Problem 89 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following reaction equilibrium \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) Initially, 1 mole of \(\mathrm{N}_{2}\) and 3 mole of \(\mathrm{H}_{2}\) are taken in a 2 litre flask. At equilibrium state, if the number of union of \(\mathrm{N}_{2}\) in \(0.6\), what is the total number of moles of all gases present in the flask? (a) \(0.8\) (b) \(1.6\) (c) \(3.2\) (d) \(6.4\)

Step-by-Step Solution

Answer

The total number of moles of all gases at equilibrium is 3.2 moles.

Step 1: Understand the Reaction

The given reaction is: \(\mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\). Initially, there are 1 mole of \(\mathrm{N}_2\) and 3 moles of \(\mathrm{H}_2\) in a 2-litre flask.

Step 2: Initial Concentrations

The initial concentrations of \(\mathrm{N}_2\) and \(\mathrm{H}_2\) are calculated by dividing the number of moles by the volume of the flask. Therefore, \([\mathrm{N}_2]_0 = \frac{1}{2} = 0.5 \text{ M}\) and \([\mathrm{H}_2]_0 = \frac{3}{2} = 1.5 \text{ M}\).

Step 3: Change in Concentrations

At equilibrium, if 0.6 moles of \(\mathrm{N}_2\) remain, then \(0.4\) moles of \(\mathrm{N}_2\) have reacted. According to the reaction stoichiometry, \(3 \times 0.4 = 1.2\) moles of \(\mathrm{H}_2\) react, and \(2 \times 0.4 = 0.8\) moles of \(\mathrm{NH}_3\) are produced.

Step 4: Equilibrium Concentrations

Calculate how many moles are present at equilibrium: \([\mathrm{N}_2] = 1 - 0.4 = 0.6 \text{ moles}\), \([\mathrm{H}_2] = 3 - 1.2 = 1.8 \text{ moles}\), and \([\mathrm{NH}_3] = 0.8 \text{ moles}\).

Step 5: Total Moles at Equilibrium

Sum the moles of all gases at equilibrium: \(0.6 + 1.8 + 0.8 = 3.2 \text{ moles}\).

Key Concepts

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It's essential for understanding how reactions occur and how energy and mass balance in a chemical system. In the context of our reaction \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\), stoichiometry helps us determine the proportional amounts of each reactant needed and how they convert into products. For example, according to the balanced equation, 1 mole of \(\text{N}_2\) reacts with 3 moles of \(\text{H}_2\) to form 2 moles of \(\text{NH}_3\). This means that the nitrogen and hydrogen react in a 1:3 ratio, which ensures that no reactants are wasted and the reaction proceeds efficiently. Without stoichiometry, we wouldn't know how much of each substance to combine.

Reaction Equilibrium

Chemical equilibrium occurs when a chemical reaction takes place at a rate where the forward reaction (reactants forming products) equals the rate of the reverse reaction (products decomposing back into reactants). At this point, the concentration of each substance remains constant over time, even though the reactions continue to occur.
In the given reaction, equilibrium helps to determine that 0.6 moles of \(\text{N}_2\) remain unreacted. By using this information, we can figure out how much \(\text{H}_2\) has reacted to form \(\text{NH}_3\). The equilibrium does not mean the reaction stops but reaches a state where there's a balance between the moles reacting and the moles regenerating. Thus, equilibrium is crucial for calculating the final quantities of reactants and products, as well as understanding the dynamics of the reaction process.

Concentration Calculation

Concentration calculations are vital for understanding the amount of each reactant and product present in a solution at any given time. This is typically expressed in terms of molarity, which is moles per liter.
For our reaction, the initial concentrations are calculated based on the initial amounts of \(\text{N}_2\) and \(\text{H}_2\) relative to the 2-liter volume of the flask. The initial concentrations are \([\text{N}_2] = 0.5 \text{ M}\) and \([\text{H}_2] = 1.5 \text{ M}\). At equilibrium, these concentrations change due to the reaction progress; knowing that 0.6 moles of \(\text{N}_2\) are left implies 0.4 moles have reacted, changing \([\text{H}_2]\) to 1.8 moles and producing 0.8 moles of \(\text{NH}_3\). Summing these values provides 3.2 total moles in the flask, which aligns with the stoichiometry and reveals the state of the system at equilibrium.