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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.

Short Answer

Expert verified
The reaction is first order; the rate constant can be found from the slope of the ln([NH₃]) vs. time plot.

Step by step solution

01

Understanding the Reaction

The given reaction is the decomposition of ammonia (NH₃) into NH₂ and H. We need to determine the order of this reaction with respect to NH₃ using the provided concentration data at different times.
02

Calculate ln(NH₃) and 1/[NH₃]

Using the initial concentration data, calculate the natural logarithm (ln) and the inverse (1/[NH₃]) for each concentration value. This will help us plot the data to determine the reaction order. For example, when the concentration of NH₃ is 8.00 × 10^-7 mol/L, ln(8.00 × 10^-7) is calculated, and similarly for 1/(8.00 × 10^-7).
03

Plot ln([NH₃]) versus Time

Create a plot of ln([NH₃]) on the y-axis against time on the x-axis. If the reaction is first order, this plot will yield a straight line, indicating a constant rate of decomposition with respect to the concentration of NH₃.
04

Plot 1/[NH₃] versus Time

Create another plot of 1/[NH₃] on the y-axis against time on the x-axis. If this reaction is second order, this plot will yield a straight line. If neither plot is a straight line, the reaction is not either first or second order.
05

Determine the Order from the Plots

By observing which plot (either ln([NH₃]) vs. time or 1/[NH₃] vs. time) produces a straight line, determine the order of the reaction. A straight line in the ln([NH₃]) plot suggests first order, while a straight line in the 1/[NH₃] plot suggests second order.
06

Find the Rate Constant from the Slope

Use the slope of the straight line from the correct plot to find the rate constant (k). If the reaction is first order, the rate constant k will be the negative of the slope from the ln([NH₃]) plot. For a second order reaction, the rate constant is the slope from the 1/[NH₃] plot.

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Key Concepts

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

Ammonia Decomposition
Ammonia decomposition is a process where ammonia molecules break down into simpler components when heated. In the given reaction, ammonia (\(\mathrm{NH}_3\)) decomposes into NH₂ and H. This reaction is significant in industrial and chemical processes where ammonia's breakdown is accelerated by heat.
When analyzing ammonia decomposition, scientists often monitor how the concentration of ammonia changes over time. By doing this, they can gain insights into the speed and efficiency of the reaction under different conditions. These observations are typically recorded as concentration values at specific time intervals.
Reaction Order
Reaction order is a key concept in reaction kinetics that tells us how the rate of a reaction depends on the concentration of the reactants. In simple terms, it describes the power to which the concentration of a reactant is raised in the rate equation. Reaction order can be determined experimentally by observing how changes in concentration affect the rate of reaction.
The major types of reaction orders are first order and second order. In a first-order reaction, the rate depends linearly on one reactant’s concentration. In a second-order reaction, the rate depends on the concentration of one reactant square or two different reactants linearly.
Rate Constant
The rate constant, often denoted as \( k \), is a proportionality constant in the rate equation of a chemical reaction. It provides important information about the reaction's speed under specific conditions.
The rate constant is different for each reaction and can vary with temperature or the presence of a catalyst. Unlike reactant concentrations, the rate constant is not affected by the initial concentration of reactants.
  • For a first-order reaction, the rate constant can be determined from a straight-line plot of \( \ln([\mathrm{NH}_3]) \) versus time.
  • For a second-order reaction, it would be determined from a straight-line plot of \( 1/[\mathrm{NH}_3] \) versus time.
First Order Reaction
A first-order reaction means the rate of reaction is directly proportional to the concentration of one reactant. In this case, if ammonia decomposition is first order, its rate depends on the concentration of ammonia only.
The plot of \( \ln([\mathrm{NH}_3]) \) versus time for a first-order reaction should be a straight line, showing that the natural logarithm (ln) of the concentration decreases linearly over time. The slope of this line is negative, and its magnitude provides the value of the rate constant.
Many decomposition reactions are first order because the reactant concentration directly influences the reaction rate.
Second Order Reaction
In second-order reactions, the reaction rate is proportional to either the square of one reactant's concentration or the product of the concentrations of two different reactants. If ammonia decomposition is second order, the rate involves a more complex dependence on reactant concentrations.
For a second-order reaction, plotting \( 1/[\mathrm{NH}_3] \) versus time produces a straight line. This is because the inverse of the concentration increases linearly as the reaction proceeds. The slope of this line provides the rate constant \( k \), which is a measure of the reaction speed.
Second order reactions often involve more interactions between reactant molecules, making them interesting for studying reaction mechanisms. Understanding these can help in optimizing reaction conditions in practical applications.

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Most popular questions from this chapter

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?

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}),\) decomposes to \(\mathrm{H}_{2} \mathrm{O}(\ell)\) and \(\mathrm{O}_{2}(\mathrm{g})\) in a reaction that is first order in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and has a rate constant \(k=1.06 \times 10^{-3} \mathrm{min}^{-1}\) at a given temperature. (a) How long will it take for \(15 \%\) of a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to decompose? (b) How long will it take for \(85 \%\) of the sample to decompose?

The decomposition of phosphine, \(\mathrm{PH}_{3},\) proceeds according to the equation $$\mathrm{PH}_{3}(\mathrm{g}) \rightarrow 1 / 4 \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g})$$ It is found that the reaction has the following rate equation: Rate \(=k\left[\mathrm{PH}_{3}\right] .\) The half-life of \(\mathrm{PH}_{3}\) is 37.9 seconds at \(120^{\circ} \mathrm{C}.\) (a) How much time is required for three fourths of the \(\mathrm{PH}_{3}\) to decompose? (b) What fraction of the original sample of \(\mathrm{PH}_{3}\) remains after 1.00 minute?

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L [where \(\left.L \text { is an electron-pair donor such as } P\left(C H_{3}\right)_{3}\right]\) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90,\) p. \(6927,1968 .\) ) A detailed study of the kinetics of the reaction led to the following mechanism: Slow \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of \(2 .\) Doubling the concentration of L. had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M}\) what is the concentration of the product after 5.0 minutes?

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