Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, \(1.10\) moles of \(\mathrm{NOCl}\) remained. Calculate the equilibrium constant \(K\) for the reaction, \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

Step-by-Step Solution

Answer

The equilibrium constant \(K\) is \(5.6 \times 10^{-4}\), option (b).

Step 1: Write the Reaction and Known Values

The given chemical reaction is \( 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \). We are given that initially, there are \( 1.25 \) moles of \( \mathrm{NOCl} \) in a \( 2.50 \mathrm{~L} \) chamber. At equilibrium, \( 1.10 \) moles of \( \mathrm{NOCl} \) remain.

Step 2: Calculate Initial Concentrations

The initial concentration of \( \mathrm{NOCl} \) is \( \frac{1.25 \text{ moles}}{2.50 \text{ L}} = 0.50 \text{ M} \). Since \( \mathrm{NO} \) and \( \mathrm{Cl}_2 \) are not initially present, their initial concentrations are \( 0 \text{ M} \).

Step 3: Determine Change and Equilibrium Concentrations

At equilibrium, \( 1.10 \) moles of \( \mathrm{NOCl} \) remain, equating to a concentration of \( \frac{1.10 \text{ moles}}{2.50 \text{ L}} = 0.44 \text{ M} \). Therefore, the change in concentration for \( \mathrm{NOCl} \) is \( -0.06 \text{ M} \) (i.e., \( 0.50 \text{ M} - 0.44 \text{ M} \)). The change for \( \mathrm{NO} \) is \( +0.06 \text{ M} \) and for \( \mathrm{Cl}_2 \) is \( +0.03 \text{ M} \) due to stoichiometry of the reaction.

Step 4: Write the Equilibrium Concentrations

At equilibrium, the concentration of \( \mathrm{NO} \) is \( 0.06 \text{ M} \), and the concentration of \( \mathrm{Cl}_2 \) is \( 0.03 \text{ M} \). The equilibrium concentration of \( \mathrm{NOCl} \) is \( 0.44 \text{ M} \).

Step 5: Calculate the Equilibrium Constant \( K \)

The equilibrium constant \( K \) is calculated as follows: \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2} = \frac{(0.06)^2 \times 0.03}{(0.44)^2} = \frac{0.000108}{0.1936} \approx 5.6 \times 10^{-4}\]

Step 6: Choose the Correct Answer

Comparing our calculated equilibrium constant \( K \approx 5.6 \times 10^{-4} \) with the options given, the correct answer is (b) \( 5.6 \times 10^{-4} \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is the state in a chemical reaction where the rates of the forward and the reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. This doesn't mean the reactions stop, but rather they occur simultaneously at the same pace.
For the given reaction,

• Initially, nitric oxide chloride, NOCl, starts to decompose into nitric oxide, NO, and chlorine gas, Clequiv 2.
• As NO and Clequiv 2 start forming, they can also react back to form NOCl.

The point when the amount of NO, NOCl, and Clequiv 2 in the reaction chamber remains constant, albeit with continuous interconversion, is known as chemical equilibrium.
The equilibrium constant, denoted as \( K \), gives us a snapshot of the ratio of product concentrations to reactant concentrations at this state. It is a crucial aspect of assessing how far the reaction has gone towards completion.

Mole Calculations

Mole calculations are fundamental in understanding chemical reactions, focusing on measuring substance quantities using moles. A mole is tied to Avogadro’s number, representing approximately \( 6.022 \times 10^{23} \) entities, like atoms or molecules.
In our exercise, we started with \( 1.25 \) moles of NOCl. This is termed the initial number of moles. You were then asked to calculate how many moles of the products and remaining reactants exist at equilibrium.
Here's how this breaks down:

• Use the given changes in moles to understand how many reacted and formed of each species.
• Since NOCl decreased and NO & Clequiv 2 were produced, calculate their final moles starting from their changes derived from stoichiometric coefficients.

All of this helps in computing equilibrium concentrations, essential for figuring out the reaction’s equilibrium constant, \( K \).

Stoichiometry

Stoichiometry helps connect mole calculations to real-world chemical reactions, embodying the essential proportional relationships between reactants and products in reactions. In the reaction \( 2 \text{NOCl}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Cl}_2(g) \), the stoichiometric coefficients are the numbers in front of each compound, which are vital for understanding reaction progression.

• For every two moles of NOCl that decompose, two moles of NO and one mole of Clequiv 2 are formed.
• The stoichiometric ratio is always consistent, mirroring these coefficients in terms of how much each substance increases or decreases within the reaction.

By linking the changes in concentration to these coefficients, you can easily determine how much of each reactant is used and how much of each product is formed, essential for both qualitative and quantitative analysis of chemical reactions.
Ultimately, understanding stoichiometry lets you accurately perform and predict results for equilibrium constant calculations, thus honing your grasp on chemical equilibria.