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The rate equation for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (giving \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ) is Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) The value of \(k\) is \(6.7 \times 10^{-5} \mathrm{s}^{-1}\) for the reaction at a particular temperature. (a) Calculate the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (b) How long does it take for the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration to drop to one tenth of its original value?

Short Answer

Expert verified
The half-life is approximately 10,343 seconds. It takes about 34,380 seconds for the concentration to drop to one tenth of its initial value.

Step by step solution

01

Understand the Rate Law

The rate law for the decomposition reaction is given as \( \text{Rate} = k[\mathrm{N}_2\mathrm{O}_5] \), indicating it is a first-order reaction. This means the rate of decomposition depends linearly on the concentration of \( \mathrm{N}_2\mathrm{O}_5 \).
02

Calculate the Half-Life for a First-Order Reaction

For a first-order reaction, the half-life is given by the formula:\[ t_{1/2} = \frac{0.693}{k}\]Substitute the given value of \( k = 6.7 \times 10^{-5} \text{s}^{-1} \) to find the half-life:\[ t_{1/2} = \frac{0.693}{6.7 \times 10^{-5}}\]Calculating gives:\[ t_{1/2} \approx 10343 \text{s}\]
03

Calculate Time to Reach One Tenth of Initial Concentration

The time required for a first-order reaction to reach a certain fraction of its initial concentration can be calculated using the formula:\[ t = \frac{1}{k} \ln \left(\frac{[\text{Initial}]}{[\text{Final}]\right)}\]In this case, \([\text{Final}] = \frac{1}{10}[\text{Initial}]\), so:\[ t = \frac{1}{6.7 \times 10^{-5}} \ln(10)\]\[ t \approx 34380 \text{s}\]

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Key Concepts

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

Rate Law
In chemistry, rate law is an equation that describes how the rate of a reaction depends on the concentration of the reactants. For first-order reactions like the decomposition of \(\mathrm{N}_2\mathrm{O}_5\), the rate law is formulated as \(\text{Rate} = k [A]\), where \(k\) is the rate constant and \([A]\) is the concentration of the reactant. This equation demonstrates that the rate at which \(\mathrm{N}_2\mathrm{O}_5\) decomposes is proportional to its concentration.
This is significant because it shows that as the concentration decreases, the rate of reaction decreases too. Understanding the rate law for first-order reactions helps predict how long a chemical reaction will take under specific conditions.
Key points about rate law for first-order reactions include:
  • The rate constant \(k\) is independent of concentration, but dependent on temperature.
  • The reaction order tells us that the concentration influences the rate directly.
  • The reaction is exponential in nature, meaning it follows a decay curve over time.
Half-life Calculation
The concept of half-life in chemistry refers to the time required for the concentration of a reactant to decrease by half. For first-order reactions, the half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
This formula results in a constant half-life that doesn't depend on the initial concentration.
Let's calculate the half-life for the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) given that the rate constant \(k\) is \(6.7 \times 10^{-5} \mathrm{s}^{-1}\):
Substitute the value of \(k\) into the formula:
\[ t_{1/2} = \frac{0.693}{6.7 \times 10^{-5}} \approx 10343 \text{s}\]
This means that it takes approximately 10343 seconds for the concentration of \(\mathrm{N}_2\mathrm{O}_5\) to reduce to half of its initial amount. This calculation can be applied repeatedly to determine the concentration at any given time point being reduced to half successively.
Decomposition Reaction
A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In the case of \(\mathrm{N}_2\mathrm{O}_5\) decomposition, it breaks down into \(\mathrm{NO}_2\) and \(\mathrm{O}_2\).
This process can be described by the reaction equation:
\[ \mathrm{N}_2\mathrm{O}_5 \rightarrow 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2 \]
The decomposition reaction is essential in understanding how time influences the change in concentration of the reactants. For \(\mathrm{N}_2\mathrm{O}_5\), as it decomposes, the rate of formation of \(\mathrm{NO}_2\) and \(\mathrm{O}_2\) can be monitored using rate laws.
Decomposition reactions are:
  • Endothermic or exothermic, depending on the necessity of energy to break bonds.
  • Influenced by temperature; higher temperatures typically increase the rate of decomposition.
  • Commonly observed in unstable compounds or compounds exposed to specific conditions.
Understanding decomposition reactions and their kinetics such as rate laws and half-life helps in predicting how substances behave and change over time, which is valuable in chemical manufacturing and analysis tasks.

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Most popular questions from this chapter

The reaction of \(\mathrm{H}_{2}\) molecules with \(\mathrm{F}\) atoms $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{F}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+\mathrm{H}(\mathrm{g})$$ has an activation energy of \(8 \mathrm{kJ} / \mathrm{mol}\) and an enthalpy change of \(-133 \mathrm{kJ} / \mathrm{mol} .\) Draw a diagram similar to Figure 15.13 for this process. Indicate the activation energy and enthalpy change on this diagram.

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The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After 27 minutes at \(27^{\circ} \mathrm{C},\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

Ammonia decomposes when heated according to the equation $$\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2}(\mathrm{g})+\mathrm{H}(\mathrm{g})$$ The data in the table for this reaction were collected at a high temperature. $$\begin{array}{cc}\text { Time (h) } & \text { [NH }\left._{3}\right] \text { (mol/L) } \\\\\hline 0 & 8.00 \times 10^{-7} \\\25 & 6.75 \times 10^{-7} \\\50 & 5.84 \times 10^{-7} \\\75 & 5.15 \times 10^{-7} \\\\\hline\end{array}$$ Plot ln \(\left[\mathrm{NH}_{3}\right]\) versus time and \(1 /\left[\mathrm{NH}_{3}\right]\) versus time. What is the order of this reaction with respect to NH \(_{3} ?\) Find the rate constant for the reaction from the slope.

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 \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

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