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

The rate of consumption of \(\mathrm{H}_2\) is \(0.06\ \mathrm{mol\ L^{-1}\ s^{-1}}\), which matches option (c).

Step 1: Understand the Reaction Kinetics

The reaction given is \(\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \to 2\mathrm{NH}_3(g)\). The rate of appearance of \(\mathrm{NH}_3\) is related to the consumption rates of \(\mathrm{N}_2\) and \(\mathrm{H}_2\). For each mole of \(\mathrm{N}_2\) consumed, 3 moles of \(\mathrm{H}_2\) are consumed.

Step 2: Write the Rate Expressions

The rate expression for \(\mathrm{N}_2\) is given by \( - \frac{d[\mathrm{N}_2]}{dt} = 0.02\ \mathrm{mol\ L^{-1}\ s^{-1}} \). The stoichiometry of the reaction indicates that \(\mathrm{H}_2\) will be consumed at a rate three times faster than \(\mathrm{N}_2\).

Step 3: Relate the Rates Using Stoichiometry

According to the stoichiometric coefficients: \(-\frac{d[\mathrm{N}_2]}{dt} = \frac{1}{3} \cdot -\frac{d[\mathrm{H}_2]}{dt}\). Therefore, \( -\frac{d[\mathrm{H}_2]}{dt} = 3 \cdot (-\frac{d[\mathrm{N}_2]}{dt})\).

Step 4: Substitute and Calculate

Substitute the given rate of \(\mathrm{N}_2\):\[ -\frac{d[\mathrm{H}_2]}{dt} = 3 \times 0.02\ \mathrm{mol\ L^{-1}\ s^{-1}} = 0.06\ \mathrm{mol\ L^{-1}\ s^{-1}} \].

Step 5: Match with the Options

Compare \(0.06\ \mathrm{mol\ L^{-1}\ s^{-1}}\) with the given options:- (a) \(0.02\)- (b) 50- (c) \(0.06\)- (d) \(0.04\).The correct answer is (c) \(0.06\).

Key Concepts

Stoichiometry

Stoichiometry is a crucial concept in chemistry, primarily dealing with the quantitative relationships between the reactants and products in a chemical reaction. It helps us determine how much of one substance is needed to react with a certain amount of another, or how much product can be formed in a reaction. This is essential for predicting yields and for scaling reactions from laboratory to industrial scale.

In the context of the given reaction \[ \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g) \]stoichiometry provides us with conversion factors based on the ratio of molecules or moles. Here, we see that:

• 1 mole of \(\mathrm{N}_2\) reacts with 3 moles of \(\mathrm{H}_2\)
• to produce 2 moles of \(\mathrm{NH}_3\)

This ratio, or stoichiometric coefficient, allows chemists to convert between the amounts of reactants and products. By examining these coefficients, we understand that for every mole of nitrogen consumed, three moles of hydrogen are consumed. This informs the calculation of the rate at which hydrogen is consumed compared to nitrogen.

Rate of Reaction

The rate of reaction is how fast a reaction takes place, generally expressed as the change in concentration of reactants or products per unit time. In this problem, the reaction rate concerning \(\mathrm{N}_2\) is given as:\[-\mathrm{d}\left[\mathrm{N}_2\right] / \mathrm{dt} = 0.02\ \mathrm{mol\ L^{-1}\ s^{-1}}\]This signifies the rate at which the nitrogen gas is consumed in the reaction. The negative sign denotes the decrease in concentration of \(\mathrm{N}_2\) over time.

To find the rate at which \(\mathrm{H}_2\) is used, we apply the stoichiometric relationship from the previous section. According to the stoichiometry of the reaction, hydrogen is consumed three times more rapidly than nitrogen, hence:

• The rate of consumption for \(\mathrm{H}_2\) would be three times that of \(\mathrm{N}_2\).
• This gives us the rate:\[ -\frac{d[\mathrm{H}_2]}{dt} = 3 \times 0.02 = 0.06\ \mathrm{mol\ L^{-1}\ s^{-1}}\]

Chemical Equations

Chemical equations represent the symbolic representation of a chemical reaction with reactants on the left, products on the right, and an arrow showing the direction of the reaction. They are balanced with respect to the conservation of mass and charge, meaning the same number of each type of atom is present among both reactants and products.

In the equation\[\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g)\]the molecules of \(\mathrm{N}_2\) and \(\mathrm{H}_2\) are reactants, whilst \(\mathrm{NH}_3\) is the product. This equation exemplifies a balanced equation with equal nitrogen and hydrogen atoms on both sides, confirming the principle of atomic conservation.

Besides helping chemists determine reaction balance, chemical equations can also be used to understand reaction yields, identify limiting reactants, and calculate reaction kinetics. They provide a compact way of expressing what happens in a reaction and serve as the basis for understanding reaction progress and mechanism.