Question

The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature: (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) None of these

Short answer

The percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(53.3\%\).

Step-by-step solution

Understanding Vapour Density

Vapour density is defined as half the molar mass of the substance. For \ \(\text{N}_2\text{O}_4\ \), the molar mass is calculated by \(2 \times \text{Molar mass of } \text{N} + 4 \times \text{Molar mass of } \text{O} = 2 \times 14 + 4 \times 16 = 92\ \). Hence, vapour density of \(\text{N}_2\text{O}_4\) is \(\frac{92}{2} = 46\ \).

Applying Vapour Density Concept to Dissociation

The vapour density given in the problem is 30. Since vapour density is \( \frac{Molar \, Mass}{2} \), this means that the apparent molar mass at the temperature in the problem is \(2 \times 30 = 60\ \).

Analyzing Dissociation

Initially, we have a molar mass of \(92\), and after dissociation, the apparent molar mass is \(60\). This dissociation implies some of \(\text{N}_2\text{O}_4\) dissociates into \(\text{NO}_2\). This can be represented as: \(\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2\). Let \(x\) be the degree of dissociation, then, for one mole of \(\text{N}_2\text{O}_4\), \(1 - x\) moles remain and \(2x\) moles of \(\text{NO}_2\) are formed.

Relating Degree of Dissociation to Apparent Molar Mass

The total molar quantity present after dissociation is \((1-x) + 2x = 1 + x\) moles. The effective molar mass is consequentially \(\frac{92}{1+x} = 60\ \).

Solving for Degree of Dissociation

We solve for \(x\) using \(\frac{92}{1+x} = 60\). Solving, \(1+x = \frac{92}{60}\) which simplifies to \(1+x = \frac{23}{15}\). This gives \(x = \frac{23}{15} - 1 = \frac{8}{15}\).

Calculating Percentage Dissociation

The percentage dissociation is calculated as \(x \times 100 = \frac{8}{15} \times 100 = 53.3\).

Key concepts

Percentage Dissociation

Percentage dissociation is an important concept in chemistry that helps us understand how much of a molecule breaks into its components under certain conditions. When we say a compound like \( \text{N}_2\text{O}_4 \) dissociates, it means that it splits into smaller molecules, such as \( \text{NO}_2 \) in this case. To find the percentage dissociation, we first calculate the degree of dissociation, denoted by \( x \), which represents the fraction of the original compound that has dissociated. In the original problem, the degree of dissociation was found to be \( \frac{8}{15} \). To convert this into a percentage, we multiply it by 100, resulting in a percentage dissociation of \( 53.3\% \). This tells us that more than half of the \( \text{N}_2\text{O}_4 \) molecules have split into \( \text{NO}_2 \) molecules at the specified temperature, indicating a significant level of dissociation.

Understanding the percentage dissociation helps us determine the state and behavior of a compound in a chemical equilibrium. If the percentage dissociation is high, it means the reaction heavily favors the formation of products, leading to a different set of conditions in the reaction mixture.

Molar Mass Calculation

Molar mass is a crucial concept used to understand the properties of substances in chemistry. It denotes the mass of one mole of a substance, expressed in grams per mole (g/mol). Calculating the molar mass allows us to convert between the amount of substance and the mass. The molar mass of a molecule is determined by adding together the molar masses of the individual atoms that comprise it.

For instance, in the problem concerning \( \text{N}_2\text{O}_4 \), the molar mass is determined by calculating:

• The molar mass of nitrogen (\( \text{N} \)) is 14 g/mol.
• The molar mass of oxygen (\( \text{O} \)) is 16 g/mol.

Since \( \text{N}_2\text{O}_4 \) consists of two nitrogen atoms and four oxygen atoms, the total molar mass can be calculated as:
\[ 2 \times 14 + 4 \times 16 = 92 \text{ g/mol} \]
Using the calculated molar mass, we can understand other properties, such as vapour density, and derive important information about the dissociation of the compound under different conditions. In this problem, the initial molar mass helps relate vapour density to dissociation, allowing further insights into the chemical behavior of \( \text{N}_2\text{O}_4 \).

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. In other words, the concentrations of reactants and products remain constant over time. For the dissociation of \( \text{N}_2\text{O}_4 \), the equilibrium can be depicted as:
\[ \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2 \]
Initially, when the reaction starts, \( \text{N}_2\text{O}_4 \) begins to dissociate into \( \text{NO}_2 \). As \( \text{NO}_2 \) molecules form, they also react with each other to reform \( \text{N}_2\text{O}_4 \). At equilibrium, these two rates are equal.

To analyze such a reaction, we consider the initial concentration of reactants and the change due to the reaction. Here, if \( x \) is the degree of dissociation, then at equilibrium the concentration of \( \text{N}_2\text{O}_4 \) would be \( 1-x \), and the concentration of \( \text{NO}_2 \) would be \( 2x \). The total number of moles, \( 1 + x \), leads to changes in the properties of the gaseous mixture, like vapour density.

Equilibrium is crucial for understanding the conditions under which reactions proceed. It helps in predicting the concentrations of products and reactants, assisting in the calculation of various properties like the degree of dissociation. Knowing the behavior at equilibrium enables adjustments in reaction conditions to achieve desired outcomes in chemical processes.