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Chapter 15: Problem 33

Gaseous NO, decomposes at \(573 \mathrm{K}.\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ The concentration of \(\mathrm{NO}_{2}\) was measured as a function of time. A graph of \(1 /\left[\mathrm{NO}_{2}\right]\) versus time gives a straight line with a slope of \(1.1 \mathrm{L} / \mathrm{mol} \cdot\) s. What is the rate law for this reaction? What is the rate constant?

Short Answer

Expert verified
Rate law: Rate = k [NO2]^2; Rate constant k = 1.1 L/mol路s.

Step by step solution

01

Identify the Order of Reaction

Since a graph of \( \frac{1}{[\mathrm{NO}_2]} \) versus time gives a straight line, this indicates that the reaction follows a second-order rate law with respect to \( \mathrm{NO}_2 \). For second-order reactions, the integrated rate law is linear for the plot \( \frac{1}{[A]} \) versus time.
02

Write the General Rate Law Formula

The general rate law for a second-order reaction can be expressed as \( \text{Rate} = k [\mathrm{NO}_2]^2 \). This equation shows that the rate of reaction is proportional to the square of the concentration of \( \mathrm{NO}_2 \).
03

Determine the Rate Constant

The slope of the line in the plot \( \frac{1}{[\mathrm{NO}_2]} \) versus time is equal to the rate constant \( k \) for a second-order reaction. Therefore, the rate constant \( k \) is \( 1.1 \, \mathrm{L/mol \, s} \).

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Key Concepts

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

Second-order reactions
Understanding second-order reactions involves recognizing that the rate of these reactions is proportional to the square of the concentration of one reactant. In simpler terms, as the concentration of the reactant decreases, the reaction slows down significantly because of this squared relationship. Second-order reactions often appear in systems where two molecules of the same kind collide and react.
In the case of the decomposition of NO鈧�, the reaction follows a second-order rate law. This is because when plotting the reciprocal of the concentration of NO鈧� over time, a straight line results. This linear relationship is a hallmark of second-order reactions and helps chemists identify the reaction order from experimental data.
Integrated rate law
The integrated rate law is a powerful tool used to describe how the concentration of reactants changes over time in a chemical reaction. For second-order reactions, the integrated rate law is specifically: \[ \frac{1}{[A]} = kt + \frac{1}{[A_0]} \] where
  • \([A]\) is the concentration of the reactant at time \(t\).
  • \([A_0]\) is the initial concentration of the reactant.
  • \(k\) is the rate constant.
  • \(t\) is the time elapsed.
This formula reveals that plotting \( \frac{1}{[A]} \) against time yields a straight line for second-order reactions, making it easier to determine both the reaction order and the rate constant from experimental data. The equation helps in converting complex laboratory data into comprehensible quantitative relations, allowing calculations and predictions about reaction progress.
Rate constant
The rate constant, often denoted as \(k\), is a crucial parameter in chemical kinetics, as it quantifies the speed of a reaction. For second-order reactions, the rate constant has units of \(\text{L/mol \; s}\). This is an important distinction because it differs from the units used in first-order (\(\text{s}^{-1}\)) reactions.
In the process of determining \(k\), experimental data is plotted. For second-order reactions specifically, the plot of \( \frac{1}{[\mathrm{NO}_2]} \) versus time will yield a line whose slope is equal to the rate constant. In the given reaction, the slope of this line is \(1.1 \, \text{L/mol \; s}\), directly indicating the value of \(k\). This allows scientists and students alike to understand how quickly or slowly a reaction proceeds under particular conditions, offering insights into reaction mechanisms and rates.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(A \rightarrow 2 B.\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{B}](\mathrm{mol} / \mathrm{L})} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10 -second interval from 0.0 to 40.0 seconds. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to 20.0 seconds. (c) What is the instantaneous rate, \(\Delta[\mathrm{B}] / \Delta \mathrm{t},\) when \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2},\) and the reaction has a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a 1.0 -L. flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K},\) what is the activation energy of the reaction?

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (secondorder) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+\mathrm{O}_{3} \rightarrow \mathrm{ClO}+\mathrm{O}_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \rightarrow \mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \rightarrow \mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \rightarrow \mathrm{HCl}+\mathrm{CHFCl} &3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

Ammonium cyanate, NH_NCO, rearranges in water to give urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}.\) $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ $$\begin{array}{cc}\text { Time (min) } & {\left[\mathrm{NH}_{4} \mathrm{NCO}\right](\mathrm{mol} / \mathrm{L})} \\\\\hline 0 & 0.458 \\\4.50 \times 10^{1} & 0.370 \\\1.07 \times 10^{2} & 0.292 \\\2.30 \times 10^{2} & 0.212 \\\6.00 \times 10^{2} & 0.114 \\\\\hline\end{array}$$ Using the data in the table: (a) Decide whether the reaction is first order or second order. (b) Calculate \(k\) for this reaction. (c) Calculate the half-life of ammonium cyanate under these conditions. (d) Calculate the concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) after 12.0 hours.
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