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Chapter 11: Problem 13

Using data given in the table for the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}\) calculate the average rate of reaction during each of these intervals: \(\begin{array}{ll}0.50 \mathrm{~h} . & \text { (b) } 0.50 \text { to } 1.0 \mathrm{~h} .\end{array}\) 0.00 to (a) 1.0 to (c) \(2.0 \mathrm{~h}\) (d) 2.0 to \(3.0 \mathrm{~h}\). \(4.0 \mathrm{~h}\) (f) 4.0 to \(5.0 \mathrm{~h}\). (e) 3.0 to \begin{tabular}{cccc} \hline Time (h) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} & Time \((\mathrm{h})\) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} \\ \hline 0.00 & 0.849 & 3.00 & 0.352 \\ 0.50 & 0.733 & 4.00 & 0.262 \\ 1.00 & 0.633 & 5.00 & 0.196 \\ 2.00 & 0.472 & & \\ \hline \end{tabular}

Short Answer

Expert verified
The average rates are 0.232, 0.200, 0.161, 0.120, 0.090, and 0.066 mol/L路h for intervals (a) to (f), respectively.

Step by step solution

01

Understand the Average Rate Formula

The average rate of reaction over a time interval is calculated using the formula: \ \[ \text{Average Rate} = -\frac{\Delta [\text{N}_2\text{O}_5]}{\Delta t} \] \ where \( \Delta [\text{N}_2\text{O}_5] \) is the change in concentration of \( \text{N}_2\text{O}_5 \) and \( \Delta t \) is the change in time over the interval.
02

Calculate Rate for Interval (a) 0.00 to 0.50 h

Use the concentrations at 0.00 h (0.849 mol/L) and 0.50 h (0.733 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.733 - 0.849 = -0.116 \text{ mol/L} \] \ \[ \Delta t = 0.50 - 0.00 = 0.50 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.116}{0.50} = 0.232 \text{ mol/L} \cdot \text{h} \]
03

Calculate Rate for Interval (b) 0.50 to 1.0 h

Use the concentrations at 0.50 h (0.733 mol/L) and 1.00 h (0.633 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.633 - 0.733 = -0.100 \text{ mol/L} \] \ \[ \Delta t = 1.00 - 0.50 = 0.50 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.100}{0.50} = 0.200 \text{ mol/L} \cdot \text{h} \]
04

Calculate Rate for Interval (c) 1.0 to 2.0 h

Use the concentrations at 1.00 h (0.633 mol/L) and 2.00 h (0.472 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.472 - 0.633 = -0.161 \text{ mol/L} \] \ \[ \Delta t = 2.00 - 1.00 = 1.00 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.161}{1.00} = 0.161 \text{ mol/L} \cdot \text{h} \]
05

Calculate Rate for Interval (d) 2.0 to 3.0 h

Use the concentrations at 2.00 h (0.472 mol/L) and 3.00 h (0.352 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.352 - 0.472 = -0.120 \text{ mol/L} \] \ \[ \Delta t = 3.00 - 2.00 = 1.00 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.120}{1.00} = 0.120 \text{ mol/L} \cdot \text{h} \]
06

Calculate Rate for Interval (e) 3.0 to 4.0 h

Use the concentrations at 3.00 h (0.352 mol/L) and 4.00 h (0.262 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.262 - 0.352 = -0.090 \text{ mol/L} \] \ \[ \Delta t = 4.00 - 3.00 = 1.00 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.090}{1.00} = 0.090 \text{ mol/L} \cdot \text{h} \]
07

Calculate Rate for Interval (f) 4.0 to 5.0 h

Use the concentrations at 4.00 h (0.262 mol/L) and 5.00 h (0.196 mol/L): \ \[ \Delta [\text{N}_2\text{O}_5] = 0.196 - 0.262 = -0.066 \text{ mol/L} \] \ \[ \Delta t = 5.00 - 4.00 = 1.00 \text{ h} \] \ \[ \text{Average Rate} = -\frac{-0.066}{1.00} = 0.066 \text{ mol/L} \cdot \text{h} \]

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the fascinating study of the speed or rate at which chemical reactions occur. It plays a crucial role in understanding how quickly reactants transform into products. The rate of a reaction can be influenced by several factors, including the concentration of the reactants, temperature, and the presence of catalysts.
By studying chemical kinetics, we can predict how a chemical system will behave over time and optimize conditions for desired reaction rates. This is particularly important in industrial processes where efficiency and cost can be highly dependent on reaction rates.
  • Reaction Mechanisms: Understanding the step-by-step sequence of elementary reactions.
  • Rate Laws: Mathematical relationships that relate the rate of a reaction to the concentration of reactants.
  • Catalysts: Substances that speed up reactions without being consumed.
Each of these factors is integral to comprehensively understanding chemical kinetics and enhancing the ability to control or predict chemical reactions.
Reaction Rates
Reaction rates refer to how quickly or slowly a reaction proceeds. It is typically measured as the change in concentration of reactants or products over time. Consider the reaction of nitrogen pentoxide (N2O5) decomposing to nitrogen dioxide (NO2) and oxygen (O2). The speed of this reaction is known as the reaction rate.
The average rate of reaction can be calculated using the formula:\[\text{Average Rate} = -\frac{\Delta [\text{N}_2\text{O}_5]}{\Delta t}\]where \(\Delta [\text{N}_2\text{O}_5]\) is the change in concentration of \(\text{N}_2\text{O}_5\), and \(\Delta t\) is the time interval. This equation tells us how fast \(\text{N}_2\text{O}_5\) is decreasing per unit of time.
Understanding reaction rates helps chemists to:
  • Determine the efficiency of a reaction.
  • Establish optimal conditions for maximum yield.
  • Predict how a system will react over various timeframes.
Analyzing reaction rates provides valuable insights into the dynamics of chemical processes.
Concentration Changes
In chemical reactions, monitoring the changes in concentration of the substances involved is key to understanding the reaction's progress over time. The concentration of reactants typically decreases while the concentration of products increases as the reaction proceeds.

In the given exercise involving the decomposition of \(\text{N}_2\text{O}_5\), the concentration changes are measured over specific time intervals such as 0.00 to 0.50 hours and 0.50 to 1.0 hours, etc.
To calculate these changes:
  • Record the initial and final concentrations of a specific interval.
  • Determine the difference to find \(\Delta [\text{N}_2\text{O}_5]\).
For instance, from time 0.00 hours to 0.50 hours, the concentration of \(\text{N}_2\text{O}_5\) decreases from 0.849 mol/L to 0.733 mol/L, resulting in a \(\Delta [\text{N}_2\text{O}_5]\) of -0.116 mol/L.
These concentration changes are crucial as they directly correspond to the average rate of reaction, providing a clear picture of how quickly or slowly the reactants are being consumed.

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Most popular questions from this chapter

Write the rate law for each of these elementary reactions. (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\)

For each of these rate laws, state the reaction order with respect to the hypothetical substances \(\mathrm{A}\) and \(\mathrm{B}\), and give the overall order. (a) Rate \(=k[\mathrm{~A}][\mathrm{B}]^{3}\) (b) Rate \(=k[\mathrm{~A}][\mathrm{B}]\) (c) Rate \(=k[\mathrm{~A}]\) (d) Rate \(=k[\mathrm{~A}]^{3}[\mathrm{~B}]\)

When phenacyl bromide and pyridine are both dissolved in methanol, they react to form phenacylpyridinium bromide. \(\mathrm{C}_{6} \mathrm{H}_{5}-\stackrel{\mathrm{O}}{\|} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Br}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}-\mathrm{CH}_{2} \mathrm{NC}_{5} \mathrm{H}_{5}^{+}+\mathrm{Br}^{-}\) When equal concentrations of reactants were mixed in methanol at \(35^{\circ} \mathrm{C}\), these data were obtained: \begin{tabular}{rc|rr} \hline Time \((\min )\) & [Reactant] \((\mathrm{mol} / \mathrm{L})\) & Time \((\mathrm{min})\) & \([\mathrm{Reactant}]\) \((\mathrm{mol} / \mathrm{L})\) \\ \hline 0 & 0.0385 & \(500 .\) & 0.0208 \\ \(100 .\) & 0.0330 & \(600 .\) & 0.0191 \\ \(200 .\) & 0.0288 & \(700 .\) & 0.0176 \\ \(300 .\) & 0.0255 & \(800 .\) & 0.0163 \\ \(400 .\) & 0.0220 & \(1000 .\) & 0.0143 \\ \hline \end{tabular} (a) Determine the rate law for this reaction. (b) Determine the overall order of this reaction. (c) Determine the rate constant for this reaction. (d) Determine the rate constant for this reaction when the concentration of each reactant is \(0.030 \mathrm{~mol} / \mathrm{L}\)

Nitryl fluoride is an explosive compound that can be made by oxidizing nitrogen dioxide with fluorine: \(2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}(\mathrm{~g})\) Several kinetics experiments, all done at the same temperature and involving formation of nitryl fluoride, are summarized in this table: \begin{tabular}{cccccc} \hline & \multicolumn{2}{c} { Initial Concentration (mol/L) } & & \\ \cline { 2 - 4 } \cline { 6 } Experiment & {\(\left[\mathrm{NO}_{2}\right]\)} & {\(\left[\mathrm{F}_{2}\right]\)} & {\(\left[\mathrm{NO}_{2} F\right]\)} & & Initial Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 1 & 0.0010 & 0.0050 & 0.0020 & & \(2.0 \times 10^{-4}\) \\ 2 & 0.0020 & 0.0050 & 0.0020 & & \(4.0 \times 10^{-4}\) \\ 3 & 0.0020 & 0.0020 & 0.0020 & & \(1.6 \times 10^{-4}\) \\ 4 & 0.0020 & 0.0020 & 0.0010 & & \(1.6 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for the reaction. (b) Determine what the order of the reaction is with respect to each reactant and each product. (c) Calculate the rate constant \(k\) and express it in appropriate units.

For the reaction \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) these data were obtained at \(1100 \mathrm{~K}\) : \begin{tabular}{ccc} \hline [NO] (mol/L) & {\(\left[\mathrm{H}_{2}\right](\mathrm{mol} / \mathrm{L})\)} & Initial Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline \(5.00 \times 10^{-3}\) & \(2.50 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) \\\ \(15.0 \times 10^{-3}\) & \(2.50 \times 10^{-3}\) & \(9.0 \times 10^{-3}\) \\ \(15.0 \times 10^{-3}\) & \(10.0 \times 10^{-3}\) & \(3.6 \times 10^{-2}\) \\ \hline \end{tabular}
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