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 \(K_{c}\) 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 \(K_{c}\) for the reaction is \(5.6 \times 10^{-4}\) (option b).

Step 1: Identify Initial Moles and Volume

We start with 1.25 moles of NOCl in a 2.50 L chamber, and at equilibrium, 1.10 moles of NOCl remains. The change in moles of NOCl is the difference between initial moles and equilibrium moles, which is 1.25 - 1.10 = 0.15 moles.

Step 2: Calculate Changes in Moles

Since the reaction is 2NOCl(g) → 2NO(g) + Cl2(g), for every 2 moles of NOCl that react, 2 moles of NO and 1 mole of Cl2 are produced. Therefore, 0.15 moles of NOCl reacting will produce 0.15 moles of Cl2 and 0.15 moles of NO.

Step 3: Calculate Equilibrium Concentrations

Calculate the concentrations at equilibrium. For NOCl, it's (1.10 moles / 2.50 L) = 0.44 M. For NO, it's (0.15 moles / 2.50 L) = 0.06 M. For Cl2, (0.075 moles / 2.50 L) = 0.03 M.

Step 4: Write the Equilibrium Constant Expression

The equilibrium expression for the reaction \(2\, \text{NOCl} \rightleftharpoons 2\, \text{NO} + \text{Cl}_2\) is \(K_c = \frac{[\text{NO}]^2[\text{Cl}_2]}{[\text{NOCl}]^2}\).

Step 5: Substitute Equilibrium Concentrations into Expression

Substitute the equilibrium concentrations into the expression: \(K_c = \frac{(0.06)^2(0.03)}{(0.44)^2} = \frac{0.000108}{0.1936}\).

Step 6: Calculate the Equilibrium Constant

Calculate the value of \(K_c\). Performing the division gives \(K_c = \frac{0.000108}{0.1936} \approx 5.6 \times 10^{-4}\).

Step 7: Choose the Correct Option

From the provided options, (b) \(5.6 \times 10^{-4}\) is our calculated value.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K_c \), is an essential concept in understanding chemical equilibrium. It is a numerical value that describes the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. The equilibrium expression depends on the balanced chemical equation and reflects the chemical species involved. For gaseous reactions, we talk about concentrations in terms of molarity (moles per liter).

In the context of our reaction, \( 2 \text{NOCl(g)} \rightleftharpoons 2 \text{NO(g)} + \text{Cl}_2(g) \), the equilibrium constant expression is written as:

• \( K_c = \frac{[\text{NO}]^2[\text{Cl}_2]}{[\text{NOCl}]^2} \)

Understanding this constant helps predict the reaction's behavior at equilibrium and how it shifts when conditions change. A small \( K_c \) value, like the one calculated, indicates that at equilibrium, the concentration of reactants is higher than that of products.

Reaction Quotient

The reaction quotient, \( Q \), is a crucial tool that allows chemists to predict the direction in which a reaction mixture will proceed to reach equilibrium. \( Q \) is calculated in the same way as the equilibrium constant \( K_c \), but it uses the initial concentrations or pressures of the reactants and products rather than their equilibrium values.

If:

• \( Q < K_c \): The reaction will proceed forward, forming more products.
• \( Q = K_c \): The reaction is at equilibrium, with no net change in concentrations.
• \( Q > K_c \): The reaction will shift backward, forming more reactants.

In our solved problem, we didn't calculate \( Q \) because we were given the equilibrium concentrations directly. However, knowing \( Q \) helps understand how far a reaction is from equilibrium and the direction it needs to go.

Mole Calculations

Mole calculations are foundational in solving chemical equilibrium problems. They allow the determination of changes in the quantities of reactants and products when a chemical reaction proceeds. Knowing the initial amounts and using simple subtraction can give insights into how much a reactant is consumed or a product is formed.

For the reaction \( 2 \text{NOCl(g)} \rightleftharpoons 2 \text{NO(g)} + \text{Cl}_2(g) \), initially, we started with 1.25 moles of NOCl. After equilibrium:

• 0.15 moles are consumed, calculated by \(1.25 - 1.10 = 0.15\).
• Thus, 0.15 moles of \( \text{NO} \) and 0.075 moles of \( \text{Cl}_2 \) are formed.

These values are integral to finding the equilibrium concentrations, leading us to the equilibrium constant.

Reaction Stoichiometry

Reaction stoichiometry involves using the coefficients from the balanced chemical equation to relate the moles of reactants to the moles of products. It is a straightforward yet powerful tool in chemistry that facilitates mole calculations and helps understand the proportionality of reactants and products.

In the chemical equation, \(2 \text{NOCl(g)} \rightarrow 2 \text{NO(g)} + \text{Cl}_2(g)\), we see that for every 2 moles of NOCl that break down, 2 moles of NO and 1 mole of \(\text{Cl}_2\) are produced.

• This balanced reaction tells us that NOCl and NO are formed in a 1:1 mole ratio, while Cl\(_2\) is produced in a 1:2 ratio of NOCl.
• Thus, understanding these relationships is fundamental in determining how changes in one substance affect the others during the reaction process.

Stoichiometry not only supports solving equilibrium calculations but also provides a framework to predict the amounts of substances involved in any chemical reaction.