Question

The degree of dissociation of N$_2$O$_4$ into NO$_2$ at 1 atm 40°C is 0.310. Calculate its K$_p$ at 40°C. Also report the degree of dissociation at 10 atm pressure at same temperature.

Answer 1

K$_p$ = 0.425 atm, degree of dissociation = 0.103

Answer 2

The problem involves the dissociation of N$_2$O$_4$ into NO$_2$ and requires calculating the equilibrium constant K$_p$ at a given temperature and then the degree of dissociation at a different pressure.

The reaction is:

$\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})$

Part 1: Calculate K$_p$ at 40°C

Let's assume we start with 1 mole of N$_2$O$_4$.

Initial moles:

$\text{N}_2\text{O}_4 = 1$ $\text{NO}_2 = 0$

At equilibrium, if the degree of dissociation is $\alpha$:

Moles of $\text{N}_2\text{O}_4 = (1 - \alpha)$ Moles of $\text{NO}_2 = 2\alpha$

Total moles at equilibrium, $n_{\text{total}} = (1 - \alpha) + 2\alpha = 1 + \alpha$

The partial pressure of each gas is given by:

$P_{\text{N}_2\text{O}_4} = \left(\frac{1 - \alpha}{1 + \alpha}\right) P_{\text{total}}$ $P_{\text{NO}_2} = \left(\frac{2\alpha}{1 + \alpha}\right) P_{\text{total}}$

The equilibrium constant $K_p$ is defined as:

$K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$

Substitute the expressions for partial pressures:

$K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} P_{\text{total}}\right)^2}{\left(\frac{1 - \alpha}{1 + \alpha}\right) P_{\text{total}}}$ $K_p = \frac{\frac{4\alpha^2}{(1 + \alpha)^2} P_{\text{total}}^2}{\frac{1 - \alpha}{1 + \alpha} P_{\text{total}}}$ $K_p = \frac{4\alpha^2 P_{\text{total}}}{(1 - \alpha)(1 + \alpha)}$ $K_p = \frac{4\alpha^2 P_{\text{total}}}{1 - \alpha^2}$

Given data for the first scenario:

$\alpha = 0.310$ $P_{\text{total}} = 1 \text{ atm}$

Substitute these values into the $K_p$ expression:

$K_p = \frac{4 \times (0.310)^2 \times 1}{1 - (0.310)^2}$ $K_p = \frac{4 \times 0.0961}{1 - 0.0961}$ $K_p = \frac{0.3844}{0.9039}$ $K_p \approx 0.4253 \text{ atm}$

Part 2: Calculate the degree of dissociation at 10 atm pressure at the same temperature

The equilibrium constant $K_p$ depends only on temperature. Since the temperature is the same (40°C), the $K_p$ value remains constant:

$K_p = 0.4253 \text{ atm}$

Now, we need to find the new degree of dissociation, let's call it $\alpha'$, when $P_{\text{total}} = 10 \text{ atm}$.

Using the same $K_p$ expression:

$K_p = \frac{4(\alpha')^2 P_{\text{total}}}{1 - (\alpha')^2}$ $0.4253 = \frac{4(\alpha')^2 \times 10}{1 - (\alpha')^2}$ $0.4253 = \frac{40(\alpha')^2}{1 - (\alpha')^2}$

Rearrange the equation to solve for $\alpha'$:

$0.4253 (1 - (\alpha')^2) = 40(\alpha')^2$ $0.4253 - 0.4253(\alpha')^2 = 40(\alpha')^2$ $0.4253 = 40(\alpha')^2 + 0.4253(\alpha')^2$ $0.4253 = (40 + 0.4253)(\alpha')^2$ $0.4253 = 40.4253(\alpha')^2$ $(\alpha')^2 = \frac{0.4253}{40.4253}$ $(\alpha')^2 \approx 0.01052$

Take the square root to find $\alpha'$:

$\alpha' = \sqrt{0.01052}$ $\alpha' \approx 0.10257$

Rounding to three significant figures (consistent with the given $\alpha$ value), we get:

$\alpha' \approx 0.103$

This result is consistent with Le Chatelier's principle: increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (reactants), thus decreasing the degree of dissociation. (0.103 < 0.310).

Explanation of the solution:

1. Define equilibrium: For the dissociation N$_2$O$_4$(g) $\rightleftharpoons$ 2NO$_2$(g), if $\alpha$ is the degree of dissociation, the equilibrium moles for 1 initial mole of N$_2$O$_4$ are $(1-\alpha)$ for N$_2$O$_4$ and $2\alpha$ for NO$_2$. Total moles are $(1+\alpha)$.

2. Express partial pressures: Partial pressure of a gas is its mole fraction multiplied by the total pressure.

   $P_{\text{N}_2\text{O}_4} = \frac{1-\alpha}{1+\alpha} P_{\text{total}}$

   $P_{\text{NO}_2} = \frac{2\alpha}{1+\alpha} P_{\text{total}}$

3. Formulate K$_p$ expression: $K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} = \frac{4\alpha^2 P_{\text{total}}}{1-\alpha^2}$.

4. Calculate K$_p$: Substitute the given $\alpha = 0.310$ and $P_{\text{total}} = 1 \text{ atm}$ into the $K_p$ expression to find $K_p \approx 0.425 \text{ atm}$.

5. Calculate new $\alpha'$: Since $K_p$ is temperature-dependent, it remains constant at 40°C. Use the calculated $K_p$ and the new total pressure $P_{\text{total}} = 10 \text{ atm}$ in the $K_p$ expression to solve for the new degree of dissociation $\alpha'$. This yields $\alpha' \approx 0.103$.