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Chapter 13: Problem 59

The rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ is \(3.4 \times 10^{-5} \mathrm{s}^{-1}\) at \(298 \mathrm{K}\). What is the rate law expression for the reaction at \(298 \mathrm{K} ?\)

Short Answer

Expert verified
Answer: Rate = 3.4 脳 10^{-5} [N2O5]

Step by step solution

01

Identify the Rate Law Expression

A rate law expression relates the rate of a reaction to the concentration of the reactants and the rate constant. The general form of a rate law expression for a reaction with reactants A and B is: Rate = k [A]^m [B]^n where Rate is the rate of reaction, k is the rate constant, [A] and [B] are the concentrations of reactants A and B, and m and n are their respective orders of reaction. For the given reaction, we have only one reactant, N2O5. Thus, our rate law expression will take the form: Rate = k [N2O5]^m
02

Determine the Order 'm' of the Reaction

The order 'm' of the reaction would be given in the problem or can be deduced from experimental data. However, since no information is provided regarding the order of the reaction, it is safe to assume that the reaction is of first order with respect to N2O5. (A common simplifying assumption if no other data is available). With this assumption, our rate law expression becomes: Rate = k [N2O5]^1
03

Substitute the Rate Constant

Now we plug the given rate constant into the rate law expression: Rate = (3.4 脳 10^{-5} s^{-1}) [N2O5]
04

Express the Rate Law in Final Form

Finally, express the rate law expression in its final form at 298 K: Rate = 3.4 脳 10^{-5} [N2O5] Here, the rate of reaction is directly proportional to the concentration of N2O5, with a proportionality constant equal to the rate constant at 298 K.

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Key Concepts

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

Reaction Order
The concept of reaction order is crucial when discussing rate laws in chemistry. Reaction order refers to the exponent to which the concentration of a reactant is raised in the rate law expression. It determines how the rate is affected by the concentration of reactants.
The reaction order can be:
  • Zero Order: The rate is independent of the concentration of the reactant.
  • First Order: The rate is directly proportional to the concentration of the reactant.
  • Second Order: The rate is proportional to the square of the concentration of the reactant.
For the decomposition of \( \mathrm{N_2O_5} \), we typically assume a first-order reaction unless specified otherwise. This means the rate of reaction depends linearly on the concentration of \( \mathrm{N_2O_5} \).
Rate Constant
The rate constant, represented by \( k \), is a crucial part of the rate law expression. It provides the proportionality factor that relates the rate of reaction to the concentrations of reactants. This constant is specific to a given reaction at a particular temperature.
Several factors influence the rate constant:
  • Temperature: Increases usually result in higher rate constants, as the kinetic energy of molecules increases.
  • Catalysts: Catalysts can alter the rate constant by lowering the activation energy of the reaction.
For the decomposition of \( \mathrm{N_2O_5} \), the given rate constant is \( 3.4 \times 10^{-5} \ \mathrm{s}^{-1} \) at \( 298 \ \mathrm{K} \). This value indicates how fast the reaction proceeds at this specific temperature.
First-Order Reaction
A first-order reaction is one where the rate depends linearly on the concentration of one reactant. In mathematical terms, for a reaction where \( A \rightarrow \text{products} \), the rate law can be described as:\[\text{Rate} = k[A]^1\]This indicates the rate is directly proportional to the concentration of \( A \).
Characteristics of first-order reactions include:
  • The half-life is constant and independent of initial concentration.
  • A plot of the natural logarithm of concentration versus time yields a straight line.
For the reaction \( 2\mathrm{N_2O_5} \rightarrow 4\mathrm{NO_2} + \mathrm{O_2} \), assuming it's first-order with respect to \( \mathrm{N_2O_5} \), simplifies understanding and predicting reaction behavior, especially when experimental data is sparse.

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Most popular questions from this chapter

Hydrogen peroxide decomposes spontaneously into water and oxygen gas via a first-order reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ but in the absence of catalysts this reaction proceeds very slowly. If a small amount of a salt containing the \(\mathrm{Fe}^{3+}\) ion is added to a \(0.437 M\) solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in water, the reaction proceeds with a half-life of 17.3 min. What is the concentration of the solution after 10.0 min under these conditions?

Values of the rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) gas at four temperatures are as follows: $$\begin{array}{cc} T(\mathrm{K}) & k\left(\mathrm{s}^{-1}\right) \\ \hline 658 & 2.14 \times 10^{5} \\ \hline 673 & 3.23 \times 10^{5} \\ \hline 688 & 4.81 \times 10^{5} \\ \hline 703 & 7.03 \times 10^{5} \\ \hline \end{array}$$ a. Determine the activation energy of the decomposition reaction. b. Calculate the value of the rate constant at \(300 \mathrm{K}.\)

On average, someone who falls through the ice covering a frozen lake is less likely to experience anoxia (lack of oxygen \()\) than someone who falls into a warm pool and is underwater for the same length of time. Why?

The rate constant for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{3}(g)+\mathrm{O}_{2}(g) $$ was determined over a temperature range of \(40 \mathrm{K},\) with the following results: $$\begin{array}{cc} T(\mathrm{K}) & k\left(M^{-1} \mathrm{s}^{-1}\right) \\ 203 & 4.14 \times 10^{5} \\ \hline 213 & 7.30 \times 10^{5} \\ \hline 223 & 1.22 \times 10^{6} \\ \hline 233 & 1.96 \times 10^{6} \\ \hline 243 & 3.02 \times 10^{6} \\ \hline \end{array}$$ a. Determine the activation energy for the reaction. b. Calculate the rate constant of the reaction at \(300 \mathrm{K}.\)

Why can't an elementary step in a mechanism have a rate law that is zero order in a reactant?
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