Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is a first-order reaction. At \(730^{\circ} \mathrm{C}\) the half-life of the reaction is \(3.58 \times 10^{3}\) min. If the initial pressure of \(\mathrm{N}_{2} \mathrm{O}\) is 2.10 atm at \(730^{\circ} \mathrm{C},\) calculate the total gas pressure after one half-life. Assume that the volume remains constant.

Short answer

The total gas pressure after one half-life is 2.625 atm.

Step-by-step solution

Understand the Problem

The problem involves the decomposition of nitrous oxide (\( \mathrm{N}_{2} \mathrm{O} \)) into nitrogen (\( \mathrm{N}_{2} \)) and oxygen (\( \mathrm{O}_{2} \)). The total gas pressure after one half-life is to be calculated. This occurs in a constant volume at a given initial pressure and temperature.

Reaction and Stoichiometry

The reaction is given by \( 2 \mathrm{N}_{2} \mathrm{O} \rightarrow 2 \mathrm{N}_{2} + \mathrm{O}_{2} \). For every 2 moles of \( \mathrm{N}_{2} \mathrm{O} \) that decompose, 2 moles of \( \mathrm{N}_{2} \) and 1 mole of \( \mathrm{O}_{2} \) are produced.

Determine Remaining Reactant

After one half-life, the amount of \( \mathrm{N}_{2} \mathrm{O} \) left is half of its initial amount due to the first-order kinetics. Therefore, the remaining \( \mathrm{N}_{2} \mathrm{O} \) pressure is \( \frac{2.10}{2} = 1.05 \) atm.

Calculate Total Pressure Increase

At the half-life, the decomposed \( \mathrm{N}_{2} \mathrm{O} \) is 1.05 atm. This produces 1.05 atm of \( \mathrm{N}_{2} \) and 0.525 atm of \( \mathrm{O}_{2} \) from the stoichiometry: \( 1 \) mole of \( \mathrm{O}_{2} \) for every 2 moles \( \mathrm{N}_{2} \mathrm{O} \).

Calculate Total Gas Pressure

The total pressure after one half-life is the sum of the pressures of all gases present: \( 1.05 \) atm (remaining \( \mathrm{N}_{2} \mathrm{O} \)) + \( 1.05 \) atm (produced \( \mathrm{N}_{2} \)) + \( 0.525 \) atm (produced \( \mathrm{O}_{2} \)). Total pressure = \( 2.625 \) atm.

Key concepts

Understanding Chemical Kinetics in First-Order Reactions

In chemical kinetics, the focus is on the rate at which a chemical reaction proceeds. A first-order reaction is one where the rate depends linearly on the concentration of a single reactant. For the decomposition of nitrous oxide \( (\mathrm{N}_{2}\mathrm{O}) \), this means that its rate decreases as \( \mathrm{N}_{2} \mathrm{O} \) decreases. The concept of half-life, commonly used in first-order reactions, is the time it takes for half of the reactant to decompose. For \( \mathrm{N}_{2}\mathrm{O} \) at \(730^{\circ} \mathrm{C} \), this half-life is \(3.58 \times 10^{3}\) minutes. The half-life remains constant throughout the reaction, which helps in predicting when half of any current amount of \( \mathrm{N}_{2} \mathrm{O} \) will decompose. This understanding is crucial in determining how much reactant remains and how much product forms after a given time, like one half-life.

The Role of Gas Pressure in Chemical Reactions

Gas pressure is a key factor in gas-phase reactions. It reflects the force exerted by gas molecules hitting the walls of a container. In the decomposition reaction of \( \mathrm{N}_{2} \mathrm{O} \), pressure changes help monitor the progress of the reaction. At the start, the pressure is determined solely by \( \mathrm{N}_{2} \mathrm{O} \). As the reaction proceeds, \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \) are formed, altering the pressure. After one half-life, the pressure due to \( \mathrm{N}_{2} \mathrm{O} \) is reduced to half its initial value because half of it has decomposed. The produced gases \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \) increase the total pressure. Thus, the total pressure at this point is the sum of the pressures exerted by \( \mathrm{N}_{2} \mathrm{O} \), \( \mathrm{N}_{2} \), and \( \mathrm{O}_{2} \). This principle shows how pressure data can be applied to deduce the extent of reaction progress without directly measuring concentrations of individual gases.

Interpreting Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. For \( \mathrm{N}_{2} \mathrm{O} \) decomposition, the balanced chemical equation is \(2 \mathrm{N}_{2} \mathrm{O} \rightarrow 2 \mathrm{N}_{2} + \mathrm{O}_{2}\). This indicates that two moles of \( \mathrm{N}_{2} \mathrm{O} \) produce two moles of \( \mathrm{N}_{2} \) and one mole of \( \mathrm{O}_{2} \). Stoichiometry enables us to calculate the amounts of products formed from a known quantity of reactants. After one half-life in our reaction, 1.05 atm of \( \mathrm{N}_{2} \mathrm{O} \) decomposes to produce equivalent 1.05 atm of \( \mathrm{N}_{2} \) and a corresponding \( 0.525 \) atm of \( \mathrm{O}_{2} \). Understanding stoichiometry is essential as it not only facilitates calculations of pressure changes in gas-phase reactions but also plays a critical role in scaling reactions for industrial applications.