For the zero-order decomposition of ammonia on tungsten $$\mathrm{NH}_{3}(\mathrm{~g}) \stackrel{\mathrm{w}}{\longrightarrow} \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)$$ the rate constant is \(2.08 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). (a) What is the half-life of a \(0.250 \mathrm{M}\) solution of ammonia? (b) How long will it take for the concentration of ammonia to drop from \(1.25 M\) to \(0.388 M ?\)

Short Answer

Expert verified
$$t_{1/2} ≈ 600.96s $$ Hence, the half-life of ammonia in this reaction is approximately 600.96 seconds. #tag_title# (Find time for concentration to drop to 0.175 M (b))#tag_content# To find the time it takes for the concentration of ammonia to drop to 0.175 M, we can use the zero-order reaction formula: $$[A]_t = [A]_0 - kt$$ where \([A]_t\) is the final concentration, \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time. We are given \([A]_0 = 0.250 \mathrm{M}\), \([A]_t = 0.175 \mathrm{M}\), and \(k = 2.08 \times 10^{-4} \mathrm{mol/L*s}\). Plugging these values into the formula, we can solve for \(t\): $$0.175 = 0.250 - (2.08 \times 10^{-4})t$$ Rearranging and solving for \(t\): $$t ≈ 360.58s$$ Thus, it takes approximately 360.58 seconds for the concentration of ammonia to drop to 0.175 M.

Step by step solution

01

(Find half-life of ammonia (a))

To find the half-life of the reaction, we'll need to obtain the equation's required values. The given rate constant, k, equals \(2.08 \times 10^{-4}\) mol/L*s. For a zero-order reaction, we can calculate the half-life using the formula: $$t_{1/2} = \frac{[A]_0}{2k} $$ where \(t_{1/2}\) is the half-life and \([A]_0\) is the initial concentration. Let's calculate the half-life given the initial concentration of \(0.250 \mathrm{M}\): $$t_{1/2} = \frac{0.250}{2(2.08 \times 10^{-4})} $$

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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 study of the speed or rate at which chemical reactions occur. This field of chemistry is essential because it not only helps us understand how fast a reaction takes place but also reveals the sequence of steps, or mechanism, that take place during the reaction. These steps can be very complex, involving the breaking and formation of multiple bonds.

The rate of a reaction and how it changes with varying conditions such as temperature, pressure, or concentration of reactants, is a central concept in kinetics. For educators aiming to deliver this concept effectively, it is crucial to introduce practical examples that students can relate with, such as the rusting of iron or the browning of fruit. Then, draw the connection to the mathematical description of these rates through rate laws and constants. Simplified visual aids like graphs showing reactant concentration versus time can be incredibly helpful for visual learners.
Reaction Rate Constant
The reaction rate constant, often represented by the symbol 'k,' is a proportionality constant in the rate law equation of a chemical reaction. It plays a crucial role in kinetics as it links the rate of reaction to the concentration of the reactants. For zero-order reactions, the rate is independent of the concentration of the reactants, which means the reaction rate constant 'k' has units of concentration per time, typically mol/L·s or M/s.

It is important to impress upon students that the magnitude of 'k' can give us an idea of how fast a reaction proceeds; a larger 'k' indicates a faster reaction. Understanding this concept is greatly enhanced by solving problems where they calculate and interpret the value of 'k,' like in the ammonia decomposition example. Providing a thorough explanation of how 'k' fits within the context of the rate equation and how it can be experimentally determined can significantly aid in demystifying this topic.
Ammonia Decomposition
Ammonia decomposition on a surface like tungsten is a typical example of a zero-order reaction in chemical kinetics. In this process, ammonia molecules break down into nitrogen and hydrogen gases. Notably, in zero-order reactions, the rate of reaction remains constant; it does not depend on the concentration of the reactant. This characteristic simplifies calculations related to the reaction process, such as determining the half-life.

When educators introduce this concept, detailing the conditions under which a reaction can be zero-order—such as the reaction occurring on a catalytic surface where active sites are fully saturated—can help students visualize and better understand the process. Moreover, using real-life industrial examples where ammonia decomposition is relevant, such as in the production of fertilizers or the Haber process, can link these theoretical concepts to real-world applications, thus helping students appreciate the importance of studying such reactions.

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Most popular questions from this chapter

The decomposition of \(\mathrm{Y}\) is a zero-order reaction. Its half-life at \(25^{\circ} \mathrm{C}\) and \(0.188 M\) is 315 minutes. (a) What is the rate constant for the decomposition of Y? (b) How long will it take to decompose a \(0.219 \mathrm{M}\) solution of \(\mathrm{Y}\) ? (c) What is the rate of the decomposition of \(0.188 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) ? (d) Does the rate change when the concentration of \(\mathrm{Y}\) is increased to \(0.289 \mathrm{M}\) ? If so, what is the new rate?

Consider the following hypothetical reaction: $$\mathrm{X}(g) \longrightarrow \mathrm{Y}(g)$$ A 200.0-mL flask is filled with \(0.120\) moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{X} & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline\end{array}$$ (a) Make a similar table for the appearance of Y. (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two twenty-minute intervals. (c) What is the average rate of appearance of \(\mathrm{Y}\) between the 20 - and 60-minute intervals?

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

Hypofluorous acid, HOF, is extremely unstable at room temperature. The following data apply to the decomposition of HOF to \(\mathrm{HF}\) and \(\mathrm{O}_{2}\) gases at a certain temperature. $$\begin{array}{cl}\hline \text { Time (min) } & \text { [HOF] } \\\\\hline 1.00 & 0.607 \\\2.00 & 0.223 \\\3.00 & 0.0821 \\\4.00 & 0.0302 \\\5.00 & 0.0111 \\\\\hline \end{array}$$ (a) By plotting the data, show that the reaction is first-order. (b) From the graph, determine \(k\). (c) Using \(k\), find the time it takes to decrease the concentration to \(0.100 \mathrm{M}\) (d) Calculate the rate of the reaction when [HOF] \(=0.0500 M\).

Experimental data are listed for the hypothetical reaction $$\mathrm{x} \longrightarrow \mathrm{Y}+\mathrm{Z}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\ {[\mathrm{X}]} & 0.0038 & 0.0028 & 0.0021 & 0.0016 & 0.0012 & 0.00087 \\\\\hline\end{array}$$ (a) Plot these data as in Figure \(11.2 .\) (b) Draw a tangent to the curve to find the instantaneous rate at \(40 \mathrm{~s}\). (c) Find the average rate over the 10 to 50 s interval. (d) Compare the instantaneous rate at \(40 \mathrm{~s}\) with the average rate over the \(40-\mathrm{s}\) interval.

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