Question

\(9.2 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\). At equilibrium, \(50 \%\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (g) is dissociated. What is the equilibrium constant (in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right) ?\) (molecular weight of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 92 ): (a) \(0.1\) (b) \(0.2\) (c) \(0.4\) (d) 2

Short answer

The equilibrium constant is 0.2 mol L^{-1}; option (b).

Step-by-step solution

Calculate Initial Moles of N2O4

First, calculate the initial moles of \( \mathrm{N}_2 \mathrm{O}_4 \). Given the mass is \( 9.2 \text{ g} \) and the molecular weight is \( 92 \), use the formula \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \). So, initial moles of \( \mathrm{N}_2 \mathrm{O}_4 \) are \( \frac{9.2}{92} = 0.1 \text{ moles} \).

Determine Moles at Equilibrium

Since \( 50\% \) of \( \mathrm{N}_2 \mathrm{O}_4 \) is dissociated, \( 0.5 \times 0.1 = 0.05 \text{ moles} \) of \( \mathrm{N}_2 \mathrm{O}_4 \) are dissociated. Therefore, remaining moles of \( \mathrm{N}_2 \mathrm{O}_4 \) at equilibrium are \( 0.1 - 0.05 = 0.05 \text{ moles} \). The dissociation results in the creation of \( 2 \times 0.05 = 0.1 \text{ moles} \) of \( \mathrm{NO}_2 \).

Calculate Concentrations at Equilibrium

The volume of the container is \( 1 \text{ litre} \). So the concentration of \( \mathrm{N}_2 \mathrm{O}_4 \) is \( \frac{0.05}{1} = 0.05 \text{ mol L}^{-1} \) and the concentration of \( \mathrm{NO}_2 \) is \( \frac{0.1}{1} = 0.1 \text{ mol L}^{-1} \).

Write Expression for Equilibrium Constant

The equilibrium expression for the reaction \( \mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2 \) is given by \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2 \mathrm{O}_4]} \).

Plug Values into Equilibrium Expression

Substitute the concentrations at equilibrium into the expression for the equilibrium constant: \( K_c = \frac{(0.1)^2}{0.05} \).

Solve for Equilibrium Constant

Calculate \( K_c \) using the expression: \( K_c = \frac{0.01}{0.05} = 0.2 \text{ mol L}^{-1} \).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a special number that tells us how well products and reactants balance once a chemical reaction has reached equilibrium. In the given exercise, we deal with the reaction involving \( \mathrm{N}_2 \mathrm{O}_4 \) and \( \mathrm{NO}_2 \). When at equilibrium, the concentrations of these substances remain constant. The equilibrium constant is calculated using the expression:

• \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2 \mathrm{O}_4]} \)

This ratio involves the concentrations of the products over the reactants raised to the power of their stoichiometric coefficients (the numbers in front of the chemical formulas in the balanced equation). For our reaction, the square is due to the two molecules of \( \mathrm{NO}_2 \) on the product side. Understanding \( K_c \) helps predict the direction of the reaction and how much of each substance is present at equilibrium.

Mole Calculation

Mole calculation serves as a foundation for many chemistry concepts. A mole is a unit that measures the amount of a substance. In this exercise, we first compute the initial moles of \( \mathrm{N}_2 \mathrm{O}_4 \). Given its mass and molecular weight, the moles are calculated using the formula:

• \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \)

For \( 9.2 \text{ g} \) of \( \mathrm{N}_2 \mathrm{O}_4 \) with a molecular weight of \( 92 \), the moles equal \( 0.1 \). Mole calculations are crucial in understanding how much of a substance participates, forms, or remains unchanged in a chemical reaction.

Chemical Reaction

A chemical reaction describes the process where substances, known as reactants, transform into different substances, called products. In this exercise, we're looking at the reaction:

• \( \mathrm{N}_2 \mathrm{O}_4 (g) \rightleftharpoons 2 \mathrm{NO}_2 (g) \)

This symbol \( \rightleftharpoons \) signifies a reversible reaction, meaning the reactants can form products and vice versa. Such reactions are essential because they can reach a state of equilibrium, where the rates of the forward and reverse reactions are equal. This reversibility is why we calculate concentration changes and ultimately find the equilibrium constant. Chemical reactions like these help us understand matter's transformation in nature.

Dissociation

Dissociation refers to the process where a compound breaks into simpler parts. It's crucial in understanding reactions that involve compounds splitting into ions or smaller molecules. In our scenario, dissociation plays a role as:

• \( 50\% \) of \( \mathrm{N}_2 \mathrm{O}_4 \) dissociates to form \( 2 \mathrm{NO}_2 \)

Here, starting with \( 0.1 \text{ moles} \) of \( \mathrm{N}_2 \mathrm{O}_4 \), half dissociates, turning into \( 0.05 \text{ moles} \) of \( \mathrm{N}_2 \mathrm{O}_4 \) and releasing \( 0.1 \text{ moles} \) of \( \mathrm{NO}_2 \). Analyzing dissociation helps with understanding how molecular changes affect overall reaction dynamics, balancing how substances transform and exist at equilibrium.