Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ The rate of this reaction in terms of \(\mathrm{N}_{2}\) at \(\mathrm{T}\) is \(-\mathrm{d}\left[\mathrm{N}_{2}\right] /\) \(\mathrm{dt}=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) What is the value of \(\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}\) (in units of \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) at the same temperature? (a) \(0.02\) (b) 50 (c) \(0.06\) (d) \(0.04\)

Step-by-Step Solution

Answer

(c) 0.06

Step 1: Write the Reaction Rate Expression

For the given chemical reaction, the rate of reaction can be expressed in terms of various reactants and products. According to stoichiometry, the rate with respect to each component can be expressed as: \[ -\frac{d[\mathrm{N}_2]}{dt} = \frac{1}{3} \times -\frac{d[\mathrm{H}_2]}{dt} = \frac{1}{2} \times \frac{d[\mathrm{NH}_3]}{dt} \]

Step 2: Relate  d[H2]/dt  to the Given Rate

Since you are given \(-\frac{d[\mathrm{N}_2]}{dt} = 0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\), you can relate it to \(-\frac{d[\mathrm{H}_2]}{dt}\) based on the stoichiometry of the reaction: \[ -\frac{d[\mathrm{H}_2]}{dt} = 3 \times -\frac{d[\mathrm{N}_2]}{dt} \]

Step 3: Calculate  d[H2]/dt

Substitute the known rate for \(-\frac{d[\mathrm{N}_2]}{dt}\) into the equation: \[ -\frac{d[\mathrm{H}_2]}{dt} = 3 \times 0.02 = 0.06 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \]

Key Concepts

Understanding Rate of Reaction

The rate of a reaction indicates how fast or slow a reaction occurs. It is an essential part of chemical kinetics, providing insight into the speed at which reactants turn into products. The rate is usually expressed as the change in concentration of a reactant or product per unit of time. For a reaction like the one between nitrogen gas \[ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \]:

• The rate can be written concerning each substance involved in the reaction.
• For reactants, this is a negative change since they decrease over time, e.g. \(-\frac{d[\mathrm{N}_2]}{dt}\).
• For products, it is positive as they are forming during the reaction, e.g. \(\frac{d[\mathrm{NH}_3]}{dt}\).

Understanding the rate helps chemists and engineers develop better processes by controlling how fast a reaction should proceed in real-life applications. By managing the reaction rate, industries can optimize product yield and energy usage.

Stoichiometry in Reactions

Stoichiometry is the part of chemistry that deals with understanding the relationships between the quantities of reactants and products in chemical reactions. It is essential to determine how different substances will react with each other. For instance, in our example reaction:\[ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \]n:

• The coefficients in the equation (1 for \(\mathrm{N}_{2}\), 3 for \(\mathrm{H}_{2}\), and 2 for \(\mathrm{NH}_{3}\)) represent the stoichiometric relationships.
• These numbers tell us that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.
• Understanding these relationships allows calculation of quantities of each reactant needed or the amount of product formed, based on the quantities of other substances involved in the reaction.

Stoichiometry also aids in relating the rate of change in concentration of one substance in a reaction to another, using the stoichiometric coefficients as shown in our exercise solution.

Reaction Rate Expression Explained

The reaction rate expression provides a mathematical framework to relate the rates of change of concentrations of reactants and products, constrained by their stoichiometry. Consider the expression given in the problem:\[-\frac{d[\mathrm{N}_2]}{dt} = \frac{1}{3} \times -\frac{d[\mathrm{H}_2]}{dt} = \frac{1}{2} \times \frac{d[\mathrm{NH}_3]}{dt}\]This equation tells us:

• The rate of decrease of nitrogen is proportionately split among the rates related to hydrogen and ammonia, utilizing their stoichiometric coefficients.
• The factor of \(\frac{1}{3}\) for \(\mathrm{H}_2\) and \(\frac{1}{2}\) for \(\mathrm{NH}_3\) comes from their coefficients in the balanced chemical equation, determining their fixed ratio at which they relate to \(\mathrm{N}_2\).

So when the problem provides the rate for nitrogen, you can use this expression to find the rate for hydrogen or ammonia by substituting the known values. This calculated rate explains how fast each component is changing during the reaction and is integral for understanding the dynamics of the system.