Question

N₂O₄ dissociates as N₂O₄(g) ⇌ 2NO₂(g) at 273 K and 2 atm pressure. The equilibrium mixture has a density of 41. What will be the degree of dissociation;

• A. 14.2%
• B. 16.2%
• C. 12.2%
• D. None of these

Answer 1

Answer: (C)

12.2%

Answer 2

The dissociation reaction is given by: N₂O₄(g) ⇌ 2NO₂(g)

The molar mass of N₂O₄ ($M_0$) is $2 \times 14 + 4 \times 16 = 92$ g/mol. The molar mass of NO₂ ($M_1$) is $14 + 2 \times 16 = 46$ g/mol.

Let $\alpha$ be the degree of dissociation of N₂O₄. If we start with 1 mole of N₂O₄: At equilibrium, moles of N₂O₄ = $1 - \alpha$ At equilibrium, moles of NO₂ = $2\alpha$ The total number of moles at equilibrium = $(1 - \alpha) + 2\alpha = 1 + \alpha$.

The average molar mass ($M_{avg}$) of the equilibrium mixture can be calculated as: $M_{avg} = \frac{\text{Total mass of mixture}}{\text{Total moles of mixture}}$ $M_{avg} = \frac{(1-\alpha)M_0 + 2\alpha M_1}{1+\alpha}$ Substituting $M_0 = 92$ and $M_1 = 46$: $M_{avg} = \frac{(1-\alpha)92 + 2\alpha(46)}{1+\alpha} = \frac{92 - 92\alpha + 92\alpha}{1+\alpha} = \frac{92}{1+\alpha}$

The problem states that "The equilibrium mixture has a density of 41". In the context of such problems, when an absolute density value leads to physically impossible results, it is often implied to be the vapour density relative to hydrogen ($H_2$). Assuming "density of 41" means vapour density (VD) = 41: Vapour density is defined as the ratio of the molar mass of the gas to the molar mass of hydrogen ($H_2$, which is 2 g/mol). $VD = \frac{M_{avg}}{M_{H_2}}$ $41 = \frac{M_{avg}}{2}$ $M_{avg} = 41 \times 2 = 82$ g/mol.

Now, we equate the two expressions for $M_{avg}$: $82 = \frac{92}{1+\alpha}$ $1+\alpha = \frac{92}{82} = \frac{46}{41}$ $\alpha = \frac{46}{41} - 1 = \frac{46 - 41}{41} = \frac{5}{41}$

Calculating the value of $\alpha$: $\alpha \approx 0.12195$ To express this as a percentage: $\alpha \approx 0.12195 \times 100\% \approx 12.195\%$

Rounding to one decimal place, we get 12.2%.