Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (g) at \(310 \mathrm{~K}\) is \(25 \%\) dissociated at 1 atm pressure. The percentage dissociation at \(0.1 \mathrm{~atm}\) and \(310 \mathrm{~K}\) is (a) \(25 \%\) (b) \(50 \%\) (c) \(76 \%\) (d) \(63 \%\)

Step-by-Step Solution

Answer

The percentage dissociation at 0.1 atm and 310 K is 63% (option d).

Step 1: Determine Initial Concentration and Reaction Equation

The decomposition of \(\mathrm{N}_2\mathrm{O}_4\) to \(2\mathrm{NO}_2\) can be represented as \[\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)\]. Initially, we have 1 mole of \(\mathrm{N}_2\mathrm{O}_4\). If \(\alpha = 0.25\) is the degree of dissociation at 1 atm, it dissociates as: \[ \mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2 \]. Initially: \([\mathrm{N}_2\mathrm{O}_4] = 1\), After dissociation: \([\mathrm{N}_2\mathrm{O}_4] = 1 - \alpha\) with \([\mathrm{NO}_2] = 2\alpha\).

Step 2: Calculate Equilibrium Constant (Kp) at 1 atm

At 1 atm, using the partial pressures: At equilibrium, \(P_{\mathrm{N}_2\mathrm{O}_4} = 1 - \alpha = 0.75\) atm and \(P_{\mathrm{NO}_2} = 2\cdot0.25 = 0.5\) atm. The equilibrium constant \(K_p\) is given by\[K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} = \frac{(0.5)^2}{0.75} = \frac{0.25}{0.75} = \frac{1}{3} atm\]

Step 3: Calculate Percentage Dissociation at 0.1 atm

Now, we need to calculate \(\alpha'\) at 0.1 atm using the same \(K_p\):\[ K_p = \frac{(2\alpha'\times0.1)^2}{(1-\alpha')\times0.1} = \frac{4(\alpha')^2\times0.1}{1-\alpha'} = \frac{4(\alpha')^2}{10(1-\alpha')}\] Since \(K_p = 1/3\):\[ \frac{4(\alpha')^2}{10(1-\alpha')} = \frac{1}{3}\]\[ 12(\alpha')^2 = 10(1-\alpha')\]\[ 12(\alpha')^2 + 10\alpha' - 10 = 0\] Solve this quadratic equation to find \(\alpha'\).

Step 4: Solve Quadratic Equation for α'

To solve \(12(\alpha')^2 + 10\alpha' - 10 = 0\), use the quadratic formula:\[\alpha' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a = 12, b = 10, c = -10\).Calculate:- \(b^2 = 10^2 = 100\)- \(4ac = 4 \times 12 \times (-10) = -480\)- \(\Delta = 100 + 480 = 580\)So, \(\alpha' = \frac{-10 \pm \sqrt{580}}{24}\).Since \(\alpha'\) must be positive, calculate \(\alpha' = \frac{-10 + \sqrt{580}}{24} \approx 0.63\) or \(63\%\).

Step 5: Conclusion

Thus, the percentage dissociation of \(\mathrm{N}_2\mathrm{O}_4\) at 0.1 atm and 310 K is approximately 63%, so the correct answer is (d) 63%.

Key Concepts

Dissociation of N2O4

In a chemical equilibrium, the dissociation of substances is a critical concept to comprehend. One such example is the dissociation of dinitrogen tetroxide (\(\mathrm{N}_2\mathrm{O}_4\)), which breaks down into nitrogen dioxide (\(\mathrm{NO}_2\)). This dissociation process can be represented by the reversible reaction:

• \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)\)

Initially, let's consider a scenario where you start with 1 mole of \(\mathrm{N}_2\mathrm{O}_4\). As equilibrium is achieved, the substance doesn't completely convert into products or stay entirely as reactants. Instead, a dynamic balance is met.
For example, if the degree of dissociation symbolized by \(\alpha\) is 25% at 1 atm, it indicates that 25% of the original \(\mathrm{N}_2\mathrm{O}_4\) has decomposed into \(\mathrm{NO}_2\). After the reaction reaches equilibrium, the concentration of \(\mathrm{N}_2\mathrm{O}_4\) will be \(1 - \alpha\) and that of \(\mathrm{NO}_2\) will be \(2\alpha\) because each molecule of \(\mathrm{N}_2\mathrm{O}_4\) produces two molecules of \(\mathrm{NO}_2\). Understanding how substances dissociate is essential to predict the concentrations of reactants and products at equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, specifically \(K_p\) for gaseous reactions, is a measure used to express the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients. It calculated using the partial pressures of gases involved in the equilibrium reaction:
Consider the reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)\). For this reaction, the expression of \(K_p\) is:

• \[ K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \]

At 1 atm, if 25% of \(\mathrm{N}_2\mathrm{O}_4\) is dissociated, the partial pressure of \(\mathrm{NO}_2\) becomes \(2 \times 0.25 = 0.5\) atm and that of \(\mathrm{N}_2\mathrm{O}_4\) reduces to \(1 - 0.25 = 0.75\) atm. Plugging these values into the equilibrium equation gives:

• \[ K_p = \frac{(0.5)^2}{0.75} = \frac{0.25}{0.75} = \frac{1}{3} \text{ atm} \]

\(K_p\) remains constant for a given temperature, serving as a crucial tool to find out the extent of reactions at different pressures, as seen in subsequent calculations.

Pressure Effects on Dissociation

Pressure plays a significant role in the extent of dissociation in gas-phase chemical equilibria. When dealing with gaseous reactions like the dissociation of \(\mathrm{N}_2\mathrm{O}_4\), changes in pressure can shift the equilibrium position.
Here's how pressure impacts dissociation: When the pressure of the system decreases, it tends to drive the reaction toward the production of more gas molecules. Given the reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)\), a decrease in pressure increases the volume, thus favoring the side with more moles of gas, in this case, \(\mathrm{NO}_2\).
In our example, reducing the pressure from 1 atm to 0.1 atm results in increased dissociation of \(\mathrm{N}_2\mathrm{O}_4\), as evidenced by the calculated 63% dissociation compared to the initial 25% at a higher pressure. This is all consistent with Le Chatelier's Principle, which states that the system will shift to counteract changes in pressure, increasing the production of \(\mathrm{NO}_2\) under lower pressure conditions.