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Chapter 13: Problem 59

In the presence of water, NO and \(\mathrm{NO}_{2}\) react to form nitrous acid (HNO,) by the following reaction: $$\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{HNO}_{2}(a q)$$ When the concentration of NO or \(\mathrm{NO}_{2}\) is doubled, the initial rate of reaction doubles. If the rate of the reaction does not depend on \(\left[\mathrm{H}_{2} \mathrm{O}\right],\) what is the rate law for this reaction?

Short Answer

Expert verified
Question: Determine the rate law for the reaction based on the given information about the effect of concentrations on the initial rate of the reaction. Answer: The rate law for the reaction is rate = k[NO][NO2].

Step by step solution

01

Understand the effect of concentration changes on the rate of reaction

The exercise tells us that when the concentration of NO or NO2 is doubled, the initial rate of the reaction doubles. This information will help us establish the order of the reaction with respect to NO and NO2.
02

Determine the order of the reaction with respect to NO and NO2

Let's first consider the effect of doubling the concentration of NO while keeping the concentration of NO2 constant: rate1 = k[NO]^m[NO2]^n rate2 = k(2[NO])^m[NO2]^n Since rate2 = 2*(rate1), we can write the equation as: 2*(k[NO]^m[NO2]^n) = k(2[NO])^m[NO2]^n Dividing both sides by k[NO]^m[NO2]^n, we get: 2 = 2^m ==> m = 1 Now let's consider the effect of doubling the concentration of NO2 while keeping the concentration of NO constant: rate1 = k[NO]^m[NO2]^n rate2 = k[NO]^m(2[NO2])^n Again, since rate2 = 2*(rate1), we can write the equation as: 2*(k[NO]^m[NO2]^n) = k[NO]^m(2[NO2])^n Dividing both sides by k[NO]^m[NO2]^n, we get: 2 = 2^n ==> n = 1
03

Write the rate law

Now that we know the reaction order with respect to NO (m) and NO2 (n), we can write the rate law for the reaction: rate = k[NO]^m[NO2]^n Replace m and n with the values obtained in Step 2: rate = k[NO]^1[NO2]^1 So the rate law for the reaction is: rate = k[NO][NO2]

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Key Concepts

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

Rate Law
The rate law of a chemical reaction gives us an idea of how the concentration of reactants affects the speed, or rate, of the reaction. It is essentially an expression that relates the rate of a chemical reaction to the concentration of its reactants. In this exercise, the rate law for the given reaction is expressed as:\[\text{rate} = k[\text{NO}][\text{NO}_2]\]Where:- \( k \) is the rate constant unique to each reaction at a given temperature.- \([\text{NO}]\) and \([\text{NO}_2]\) are the concentrations of the reactants.This tells us that the reaction rate changes in direct proportion to the concentrations of NO and NO鈧�. This specific relationship means that the rate of reaction will increase as the concentration of these reacting substances increases.
Reaction Order
The reaction order gives insight into how the concentration of each reactant impacts the overall reaction rate. In the exercise, the order is determined from the effect on the reaction rate when the reactant concentrations are changed. For our reaction: - The order with respect to NO is 1, due to the doubling of the reaction rate when the concentration of NO is doubled. - Similarly, the order with respect to NO鈧� is also 1, as doubling the concentration of NO鈧� doubles the rate. Thus, the overall reaction order is the sum of these individual orders, which is 2 (1 for NO + 1 for NO鈧�). This tells us that both NO and NO鈧� equally affect the speed with which products are formed. These simple first-order dependencies on NO and NO鈧� concentrations are crucial for predicting how a change in conditions can change the reaction dynamics.
Effect of Concentration on Reaction Rate
Understanding the effect of concentration is vital for manipulating reaction speeds in both industrial and laboratory settings. The exercise demonstrates that when the concentration of either NO or NO鈧� is increased, the rate of the reaction increases as well. This is consistent with the established rate law and reaction order. In chemical kinetics: - **Higher reactant concentration increases collisions:** More molecules mean a higher likelihood of collisions, hence increasing the reaction rate. - **Doubling the concentration doubles the rate:** In our reaction, the direct proportionality between concentration and rate means if you double either NO or NO鈧�, the rate doubles. Such knowledge can be applied practically, allowing chemists to control reaction conditions to favor faster product formation or to maintain a reaction safely when runaway conditions (explosive reactions) are a risk. This principle holds true as long as the reaction conditions (such as temperature and presence of a catalyst) remain constant.

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Most popular questions from this chapter

Because the units of concentration in the term \(\ln \left([\mathrm{X}] /[\mathrm{X}]_{0}\right)\) cancel out in the integrated rate law for first- order reactions (Equation 13.16 ), molar concentration can be replaced by any concentration term. With gases, for example, partial pressures may be used. The decomposition of phosphine gas \(\left(\mathrm{PH}_{3}\right)\) at \(600^{\circ} \mathrm{C}\) is first order in \(\mathrm{PH}_{3}\) with \(k=0.023 \mathrm{s}^{-1}\) $$4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)$$ If the initial partial pressure of \(\mathrm{PH}_{3}\) is 375 torr, what percent of \(\mathrm{PH}_{3}\) reacts in \(1 \mathrm{min}\) ?

The rate of the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ was determined in three experiments at \(225^{\circ} \mathrm{C} .\) The results are given in the following table: $$\begin{array}{cccc}\text { Experiment } & \left[\mathrm{NO}_{2}\right]_{0}(\mathrm{M}) & [\mathrm{CO}]_{0}(\mathrm{M}) & \begin{array}{c}\text { Initial Rate } \\\\(M / \mathrm{s})\end{array} \\\\\hline 1 & 0.263 & 0.826 & 1.44 \times 10^{-5} \\\\\hline 2 & 0.263 & 0.413 & 1.44 \times 10^{-5} \\\\\hline 3 & 0.526 & 0.413 & 5.76 \times 10^{-5} \\\\\hline\end{array}$$ a. Determine the rate law for the reaction. b. Calculate the value of the rate constant at \(225^{\circ} \mathrm{C}\) c. Calculate the rate of appearance of \(\mathrm{CO}_{2}\) when \(\left[\mathrm{NO}_{2}\right]=[\mathrm{CO}]=0.500 \mathrm{M}\)

Laughing Gas Nitrous oxide ( \(\mathrm{N}_{2} \mathrm{O}\) ) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is a potent greenhouse gas and decomposes slowly to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{N}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ If the plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}\right]\) as a function of time is linear, what is the rate law for the reaction? b. How many half-lives will it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}\) to reach \(6.25 \%\) of its original concentration?

In the following mechanism for NO formation, oxygen atoms are produced by breaking \(\mathrm{O}=\mathrm{O}\) bonds at high temperature in a fast reversible reaction. If \(\Delta[\mathrm{NO}] / \Delta t=k\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2},\) which step in the mechanism is the rate-determining step? $$\begin{aligned} (1)\quad\quad\quad\quad\quad\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{O}(g) \\ (2)\quad\quad\mathrm{O}(g)+\mathrm{N}_{2}(g) & \rightarrow \mathrm{NO}(g)+\mathrm{N}(g) \\ (3)\quad\quad\mathrm{N}(g)+\mathrm{O}(g) & \rightarrow \mathrm{NO}(g) \\ overall \quad \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{NO}(g) \end{aligned}$$

A proposed mechanism for the reaction of \(\mathrm{NO}_{2}(g)\) and \(\mathrm{CO}(g)\) is $$\begin{array}{ll}\text { Step 1 } & 2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{NO}_{3}(g) \quad \text { slow } \\ \text { Step 2 } & \mathrm{NO}_{3}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}_{2}(g)\end{array}$$ a. Write the equation for the overall reaction. b. Write the rate law predicted by the mechanism for the overall reaction. c. Identify the reactants and products of the reaction. d. Identify any intermediates in the reaction.
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