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Chapter 14: Problem 23

The decomposition of ammonia on a metal surface to form \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is a zero-order reaction (Figure \(14.7 \mathrm{c}) .\) At \(873^{\circ} \mathrm{C},\) the value of the rate constant is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). How long it will take to completely decompose 0.16 g of \(\mathrm{NH}_{3}\) in a \(1.0-\mathrm{L}\) flask?

Short Answer

Expert verified
It will take approximately 6,260 seconds to completely decompose 0.16 g of NH3.

Step by step solution

01

Convert grams to moles

First, we need to convert the mass of ammonia (NH_3 m) into moles. The molar mass of ammonia is calculated using the atomic weights: 1 nitrogen (N) = 14.01 g/mol + 3 hydrogen (H) = 3 x 1.008 g/mol = 17.034 g/mol. Given 0.16 g of NH_3 m, the number of moles of NH_3 m is calculated as follows: \[ \text{moles of } \mathrm{NH}_{3} = \frac{0.16 \text{ g}}{17.034 \text{ g/mol}} \approx 0.00939 \text{ moles} \].
02

Use the zero-order rate equation

For a zero-order reaction, the rate of reaction is constant and given by: \[ \text{Rate} = k \].Where \( k \) is the rate constant, and the concentration changes with time are expressed by: \[ \text{[NH}_3\text{]}_t = \text{[NH}_3\text{]}_0 - kt \].Here,\( \text{[NH}_3\text{]}_t \) is the concentration at time \( t \), \( \text{[NH}_3\text{]}_0 \) is the initial concentration.In this case, we decompose the whole amount, so \( \text{[NH}_3\text{]}_t \ = 0 \).
03

Calculate initial concentration

The initial concentration \( \text{[NH}_3\text{]}_0 \) is determined as follows: \[ \text{[NH}_3\text{]}_0 = \frac{\text{moles of } \mathrm{NH}_{3}}{\text{volume of flask}} = \frac{0.00939 \text{ moles}}{1.0 \text{ L}} = 0.00939 \text{ M} \].
04

Rearrange rate equation and solve for time

We rearrange the equation\[ 0 = \text{[NH}_3\text{]}_0 - kt \]to solve for time \( t \):\[ t = \frac{\text{[NH}_3\text{]}_0}{k} \].Substitute the known values \( \text{[NH}_3\text{]}_0 = 0.00939 \text{ M} \) and \( k = 1.5 \times 10^{-3} \text{ mol/L} \cdot \text{s} \):\[ t = \frac{0.00939 \text{ M}}{1.5 \times 10^{-3} \text{ mol/L} \cdot \text{s}} \approx 6.26 \times 10^{3} \text{ s} \].

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Key Concepts

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

Ammonia Decomposition
The decomposition of ammonia ( H_{3} ) on a metal surface is an important zero-order reaction. This type of reaction typically occurs on a catalyst surface, where the rate is independent of the concentration of the reactant. As ammonia decomposes, it breaks down into nitrogen ( _{2} ) and hydrogen ( _{2} ). This transformation is achieved at a higher temperature, such as 873掳C in the given example. The process is crucial in various industrial applications, including the production of hydrogen gas. Understanding the decomposition of ammonia helps us understand complex reaction pathways and energy transformations.
Rate Constant
In chemical kinetics, the rate constant () plays a pivotal role. It is a measure of how quickly a reaction proceeds. For the zero-order reaction described, the rate constant is given as \(1.5 \times 10^{-3} \text{ mol/L} \cdot \text{s}\). Here are some key points about the rate constant:
  • It is a fixed value at a specific temperature.
  • For zero-order reactions, it translates directly to the rate because changes in concentration do not affect the rate.
  • Knowing the rate constant allows us to determine the time needed for a reaction to reach completion, especially when the initial concentration is known.
This constant is critical for calculating how fast a certain amount of ammonia decomposes under specified conditions.
Reaction Kinetics
Reaction kinetics explores how and why reactions occur at specific rates. This field of study looks at various factors influencing the reaction rate, including temperature, concentration, and the presence of catalysts. In zero-order reactions:
  • The rate is constant over time.
  • The concentration of the reactant decreases linearly with time.
  • The half-life depends on the initial concentration and decreases as the reaction progresses.
By studying kinetics, chemists can predict reaction behavior, improve processes, and design better industrial applications.
Chemical Concentration Calculations
Calculating chemical concentrations involves converting amounts from one unit to another (e.g., grams to moles) and using those values to determine molarity. Here's how it applies to our example:
  • First, convert the mass of ammonia to moles using its molar mass. For NH_3, this is approximately 17.034 g/mol. Thus, 0.16 g of NH_3 equals about 0.00939 moles.
  • The initial concentration is found by dividing the moles by the volume of the container (1.0 L), giving 0.00939 M.
  • Using the concentration and rate constant, the time to fully decompose the ammonia can be calculated. Using \(t = \frac{\text{[NH}_3\text{]}_0}{k}\), we find the time to be approximately 6260 seconds.
These calculations form the backbone of understanding reaction kinetics and their real-world applications.

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Most popular questions from this chapter

The conversion of cyclopropane to propene (Example \(14.5)\) occurs with a first-order rate constant of \(2.42 \times 10^{-2} \mathrm{h}^{-1} .\) How long will it take for the concentration of cyclopropane to decrease from an initial concentration of \(0.080 \mathrm{mol} / \mathrm{L}\) to \(0.020 \mathrm{mol} / \mathrm{L} ?\)

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. (a) Determine the rate constant for this reaction if it takes 1.76 min for the concentration of \(\mathrm{NO}_{2}\) to fall from 0.250 mol/L to 0.100 mol/L. (b) If the chemical equation is written as $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ what is the value of the rate constant?

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2},\) with a half-life of \(30 .\) minutes at room temperature: $$ \mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ If the partial pressure of HOF in a 1.00-L flask is initially \(1.00 \times 10^{2} \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what are the total pressure in the flask and the partial pressure of HOF after exactly 30 minutes? After 45 minutes?

Hydrogenation reactions, processes wherein \(\mathrm{H}_{2}\) is added to a molecule, are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Explain why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal.

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step \(1: \quad\) Fast, reversible: \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step \(2: \quad\) Fast, reversible: \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+}\) Step 3: Slow \(\mathrm{XH}^{+} \rightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?
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