Which one of the following equations is correct for the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ?\) (a) \(\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (b) \(\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{1}{-3} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (c) \(\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (d) \(\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{1}{-2} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\)

Short Answer

Expert verified
Option (b) is the correct equation.

Step by step solution

01

Understand the Stoichiometry of the Reaction

Examine the balanced chemical equation for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\). This tells us that one molecule of \(\mathrm{N}_{2}\) reacts with three molecules of \(\mathrm{H}_{2}\) to produce two molecules of \(\mathrm{NH}_{3}\).
02

Relate the Rates of Consumption and Production

The rate of disappearance of \(\mathrm{H}_{2}\) is three times the rate of appearance of \(\mathrm{NH}_{3}\), due to the stoichiometry of the reaction, and vice versa. This can be expressed as \(\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} = 2\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) or \(\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} = \frac{1}{-3}\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\).
03

Compare with the Given Options

Option (b) \(\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{1}{-3} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) correctly represents the stoichiometric relationship between the rates of formation of \(\mathrm{NH}_{3}\) and consumption of \(\mathrm{H}_{2}\).

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Key Concepts

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

Stoichiometry
Stoichiometry is the cornerstone of understanding chemical reactions. It involves calculating the proportions of reactants and products involved in chemical reactions. It's like a recipe for a cake, telling you how much of each ingredient you need to mix to get the desired result.

In our exercise, the balanced chemical equation \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \) shows the exact number of nitrogen (\(N_2\)) and hydrogen (\(H_2\)) molecules reacting to form ammonia (\(NH_3\)). The coefficients in front of each molecule (1 for \(N_2\), 3 for \(H_2\), and 2 for \(NH_3\)) are the stoichiometric coefficients.

Essentially, these numbers tell us that to produce two molecules of ammonia, we need one molecule of nitrogen and three molecules of hydrogen. Understanding this stoichiometric relationship helps us to calculate the amount of reactants needed to produce a certain amount of product, creating a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and observe.
Rate of Reaction
The rate of reaction paints a picture of how swiftly a chemical reaction proceeds. Imagine watching a race where some runners are faster than others; similarly, in chemistry, some reactions reach the finish line sooner. The rate of a reaction is quantified by the change in concentration of a reactant or product per unit time.

In the given exercise, when we observe that \( \frac{1}{2}\frac{\mathrm{d}[\mathrm{NH}_{3}]}{\mathrm{dt}} = \frac{1}{-3}\frac{\mathrm{d}[\mathrm{H}_{2}]}{\mathrm{dt}} \) it's essentially comparing the rates at which ammonia is formed and hydrogen is consumed. So, if we were to picture this as a race, our \(NH_3\) molecules are appearing (finishing the race) at half the speed at which our \(H_2\) molecules are disappearing from the starting line. By understanding the rate at which substances react, we can influence chemical processes in industries, environmental systems, and even within our own bodies.
Chemical Kinetics
Diving deeper into how fast reactions occur, we land in the realm of chemical kinetics, a subfield of chemistry focusing on reaction speeds and the pathway taken during a reaction—known as the reaction mechanism.

Chemical kinetics goes beyond simply measuring the speed of reaction; it's like understanding both the speed of each car in a race and the reasons why each car has its particular speed. Factors that can influence the reaction rate include the concentration of reactants, temperature, and the presence of a catalyst.

In our exercise, the rate at which ammonia is produced and hydrogen is consumed can be studied using the principles of chemical kinetics to optimize the conditions for the fastest production of ammonia with minimal waste. Through kinetics, we can also predict how long a reaction will take and design experiments and industrial processes accordingly. Learnings in chemical kinetics are crucial in the development of new medications, materials, and energy conversion technologies.

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Most popular questions from this chapter

The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at \(25^{\circ} \mathrm{C}\) are \(3.0 \times 10^{-4} \mathrm{~s}^{-1}, 104.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(6 \times 10^{14} \mathrm{~s}^{-1}\) respectively:

For the reaction a \(\mathrm{A} \longrightarrow \mathrm{xP}\) when \([\mathrm{A}]=2.2 \mathrm{mM}\) the rate was found to be \(2.4 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1}\) On reducing concentration of \(\mathrm{A}\) to half, the rate changes to \(0.6 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1} .\) The order of reaction with respect to \(\mathrm{A}\) is: (a) \(1.5\) (b) \(2.0\) (c) \(2.5\) (d) \(3.0\)

For a first-order reaction, (a) The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{kt}}\right)\) (b) The pre-exponential factor in the Arrhenius equation has the dimensions of time \(\mathrm{t}^{-1}\). (c) The time taken for the completion of \(75 \%\) reation is thrice the t \(1 / 2\) of the reaction. (d) Both (a) and (b)

The activation energy for a simple chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \(\mathrm{E}\) in forward direction. The activation energy for reverse reaction: (a) Can be less than or more than \(\mathrm{E}_{\mathrm{a}}\) (b) Is always double of \(E_{a}\) (c) Is negative of \(\mathrm{E}_{\mathrm{a}}\) (d) Is always less than \(\mathrm{E}_{\mathrm{a}}\)

The rate constant of a first-order reaction is \(6 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration is \(0.10 \mathrm{M}\), the initial rate of reaction is: (a) \(6 \times 10^{-3} \mathrm{Ms}^{-1}\) (b) \(6 \times 10^{-1} \mathrm{Ms}^{-1}\) (c) \(6 \times 10^{-6} \mathrm{Ms}^{-1}\) (d) \(6 \times 10^{-8} \mathrm{Ms}^{-1}\)

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