Question

The vapour density of a sample of $N_2O_4$ gas is 35. What percent of $N_2O_4$ molecules are dissociated into $NO_2$?

Answer 1

31.43%

Answer 2

The problem asks for the percentage dissociation of $N_2O_4$ molecules into $NO_2$, given the observed vapour density of the gas sample.

1. Write the dissociation reaction: The dissociation of dinitrogen tetroxide ($N_2O_4$) into nitrogen dioxide ($NO_2$) is represented as: $N_2O_4 (g) \rightleftharpoons 2NO_2 (g)$

2. Calculate the theoretical molecular weight of $N_2O_4$: Molar mass of Nitrogen (N) = 14 g/mol Molar mass of Oxygen (O) = 16 g/mol Theoretical molecular weight of $N_2O_4 = (2 \times \text{Molar mass of N}) + (4 \times \text{Molar mass of O})$ $= (2 \times 14) + (4 \times 16)$ $= 28 + 64 = 92$ g/mol

3. Calculate the theoretical vapour density (D): Theoretical vapour density (D) is half of the theoretical molecular weight. $D = \frac{\text{Theoretical Molecular Weight}}{2} = \frac{92}{2} = 46$

4. Identify the observed vapour density (d): The problem states that the vapour density of the sample is 35. So, observed vapour density, $d = 35$.

5. Determine the number of moles of products from one mole of reactant (n): From the balanced chemical equation $N_2O_4 (g) \rightleftharpoons 2NO_2 (g)$, one mole of $N_2O_4$ dissociates to form two moles of $NO_2$. Therefore, $n = 2$.

6. Apply the formula for the degree of dissociation ($\alpha$): The degree of dissociation ($\alpha$) is related to the theoretical vapour density (D), observed vapour density (d), and the number of moles of products formed from one mole of reactant (n) by the formula: $\alpha = \frac{D - d}{d(n - 1)}$

Substitute the values: $D = 46$ $d = 35$ $n = 2$

$\alpha = \frac{46 - 35}{35(2 - 1)}$ $\alpha = \frac{11}{35(1)}$ $\alpha = \frac{11}{35}$

7. Calculate the percentage dissociation: $\alpha \approx 0.3142857$ Percentage dissociation $= \alpha \times 100$ Percentage dissociation $= 0.3142857 \times 100 = 31.42857\%$

Rounding to two decimal places, the percentage dissociation is $31.43\%$.

The final answer is $\boxed{\text{31.43}}$.

Explanation of the solution: The dissociation of $N_2O_4$ to $NO_2$ is $N_2O_4 \rightleftharpoons 2NO_2$.

1. Calculate theoretical molar mass of $N_2O_4$: 92 g/mol.
2. Calculate theoretical vapor density (D): $92/2 = 46$.
3. Given observed vapor density (d): 35.
4. Use the formula for degree of dissociation ($\alpha$): $\alpha = \frac{D - d}{d(n - 1)}$, where $n=2$ (moles of products from 1 mole of reactant).
5. Substitute values: $\alpha = \frac{46 - 35}{35(2 - 1)} = \frac{11}{35}$.
6. Convert to percentage: $\frac{11}{35} \times 100 \approx 31.43\%$.