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Chapter 13: Problem 136

Assume that the formation of nitrogen dioxide: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the condition described in (b), the half-life of the reaction has been estimated to be \(6.4 \times 10^{3}\) min. What would be the half-life if the initial concentration of NO were \(10 \mathrm{ppm} ?\)

Short Answer

Expert verified
The rate law for the reaction is \(Rate = k[NO]^2[O_{2}]\). Given the minimal concentration of NO, it turns into a pseudo-first order reaction, simplifying the rate law to \(Rate = k'[O_{2}]\). Despite a change in the initial concentration of NO, the half-life would remain constant at \(6.4 \times 10^{3}\) min due to the nature of a first order reaction.

Step by step solution

01

Write the Rate Law

Based on the elementary reaction provided \(2 NO(g) + O_{2}(g) -> 2 NO_{2}(g)\), the rate law is given by the equation: Rate = k[NO]^2[O_{2}], where k is the rate constant, and [NO] and [O_{2}] represent the concentrations of NO and O_{2} respectively.
02

Simplifying The Rate Law

Considering a contamination of 2.0 ppm of NO by volume, this tiny concentration essentially means that [NO] remains almost constant and doesn't change drastically compared to [O_{2}]. As such, the reaction order in NO becomes pseudo-zero, which simplifies the rate law to: Rate = k'[O_{2}], with k' being a new constant incorporating the original k along with [NO]^2.
03

Half-Life With Altered Initial Concentration

The question states that with an initial concentration of 2.0 ppm of NO, the half-life is \(6.4 \times 10^{3}\) min. However, since this reaction now behaves as pseudo-first order (with respect to O_{2}), the half-life of a first order reaction is unaffected by initial concentrations. Therefore, even if the initial concentration would be increased to 10 ppm, the half-life would remain unchanged at \(6.4 \times 10^{3}\) min.

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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 is crucial in chemical kinetics as it links the speed of a reaction, known as the reaction rate, to the concentration of reactants. The rate law for a given reaction can be expressed with the equation:
\( \text{Rate} = k[\text{A}]^m[\text{B}]^n \),
where \( k \) is the rate constant, \( m \) and \( n \) are the orders of reaction with respect to reactants A and B, respectively. In the exercise, the rate law for the elementary reaction involving nitrogen oxide and oxygen is determined to be \( \text{Rate} = k[\text{NO}]^2[O_2] \), indicating that the reaction rate depends on the concentration of NO squared and the concentration of O2.
Elementary Reactions
Elementary reactions are reactions that occur in a single step and represent a basic process where reactants convert directly into products. As they occur in a single event, the stoichiometry of the reaction directly determines the reaction order. For example, in the formation of nitrogen dioxide, \( 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g) \), the stoichiometry is used to derive the rate law, leading to a second-order dependence on NO and a first-order on O2.
Half-life
The concept of half-life is pivotal when studying the decay of substances or the progress of a reaction. It is defined as the time taken for the concentration of a reactant to reduce to half its initial value. For first-order reactions, the half-life is constant and independent of the initial concentration. This invariant nature is illustrated in our reaction, where despite the increase in NO concentration from 2.0 ppm to 10 ppm, the half-life remains unchanged for the oxygen-dependent reaction rate.
Reaction Order
The reaction order indicates how the rate of the reaction responds to changes in the concentration of reactants. In the given example, the rate law initially suggests a second-order reaction with respect to NO and first-order with respect to O2. However, due to the low concentration of NO, the reaction order with respect to NO is approximated as zero (pseudo-zero order), leading to a simplification in the rate law. This approximation is useful for reactions where one reactant's concentration is significantly lower and remains nearly constant through the reaction.
Concentration Dependence
Concentration dependence is at the heart of reaction rates and implies that the rate at which a reaction occurs depends on the molar concentration of its reactants at any given moment. In the context of the presented reaction, originally, the rate depends on both NO and O2 concentrations. However, if the concentration of NO stays relatively constant during the reaction, as it does with low ppm levels, it's no longer a deciding factor for the rate, simplifying the rate law to only include the concentration of O2.

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Most popular questions from this chapter

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

Is the rate constant \((k)\) of a reaction more sensitive to changes in temperature if \(E_{\mathrm{a}}\) is small or large?

The following expression shows the dependence of the half-life of a reaction \(\left(t_{\frac{1}{2}}\right)\) on the initial reactant concentration [A] \(_{0}:\) $$ t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{~A}]_{0}^{n-1}} $$ where \(n\) is the order of the reaction. Verify this dependence for zero-, first-, and second-order reactions.

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for A and 18.0 min for B. If the concentrations of \(A\) and \(B\) are equal initially, how long will it take for the concentration of A to be four times that of \(\mathrm{B}\) ?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$
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