a solute per liter of solution. When the concentration of reactants changes, it can significantly impact the reaction rate:
  • An increase in concentration usually leads to more frequent collisions between reactant molecules, thereby speeding up the reaction.
  • If a reactant's order is zero, changes in its concentration do not affect the overall rate.
Understanding how reactant concentration affects rate is vital in controlling reactions in industrial processes and laboratory settings.
Zero-Order Reaction
In a zero-order reaction, the rate is independent of the concentration of the reactants. This means that the rate remains constant as long as there is some amount of reactant present. A typical rate law for a zero-order reaction looks like this:\[ \text{rate} = k \]where the rate continues at a constant pace determined by k (rate constant), regardless of the concentrations of the reactants. Zero-order reactions are less common and often observed where a catalyst or specific surface area limits the reaction. They provide valuable information about processes where only catalysts or other non-concentration factors affect the rate.
Overall Reaction Order
The overall reaction order is the sum of the powers of the concentration terms in a rate law. It depicts the influence of all reactants' concentrations on the reaction rate as a whole. Calculating it involves adding up the individual orders, like so:
  • For rate law \( \text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n \), the overall reaction order is \( m+n \).
The overall reaction order helps chemists to understand complex reactions and predict how changes in reactant concentrations will affect the rate. High-order reactions tend to be sensitive to concentration changes, while low-order ones are less so. Understanding this can be critical in industries where precise reaction rates define the quality and yield of a product.

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Most popular questions from this chapter

Write the equation relating the half-life of a secondorder reaction to the rate constant. How does it differ from the equation for a first-order reaction?

Specify which of the following species cannot be isolated in a reaction: activated complex, product, intermediate.

The rate constants of some reactions double with every \(10^{\circ}\) rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

The rate at which tree crickets chirp is \(2.0 \times 10^{2}\) per minute at \(27^{\circ} \mathrm{C}\) but only 39.6 per minute at \(5^{\circ} \mathrm{C}\). From these data, calculate the "activation energy" for the chirping process. (Hint: The ratio of rates is equal to the ratio of rate constants.)

Polyethylene is used in many items, including 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 together. (Ethylene is the basic unit, or monomer, for polyethylene.) The initiation step is: \(\mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot\) (initiation) The \(\mathrm{R}\). species (called a radical) reacts with an ethylene molecule (M) to generate another radical: $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} $$ The reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: \(\mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad\) (propagation) This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine: $$ \mathbf{M}^{\prime} \cdot+\mathbf{M}^{\prime \prime} \cdot \stackrel{k_{\mathrm{t}}}{\longrightarrow} \mathbf{M}^{\prime}-\mathbf{M}^{\prime \prime} \quad \text { (termination) } $$ The initiator frequently used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]\) : $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} $$ 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 preceding 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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