Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)

Step-by-Step Solution

Answer

Option (a) is the correct relationship for rate expressions.

Step 1: Understand the Reaction

The chemical reaction is \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \). This indicates that 1 mole of \( \mathrm{N}_{2} \) reacts with 3 moles of \( \mathrm{H}_{2} \) to produce 2 moles of \( \mathrm{NH}_{3} \).

Step 2: Determine Rate Expression for \( N_{2} \)

The rate of disappearance of \( \mathrm{N}_{2} \) is given by \( -\mathrm{d}[\mathrm{N}_{2}]/\mathrm{dt} \). Since only 1 mole of \( \mathrm{N}_{2} \) participates, the rate is simply \( -\mathrm{d}[\mathrm{N}_{2}]/\mathrm{dt} = \text{rate} \).

Step 3: Determine Rate Expression for \( H_{2} \)

For 3 moles of \( \mathrm{H}_{2} \), the rate is \( -\frac{1}{3} \mathrm{d}[\mathrm{H}_{2}]/\mathrm{dt} = \text{rate} \). This accounts for the stoichiometry of the reaction where 3 moles of \( \mathrm{H}_{2} \) are involved for every mole of \( \mathrm{N}_{2} \).

Step 4: Determine Rate Expression for \( NH_{3} \)

The formation rate of \( \mathrm{NH}_{3} \) is given by \( \frac{1}{2} \mathrm{d}[\mathrm{NH}_{3}]/\mathrm{dt} = \text{rate} \). This takes into account that 2 moles are formed.

Step 5: Compare with Given Options

Option (a) presents: rate \( = -\mathrm{d}[\mathrm{N}_{2}]/\mathrm{dt} = -\frac{1}{3} \mathrm{d}[\mathrm{H}_{2}]/\mathrm{dt} = \mathrm{d}[\mathrm{NH}_{3}]/\mathrm{dt} \), which matches the correct relationship among the rate expressions considering stoichiometry.

Key Concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is reached when the rates of the forward and reverse reactions are equal, resulting in the concentrations of reactants and products remaining constant over time. This does not mean that the reactants and products are in equal quantities, but rather that their rates of change balance each other out. The concept of equilibrium is crucial because many reactions do not go to completion but rather reach a point where both forward and reverse reactions continue to occur at the same rate.
Chemical equilibrium is characterized by the equilibrium constant, denoted as \( K \), which is calculated using the concentrations of the products and reactants at equilibrium. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is given by

• \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

It provides a measure of the "position" of the equilibrium. A large \( K \) value indicates that, at equilibrium, products are favored, whereas a small \( K \) value suggests that reactants are predominant. However, for the given reaction of nitrogen and hydrogen forming ammonia, the exercise specifically deals with reaction rates not equilibria.

Reaction Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict how much product will be formed from a given amount of reactant and vice versa. In the given reaction, \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \), stoichiometry tells us that:

• One mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
• The specific ratio helps in understanding how changes in one part of the reaction impact another.

The stoichiometry impacts the rate relationship in a reaction. Since 3 moles of \( \mathrm{H}_{2} \) are needed for every mole of \( \mathrm{N}_{2} \) used, the reaction rate involving \( \mathrm{H}_{2} \) must account for this by dividing by 3. Similarly, with 2 moles of \( \mathrm{NH}_{3} \) produced per mole of nitrogen, the rate of formation of \( \mathrm{NH}_{3} \) is double that of the nitrogen disappearance.
Understanding stoichiometry is vital in fields like chemistry and engineering, as it allows for the optimization of reactant usage and minimizes waste.

Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that affect these rates. For the given reaction involving nitrogen and hydrogen, the focus is not just on the balanced equation but on how fast the reactants turn into products over time.

• The rate can be defined in terms of the change in concentration of either reactants or products with time.
• Kinetic studies help determine the rate laws and how different conditions like temperature and pressure affect these rates.

For example, the rate of reaction \( -\mathrm{d}[ ext{Reactant}]/\mathrm{dt} \) is a way to express how the concentration of a reactant decreases per unit of time. In this exercise:

• For \( \mathrm{N}_{2} \), the rate is \( -\mathrm{d}[ ext{N}_2]/\mathrm{dt} \).
• For \( \mathrm{H}_{2} \), it's \( -\frac{1}{3} \mathrm{d}[ ext{H}_2]/\mathrm{dt} \) to account for stoichiometry.
• For \( \mathrm{NH}_{3} \), it's \( \frac{1}{2} \mathrm{d}[ ext{NH}_3]/\mathrm{dt} \).

Reaction kinetics is essential not only for predicting how fast a reaction will proceed but also for designing processes in industries, affecting yields, and optimizing conditions to achieve desired outcomes efficiently.