Problem 6 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For a reaction \(2 \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2}+3 \mathrm{H}_{2}\), it is observed that \(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t}=K_{1}\left[\mathrm{NH}_{3}\right] ;+\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=K_{2}\left[\mathrm{NH}_{3}\right]\) and \(+\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t}=K_{3}\left[\mathrm{NH}_{3}\right]\) The correct relation between \(K_{1}, K_{2}\) and \(K_{3}\) is (a) \(K_{1}=K_{2}=K_{3}\) (b) \(2 K_{1}=3 K_{2}=6 K_{3}\) (c) \(3 K_{1}=6 K_{2}=2 K_{3}\) (d) \(6 K_{1}=3 K_{2}=2 K_{3}\)

Step-by-Step Solution

Answer

The correct relation between the rate constants is (c) \(3 K_{1} = 6 K_{2} = 2 K_{3}\).

Step 1: Identify Given Reaction and Rates

The given chemical reaction is: \(2 \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2} + 3 \mathrm{H}_{2}\). It is observed that the rate of disappearance of ammonia (\(\mathrm{NH}_{3}\)) is \(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d}t}\), the rate of appearance of nitrogen (\(\mathrm{N}_{2}\)) is \(+\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d}t}\), and the rate of appearance of hydrogen (\(\mathrm{H}_{2}\)) is \(+\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d}t}\). Each is proportional to the concentration of ammonia (\(\mathrm{NH}_{3}\)), with rate constants \(K_{1}, K_{2},\) and \(K_{3}\) respectively.

Step 2: Write Rate Equations Based on Stoichiometry

According to stoichiometry of the balanced chemical equation, 2 moles of ammonia yields 1 mole of nitrogen and 3 moles of hydrogen. Therefore, the rates of formation of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are related to the rate of disappearance of \(\mathrm{NH}_{3}\) by the ratios of their stoichiometric coefficients. The rate equations are: \(-\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d}t} = \frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d}t} = \frac{1}{3}\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d}t}\).

Step 3: Apply the Stoichiometric Ratios to the Rate Constants

Given that formation rates of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are proportional to the disappearance rate of \(\mathrm{NH}_{3}\), the following relations can be established using the stoichiometric coefficients: \(K_{1} = 2K_{2}\) and \(K_{1} = \frac{2}{3}K_{3}\).

Step 4: Find the Correct Relationship Between the Rate Constants

From the previous step, we can conclude that: \(2K_{2} = \frac{2}{3}K_{3}\), which simplifies to \(3K_{2} = K_{3}\). Now, combine this with the established relation \(K_{1} = 2K_{2}\) to get \(K_{1} = 2K_{2} = \frac{2}{3}K_{3}\), which after multiplying through by 3 to clear the fraction becomes \(3K_{1} = 6K_{2} = 2K_{3}\).

Key Concepts

Reaction Rates

In chemical kinetics, reaction rates refer to the speed at which reactants are converted into products. It's essential for students to understand that reaction rates can be affected by various factors, including concentration, temperature, and the presence of a catalyst.

In the given problem, the reaction rate is measured in terms of the time rate change of the concentration of ammonia (\textbf{NH}\(_3\)). Specifically, we look at the decrease in \textbf{NH}\(_3\) concentration, which is represented by the negative differential \textbf{-d[\textbf{NH}\(_3\)]/dt}. Understanding this concept is crucial, as it's a direct measure of how quickly the reaction is occurring. In kinetics problems, getting a grasp on the significance of these negative and positive signs in rate expressions is the first step. These indicate whether a species is being consumed (negative sign) or produced (positive prices) over time.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced during reactions.

In the provided exercise, stoichiometry plays a critical role in linking the disappearance of ammonia to the formation of nitrogen and hydrogen. It's used to set up a proportional relationship between the respective rates of reactants and products based on their coefficients in the balanced equation. This means for every 2 moles of \textbf{NH}\(_3\) used, 1 mole of \textbf{N}\(_2\) and 3 moles of \textbf{H}\(_2\) are produced. Translating these stoichiometric ratios into rate expressions is vital in solving the problem.

Rate Law

Rate law is an expression that relates the reaction rate to the concentration of the reactants. It is generally represented as rate = k[\textbf{A}]\(^m\)[\textbf{B}]\(^n\)..., where 'k' is the rate constant, and 'm' and 'n' are the orders of the reaction with respect to reactants \textbf{A} and \textbf{B}, respectively. The rate law is determined experimentally and provides insight into the mechanism of the reaction.

In this case, the rate law is assumed to be directly proportional to the concentration of \textbf{NH}\(_3\). Thus, the rate constant 'k' for each substance (\textbf{K}\(_1\), \textbf{K}\(_2\), and \textbf{K}\(_3\)) will allow us to establish a relationship that uniquely defines the reaction rate. Understanding how to use the rate law and the significance of the rate constant is central to solving kinetic problems.

Differential rate expressions involve calculus, specifically the use of derivatives to describe rates of reaction. They represent how the concentration of a substance changes over time. Differential rate expressions are instrumental in deriving integrated rate laws and understanding reaction dynamics.

The problem presented includes differential expressions for both the consumption of ammonia and the production of nitrogen and hydrogen. By understanding that the rates of formation for these products must be in direct proportion to their stoichiometric coefficients, we use these expressions to find the relationship between the rate constants. Mastering differential rate expressions is crucial for students as it provides a window into the instantaneous rate at any given point in a chemical reaction.