Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At room temperature the following reactions proceed nearly to completion : $$ 2 \mathrm{NO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4} $$ The dimer, \(\mathrm{N}_{2} \mathrm{O}_{4}\), solidifies at \(262 \mathrm{~K}\). A \(250 \mathrm{~mL}\) flask and a \(100 \mathrm{~mL}\). flask are separated by a stop-cock. At \(300 \mathrm{~K}\), the nitric oxide in the larger flask exerts a pressure of \(1.053 \mathrm{~atm}\). and the smaller one contains oxygen at \(0.789\) atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled at \(220 \mathrm{~K}\). Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at \(220 \mathrm{~K}\). (Assume the gases to behave ideally).

Step-by-Step Solution

Answer

The pressure of gas remaining at 220 K is 0.222 atm, consisting solely of NO gas.

Step 1: Calculate Moles of Initial Gases

First, we calculate the moles of each gas present initially using the ideal gas law: \(PV = nRT\). For NO in the larger flask:\[n(\text{NO}) = \frac{P \cdot V}{R \cdot T} = \frac{1.053 \text{ atm} \times 0.250 \text{ L}}{0.0821 \text{ L atm/mol K} \times 300 \text{ K}} \approx 0.0107 \text{ mol}\]For \(\text{O}_2\) in the smaller flask:\[n(\text{O}_2) = \frac{0.789 \text{ atm} \times 0.100 \text{ L}}{0.0821 \text{ L atm/mol K} \times 300 \text{ K}} \approx 0.00321 \text{ mol}\]

Step 2: Determine Limiting Reagent and Reaction Completion

The stoichiometry of our reaction \(2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \rightarrow \text{N}_2\text{O}_4\) shows 2 moles of NO react with 1 mole of \(\text{O}_2\). With 0.0107 mol of NO and 0.00321 mol of \(\text{O}_2\), the limiting reagent is \(\text{O}_2\). The whole \(\text{O}_2\) will be used up, reacting with \(2 \times 0.00321 \text{ mol}\) of NO which is 0.00642 mol. So we will have \(0.0107 \text{ mol} - 0.00642 \text{ mol} = 0.00428 \text{ mol NO left.}\)

Step 3: Calculate Moles of N2O4 Formed

With complete reaction, 0.00321 mol of \(\text{O}_2\) will produce an equal amount in moles of \(\text{NO}_2\), resulting in half as much \(\text{N}_2\text{O}_4\) (because of dimer formation): \[n(\text{N}_2\text{O}_4) = \frac{0.00321}{2} = 0.001605 mol.\]

Step 4: Find Total Volume and Conditions at 220 K

The total volume is the sum of the two flasks, so \(250 \text{ mL} + 100 \text{ mL} = 350 \text{ mL} = 0.350 \text{ L}.\) At 220 K, the remaining gases will exert a new pressure. Only NO gas contributes to the pressure, as \(\text{N}_2\text{O}_4\) is a solid. Total moles remaining are 0.00428 mol (NO).

Step 5: Calculate Pressure at 220 K

Using ideal gas law for NO gas: \[P = \frac{nRT}{V} = \frac{(0.00428 \, \text{mol}) \times (0.0821 \, \text{L atm/mol K}) \times 220 \, \text{K}}{0.350 \, \text{L}} \approx 0.222 \text{ atm}.\]

Step 6: State Final Composition

At 220 K, the only gas contributing to the pressure is unreacted NO, since \(\text{N}_2\text{O}_4\) solidifies. The final gas composition is only NO with partial pressure 0.222 atm. No \(\text{O}_2\) or gaseous \(\text{N}_2\text{O}_4\) remains because \(\text{O}_2\) reacted completely and \(\text{N}_2\text{O}_4\) is solid.

Key Concepts

Stoichiometry

Stoichiometry is a core concept in chemistry that involves the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It provides the quantitative relationship between different substances involved. In the given reaction, the stoichiometry is crucial to find out how much of each reactant is needed and how much product is formed. The chemical equation for our reaction is:

• \(2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \rightarrow \text{N}_2\text{O}_4\)

This equation tells us that two moles of nitric oxide (NO) react with one mole of oxygen (\(\text{O}_2\)) to eventually form nitrogen dioxide (\(\text{NO}_2\)), which then dimerizes to form dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)).
By understanding the stoichiometry, we can determine that one mole of \(\text{O}_2\) is required to react completely with two moles of NO, enforcing the importance of mole-to-mole ratios in chemical equations. This step ensures that the reaction is fully understood in terms of how chemicals interact on a molecular level.

Limiting Reagent

The limiting reagent is the reactant that is completely used up first in a chemical reaction, thus determining the amount of product that can be formed. It is a foundational concept in stoichiometry and helps us predict the maximum yield of a reaction. In our example, we calculated the moles of each initial reactant using the Ideal Gas Law. With 0.0107 mol of NO and 0.00321 mol of \(\text{O}_2\), we need to identify which reactant will run out first.

• The stoichiometric ratio from the balanced equation shows that \(2\text{ NO}: 1\text{ O}_2\).
• We need 2 moles of NO for every mole of \(\text{O}_2\).
• Given our quantities, \(\text{O}_2\) will limit the reaction as it will be completely consumed reacting with some of the NO.

Once the \(\text{O}_2\) is exhausted, no further reaction can occur, leaving unreacted NO. This concept is critical in chemical manufacturing and laboratory to ensure the efficient use of chemicals and predict by-products.

Chemical Reaction Dynamics

Chemical reaction dynamics explore the processes and conditions under which chemical reactions occur. This includes aspects such as temperature, pressure, and reaction intermediates, particularly relevant in our reaction where the behavior of the gases under varying conditions is evaluated. Initially, the gases are at 300 K and atmospheric pressure vary by reactant:

• NO in the larger flask exerts a pressure of 1.053 atm.
• \(\text{O}_2\) in the smaller flask is at 0.789 atm.

When the gases mix in the combined 350 mL volume, dynamics change as they react. After the reaction, when the system is cooled to 220 K, different chemical behavior emerges because \(\text{N}_2\text{O}_4\) solidifies and doesn't contribute to the pressure. Only unreacted NO influences the final pressure.

• The concept of ideal gases simplifies calculations but assumes no intermolecular forces.
• The reaction proceeding to completion at room temperature highlights dynamic conditions influencing reaction pathways.

Chemical Reaction Dynamics consider how environmental changes, like cooling, affect the products and gas pressure, reinforcing how reactions are context-sensitive.