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

Nitrosyl bromide (NOBr) decomposes to nitrogen oxide and bromine. Use the following data to determine the order of the decomposition reaction of nitrosyl bromide. $$\begin{array}{cccccc}\hline \text { Time (s) } & 0 & 6 & 12 & 18 & 24 \\ \text { [NOBr] } & 0.0286 & 0.0253 & 0.0229 & 0.0208 & 0.0190 \\\\\hline\end{array}$$

The first-order rate constant for the decomposition of a certain drug at \(25^{\circ} \mathrm{C}\) is \(0.215 \mathrm{month}^{-1}\) (a) If \(10.0 \mathrm{~g}\) of the drug is stored at \(25^{\circ} \mathrm{C}\) for one year, how many grams of the drug will remain at the end of the year? (b) What is the half-life of the drug? (c) How long will it take to decompose \(65 \%\) of the drug?

When a base is added to an aqueous solution of chlorine dioxide gas, the following reaction occurs: $$2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}$$ The reaction is first-order in \(\mathrm{OH}^{-}\) and second-order for \(\mathrm{ClO}_{2}\). Initially, when \(\left[\mathrm{ClO}_{2}\right]=0.010 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.030 \mathrm{M}\), the rate of the reaction is \(6.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). What is the rate of the reaction when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{ClO}_{2}\) and \(95.0 \mathrm{~mL}\) of \(0.155 \mathrm{M} \mathrm{NaOH}\) are added?

The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).

A catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ}\) to \(206 \mathrm{~kJ} .\) By what factor would you expect the reaction-rate constant to increase at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factors \((\mathrm{A})\) are the same for both reactions. (Hint: Use the formula \(\ln k=\ln \mathrm{A}-E_{2} / R T .\) )

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