Problem 75 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If \(0.25\) per cent of nitrogen is converted to ammonia by the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ calculate the equilibrium constant \(\left(K_{c}\right)\) in concentration units. What will be the value of \(K_{c}\) for the following equilibrium? $$ \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{ag}) $$

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) for the first reaction is 0.000018, and for the adjusted reaction, \(K_c'\) is 0.0042.

Step 1: Initial Moles and Volume

You start with 1 mole of nitrogen and 3 moles of hydrogen in a 4-litre container. This information sets the framework for calculating initial concentrations. Initial concentration of nitrogen, \([N_2]_0\), is \(\frac{1 \text{ mole}}{4 \text{ L}} = 0.25 \text{ M}\), and initial concentration of hydrogen, \([H_2]_0\), is \(\frac{3 \text{ moles}}{4 \text{ L}} = 0.75 \text{ M}\). Ammonia, initially, is absent so its concentration is \([NH_3]_0 = 0 \text{ M}\).

Step 2: Change in Concentrations due to Reaction

According to the reaction, 0.25% of nitrogen is converted into ammonia. Hence, the moles of nitrogen used are \(0.25\% \times 1 = 0.0025 \text{ moles}\). This reaction also uses three times as much hydrogen, meaning \(3 \times 0.0025 = 0.0075\text{ moles of } H_2\) are used, producing \(2 \times 0.0025 = 0.005 \text{ moles of } NH_3\).

Step 3: Concentration at Equilibrium

From Step 2, we deduced that 0.0025 moles of \(N_2\), 0.0075 moles of \(H_2\) are consumed, while 0.005 moles of \(NH_3\) are formed. Thus, the concentrations at equilibrium are: \([N_2] = 0.25 - 0.0025 = 0.2475\text{ M}\), \([H_2] = 0.75 - 0.0075 = 0.7425\text{ M}\), \([NH_3] = 0.005\text{ M}\).

Step 4: Equilibrium Constant \(K_c\) for First Reaction

The expression for the equilibrium constant \(K_c\) for the reaction \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\) is: \[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]Substituting the equilibrium concentrations, we get: \[ K_c = \frac{(0.005)^2}{(0.2475)(0.7425)^3} \approx 0.000018 \]

Step 5: Adjusting \(K_c\) for Second Reaction

For the reaction \( \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g)\), the balanced equation is different, essentially half of the original reaction. The equilibrium constant for this reaction is related to the initial \(K_c\) by raising it to the power \(\frac{1}{2}\), i.e., \(K_c' = \sqrt{K_c}\).Thus, \[ K_c' = \sqrt{0.000018} \approx 0.0042\]

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the concentrations of reactants and products remain constant over time. This happens because the rate of the forward reaction equals the rate of the reverse reaction. When we talk about equilibrium in chemical reactions, we're implying that the system has reached a balance where the concentration levels don't change anymore, even though reactions are still occurring.
This concept is crucial because it determines how much of a product will be available at any given point once equilibrium is achieved. Understanding chemical equilibrium helps predict the direction in which a reaction will proceed and how it will behave under different conditions such as changes in concentration, temperature, or pressure. In our reaction example:

• The nitrogen and hydrogen gases reach an equilibrium with the ammonia gas.
• The equilibrium constant, denoted as \( K_c \), quantifies the ratio of product concentration to reactant concentration at this equilibrium stage.

These equilibrium states can be shifted by altering the conditions, which is known as Le Chatelier's principle.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationships between the substances involved in a chemical reaction. It is essentially the calculation of the relative quantities of reactants and products in chemical reactions. In our exercise, stoichiometry guides us on how many moles of nitrogen and hydrogen are consumed and how many moles of ammonia are produced.
For the given reaction:

• One mole of nitrogen \( \text{N}_2 \) reacts with three moles of hydrogen \( \text{H}_2 \) to produce two moles of ammonia \( \text{NH}_3 \).
• Knowing this stoichiometric ratio allows us to determine how changes in the molar amounts of reactants affect the production of products.
• In this case, a 0.003 moles change in nitrogen results in a 0.005 moles increase in ammonia, following the molecular coefficients of the balanced equation.

This relation supports the completion of concentration changes needed in calculations, as it helps quantify the exact consumption or generation of a species in the reaction.

Concentration Calculations

Concentration calculations are a vital part of chemical equilibrium problems as they allow us to understand the quantity of substances in a given reaction mixture. Concentration is typically expressed in molarity (\( M \)), which is moles per liter. In our exercise:

• We began with concentrations found by dividing the moles of each component by the volume of the container (4 liters, in this case).
• Initially, nitrogen and hydrogen concentrations are 0.25 M and 0.75 M, respectively.
• After some reaction progress, equilibrium concentrations are calculated by subtracting the amount of reactants that have been converted to products.
• For ammonia, since it is formed, the concentration increases to 0.005 M.

These concentrations are then used in the equilibrium constant formula \( K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \) to calculate \( K_c \).
Understanding concentration calculations ensures that students can practically apply equilibrium concepts to find how much of a substance is present at any stage of a reaction, which is essential for real-world chemical applications.