Question

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is: (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

Short answer

The resultant pressure is 1.2 atm (option a).

Step-by-step solution

Understand the decomposition reaction

The chemical reaction for the decomposition of \(\mathrm{N}_2\mathrm{O}_4\) is as follows: \[\mathrm{N}_2\mathrm{O}_4(g) \rightarrow 2\mathrm{NO}_2(g)\] If 20% by mass decomposes, this indicates a decomposition of 0.2 moles out of the initial 1 mole.

Moles of substances involved

Initially, we have 1 mole of \(\mathrm{N}_2\mathrm{O}_4\). After decomposition, 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) convert into 0.2*2 = 0.4 moles of \(\mathrm{NO}_2\). Hence, after decomposition, 0.8 moles of \(\mathrm{N}_2\mathrm{O}_4\) and 0.4 moles of \(\mathrm{NO}_2\) are present.

Calculate total moles of gas

Total moles of gas post-decomposition = moles of \(\mathrm{N}_2\mathrm{O}_4\) + moles of \(\mathrm{NO}_2\) = 0.8 + 0.4 = 1.2 moles.

Using ideal gas law concept

Initially, at 1 atm pressure and 1 mole, 1 atm corresponds to the initial condition. When the gas's total number of moles increases, we will find the new pressure proportionally since the volume and temperature can be assumed proportional under ideal conditions\: \( P_2 = P_1 \times \frac{\text{Total moles after decomposition}}{\text{Initial moles}} = 1 \times \frac{1.2}{1} = 1.2 \:\text{atm} \).

Conclusion

Thus, the resultant pressure after heating the container to \(600\:\mathrm{K}\) will be \(1.2\:\mathrm{atm}\).

Key concepts

Decomposition Reaction

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the given exercise, the compound \(\mathrm{N}_2\mathrm{O}_4\) undergoes decomposition to form \(\mathrm{NO}_2\). This reaction can be represented as:\[\mathrm{N}_2\mathrm{O}_4(g) \rightarrow 2\mathrm{NO}_2(g)\]The reaction indicates that one mole of \(\mathrm{N}_2\mathrm{O}_4\) produces two moles of \(\mathrm{NO}_2\). When dealing with decomposition reactions:

• Pay attention to the stoichiometry, which provides the mole ratio between reactants and products.
• Understand which compounds participate and how they transform into different products.
• Recognize that such reactions typically require some form of energy input, like heating, to proceed.

In this exercise, by heating to 600K, 20% of \(\mathrm{N}_2\mathrm{O}_4\) decomposes into \(\mathrm{NO}_2\), showcasing the practical application of decomposition in changing chemical states.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed as:\[\text{PV} = \text{nRT}\]where

• \(P\) stands for pressure,
• \(V\) is volume,
• \(n\) denotes the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) represents temperature in Kelvin.

This equation is useful for predicting how gases will react to changes in temperature, pressure, or volume. In cases where temperature and/or volume changes are not specified, they are typically assumed to be constant or can be considered in a comparable initial and final state, which allows solving for pressure changes. In the exercise, although precise temperature and volume values aren't provided, it's highlighted that as the moles change, the pressure adjusts proportionally, which follows the ideal gas law's logic.

Mole Concept

The mole concept is crucial for understanding chemical reactions, including decomposition. A mole is a unit that measures the amount of substance, and Avogadro’s number, \(6.022 \times 10^{23}\), defines the number of representative particles in a mole.In the given problem:

• We start with 1 mole of \(\mathrm{N}_2\mathrm{O}_4\).
• 20% decomposes, meaning 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) are converted into \(\mathrm{NO}_2\).
• The decomposition equation shows that 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) produce 0.4 moles of \(\mathrm{NO}_2\) due to stoichiometry.
• Thus, after decomposition, we have 0.8 moles of \(\mathrm{N}_2\mathrm{O}_4\) and 0.4 moles of \(\mathrm{NO}_2\) in the container.

Understanding these conversions is key to predicting the outcome of chemical processes and is foundational for calculating reactants and products in balanced equations.

Gas Pressure Calculation

Gas pressure calculation is pivotal when dealing with reactions conducted in closed systems, as gases expand or contract with reactions. This pressure indicates how gases behave and helps forecast expected conditions post-reaction.In this exercise:

• The initial pressure is 1 atm with 1 mole of gas present (\(\mathrm{N}_2\mathrm{O}_4\)).
• When 20% decomposes and produces \(\mathrm{NO}_2\), the total moles of gas post-reaction become 1.2 moles.
• The ideal gas law implies that, keeping temperature and volume proportions constant, pressure scales with the number of moles.
• Hence, new pressure \(P_2\) equals initial pressure times the ratio of total moles after decomposition over initial moles: \[ P_2 = 1 \times \frac{1.2}{1} = 1.2\,\text{atm} \]

This calculation shows how chemical changes can affect pressure readings and is based on the assumption that gases behave ideally during such transformations.