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In the synthesis of ammonia, if \(-\Delta\left[\mathrm{H}_{2}\right] / \Delta t=\) \(4.5 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min},\) what is \(\Delta\left[\mathrm{NH}_{3}\right] / \Delta t ?\) $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$

Short Answer

Expert verified
The rate of formation of \( \mathrm{NH}_{3} \) is \( 3.0 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min} \).

Step by step solution

01

Identify the Chemical Equation and Rates

The given chemical reaction is \( \mathrm{N}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \). We are provided with the rate of consumption of \( \mathrm{H}_{2} \), which is \(-\Delta \left[ \mathrm{H}_{2} \right] / \Delta t = 4.5 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min} \). We need to find the rate of formation of \( \mathrm{NH}_{3} \), i.e., \( \Delta \left[ \mathrm{NH}_{3} \right] / \Delta t \).
02

Use Stoichiometry to Relate Rates

According to the balanced equation, 3 moles of \( \mathrm{H}_{2} \) are needed to form 2 moles of \( \mathrm{NH}_{3} \). Hence, the rate of formation of \( \mathrm{NH}_{3} \) is \( \frac{2}{3} \) times the rate of consumption of \( \mathrm{H}_{2} \).
03

Calculate the Rate of Formation of Ammonia

To find \( \Delta \left[ \mathrm{NH}_{3} \right] / \Delta t \), use the rate relationship: \( \Delta \left[ \mathrm{NH}_{3} \right] / \Delta t = \frac{2}{3} \left(- \Delta \left[ \mathrm{H}_{2} \right] / \Delta t\right) \). Substituting the given rate of \( \mathrm{H}_{2} \), we have: \[ \Delta \left[ \mathrm{NH}_{3} \right] / \Delta t = \frac{2}{3} \times 4.5 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min} \]Calculating gives: \[ \Delta \left[ \mathrm{NH}_{3} \right] / \Delta t = 3.0 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min} \]
04

Verify the Calculation

Double check the calculation by multiplying \( \frac{2}{3} \) with \( 4.5 \times 10^{-4} \), which should result in \( 3.0 \times 10^{-4} \). This confirms that the rate of formation of \( \mathrm{NH}_{3} \) is correct.

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict how much of a substance is needed or produced. In the context of the ammonia synthesis reaction, stoichiometry tells us that for every 1 mole of nitrogen (_2), 3 moles of hydrogen (_2) are required to produce 2 moles of ammonia (H_3).
This relationship is crucial because it helps us understand the proportion of reactants needed and the amount of product formed. By using stoichiometry, chemists can calculate the rates of reactions and the quantities involved. In our worked example, we used the balanced chemical equation to determine that the rate of ammonia formation is \(\frac{2}{3}\) times the rate at which hydrogen gas is consumed.
To apply stoichiometry in reaction rates, we rely on conversion factors derived from the coefficients in a balanced chemical equation. Here, the conversion factor comes from the numbers in front of the substances: 3 for hydrogen and 2 for ammonia. Thus, stoichiometry serves as a bridge to connect the macroscopic predictions we can make about reactions with the microscopic world of atoms and molecules.
Chemical Equations
A chemical equation is a symbolic representation of a chemical reaction. It provides information about which reactants are involved and what products are formed. In our example, the balanced chemical equation is \[ \mathrm{N}_2(\mathrm{g}) + 3 \mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{g}) \].
The equation tells us that one molecule of nitrogen reacts with three molecules of hydrogen to form two molecules of ammonia. The state symbols (\(\mathrm{g}\)) indicate that all substances involved are gases. Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation, adhering to the law of conservation of mass.
To read a chemical equation effectively, you need to identify the reactants (substances that start the reaction) and products (the substances formed). Additionally, the coefficients (numbers in front of chemical formulas) tell us the ratio in which substances react or are produced. In our exercise, understanding these equations is key to solving the problem because the coefficients directly inform the stoichiometric relationships and reaction rates.
Ammonia Synthesis
Ammonia synthesis, commonly known as the Haber-Bosch process, is an industrial-scale method to produce ammonia from nitrogen and hydrogen. The general reaction is:\( \mathrm{N}_2(\mathrm{g}) + 3 \mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{g}) \).
This reaction is of immense importance because ammonia is a critical component in fertilizers, thereby playing a vital role in agriculture and food production worldwide. The process was developed by Fritz Haber and Carl Bosch in the early 20th century, revolutionizing chemical manufacturing and helping to meet the world's food demands.
The synthesis of ammonia is carried out at high pressures and temperatures, typically using an iron catalyst to speed up the reaction. The balanced chemical equation we've examined demonstrates the conversion of nitrogen and hydrogen into ammonia. Each molecule of nitrogen reacts with three molecules of hydrogen, illustrating an elegant example of efficient chemical transformation, underscoring the significance of stoichiometry and balanced equations.
Ammonia synthesis not only underscores the importance of chemical equations in predicting reaction outcomes but also highlights the practical applications of stoichiometry in large-scale chemical processes.

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Most popular questions from this chapter

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

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1: Slow \(\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2: Fast \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

Hydrogen iodide decomposes when heated, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) The rate law for this reaction is \(-\Delta[\mathrm{HI}] / \Delta t=k[\mathrm{HI}]^{2} .\) At \(443^{\circ} \mathrm{C}, k=30 . \mathrm{L} / \mathrm{mol} \cdot\) min. If the initial \(\mathrm{HI}(\mathrm{g})\) concentration is \(3.5 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) what concentration of HI (g) will remain after \(10 .\) minutes?

Nitrogen oxides, \(\mathrm{NO}_{x}\) (a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) collectively designated as \(\mathrm{NO}_{x}\) ), play an essential role in the production of pollutants found in photochemical smog. The \(\mathrm{NO}_{x}\) in the atmosphere is slowly broken down to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in a first-order reaction. The average half-life of \(\mathrm{NO}_{x}\) in the smokestack emissions in a large city during daylight is 3.9 hours. (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of NO, remains after 5.25 hours? (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times 10^{-6} \mathrm{mg} ?\)

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2},\) and the reaction has a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a 1.0 -L. flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

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