Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The resultant pressure is 1.2 atm (option d).

Step 1: Understand the Given Information

We start with 1 mole of \(\mathrm{N}_2\mathrm{O}_4(\mathrm{~g})\) and it undergoes decomposition at \(600 \mathrm{~K}\) to \(\mathrm{NO}_2(\mathrm{~g})\) such that 20% of the mass decomposes. The initial pressure is \(1.0 \mathrm{~atm}\). We must calculate the final pressure.

Step 2: Write the Decomposition Reaction

The decomposition reaction for \(\mathrm{N}_2\mathrm{O}_4\) is:\[\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\]This equation shows that for every 1 mole of \(\mathrm{N}_2\mathrm{O}_4\), 2 moles of \(\mathrm{NO}_2\) are formed.

Step 3: Determine Moles of Decomposed \(\mathrm{N}_2\mathrm{O}_4\)

20% by mass of \(\mathrm{N}_2\mathrm{O}_4\) decomposes, implying \[0.2 \times 1 \text{ mole} = 0.2 \text{ moles} \text{ of } \mathrm{N}_2\mathrm{O}_4\] decomposes to produce \(2 \times 0.2 = 0.4 \) moles of \(\mathrm{NO}_2\).

Step 4: Calculate Remaining Moles of \(\mathrm{N}_2\mathrm{O}_4\)

The remaining moles of \(\mathrm{N}_2\mathrm{O}_4\) after decomposition are:\[1 - 0.2 = 0.8 \text{ moles}\]

Step 5: Find Total Moles Present after Decomposition

Total moles after decomposition include moles of remaining \(\mathrm{N}_2\mathrm{O}_4\) and the moles of newly formed \(\mathrm{NO}_2\):\[0.8 + 0.4 = 1.2 \text{ moles}\]

Step 6: Use Gas Laws to Find Final Pressure

Using the ideal gas law under constant volume and temperature changes where initial moles and moles of gas change:\[P_2 = P_1 \times \frac{n_2}{n_1}\]Given \(P_1 = 1.0 \text{ atm}\), \(n_1 = 1.0 \text{ mole}\), \(n_2 = 1.2 \text{ moles}\), then:\[P_2 = 1.0 \times \frac{1.2}{1.0} = 1.2 \text{ atm}\]

Step 7: Select the Correct Answer

The calculated resultant pressure is \(1.2 \text{ atm}\), which corresponds to option (d).

Key Concepts

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In the context of this exercise, we observe the decomposition of dinitrogen tetroxide (NO_4 ext{ ) under increased temperature conditions. The chemical equation governing this reaction can be written as follows: \[\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\]In this reaction, one mole of NO_4 decomposes to form two moles of nitrogen dioxide (O_2 ext{ ). **Important Points to Remember:**- Decomposition reactions typically require an energy input such as heat, light, or electricity.- The reaction is reversible, as indicated by the double arrow in the equation. - Knowledge of the decomposition reaction allows us to compute other parameters, such as the amount of product formed and how the condition changes affect the equilibrium state.

Ideal Gas Law

The ideal gas law provides a fundamental equation to relate different properties of gases, often expressed as \(PV = nRT\). Here, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. In this exercise, the volume and temperature remain constant, while the pressure changes as a result of a change in the number of moles of gas due to decomposition.To calculate the final pressure (\(P_2\)), given the initial conditions where 20% of \(\mathrm{N}_2\mathrm{O}_4\) decomposes into \(\mathrm{NO}_2\), we use the relation derived from the ideal gas law: \(\[P_2 = P_1 \times \frac{n_2}{n_1}\]\)**Things to Note:** - The ideal gas law assumes that gas molecules do not interact and occupy no volume of their own.- The gas constant \(R\) is typically given as 0.0821 atm⋅L/mol⋅K.- Understanding how changes in temperature, pressure, and volume relate allows us to predict behavior changes when gases undergo chemical transformations.

Stoichiometry

Stoichiometry involves measuring and calculating the elements and compounds in a chemical reaction. Here, knowledge in stoichiometry helps us understand and quantify the decomposition of O_4 to O_2. By calculating the amount of substance that participates in the reaction, we have a clear picture of how many moles of reactants transform into products. For example, if the decomposition of O_4 is 20% by mass, it indicates that 0.2 moles of O_4 decomposed. Using the decomposition equation, we can then deduce that 0.4 moles of O_2 are produced. **Key Stoichiometric Insights:** - Always start by balancing the chemical equation. - Calculate the moles required or produced using the balanced equation. - Stoichiometry is vital for converting between moles of different substances in a reaction. - Understanding stoichiometry assists in scaling reactions for practical applications, like industrial synthesis or laboratory preparation.