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Chapter 13: Problem 7

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of \(0.066 M / \mathrm{s}\). (a) At what rate is \(\mathrm{NO}_{2}\) being formed? (b) At what rate is molecular oxygen reacting?

Short Answer

Expert verified
(a) The rate of formation of NO2 is 0.066 M/s. (b) The rate of reaction of O2 is 0.033 M/s.

Step by step solution

01

Calculate the rate of formation of NO2

In the reaction, the molar ratio of NO2 to NO is 2:2, i.e., for every 2 moles of NO that react, 2 moles of NO2 are formed. Hence, the rate of formation of NO2 is the same as the rate of reaction of NO. Therefore, the rate of formation of NO2 is also 0.066 M/s.
02

Calculate the rate of reaction of O2

The molar ratio of O2 to NO is 1:2, i.e for every 2 moles of NO that react, 1 mole of O2 is consumed. Therefore, the rate of reaction of O2 is half the rate of reaction of NO. Thus, the rate of reaction of O2 is \(0.066 M/s ÷ 2 = 0.033 M/s.\)

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Most popular questions from this chapter

Write the Arrhenius equation and define all terms.

Cyclobutane decomposes to ethylene according to the equation $$ \mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(g) $$ Determine the order of the reaction and the rate constant based on the following pressures, which were recorded when the reaction was carried out at \(430^{\circ} \mathrm{C}\) in a constant-volume vessel. $$ \begin{array}{rc} \hline \text { Time (s) } & P_{\mathrm{C}_{4} \mathrm{H}_{8}}(\mathrm{mmHg}) \\\ \hline 0 & 400 \\ 2,000 & 316 \\ 4,000 & 248 \\ 6,000 & 196 \\ 8,000 & 155 \\ 10,000 & 122 \\ \hline \end{array} $$

Explain what is meant by the rate law of a reaction.

The rate constant for the second-order reaction $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M ?\)

The rate law for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ is given by rate \(=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] .\) (a) What is the order of the reaction? (b) A mechanism involving the following steps has been proposed for the reaction: $$ \begin{array}{c} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOCl}(g) \end{array} $$ If this mechanism is correct, what does it imply about the relative rates of these two steps?
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