Question

An equilibrium mixture at 300 K contains N$_2$O$_4$ and NO$_2$ at 0.28 and 1.1 atm, respectively. If the volume of container is doubled, calculate the new equilibrium pressure of two gases.

Answer 1

P(N_2O_4) = 0.095 atm, P(NO_2) = 0.64 atm

Answer 2

The problem involves calculating new equilibrium pressures after a volume change for the dissociation of N$_2$O$_4$ into NO$_2$.

The reaction is: $\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})$

1. Calculate the equilibrium constant ($K_p$) from the initial conditions: At 300 K, the initial equilibrium pressures are $P_{\text{N}_2\text{O}_4} = 0.28 \text{ atm}$ and $P_{\text{NO}_2} = 1.1 \text{ atm}$. The expression for $K_p$ is: $K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$ Substitute the given values: $K_p = \frac{(1.1)^2}{0.28} = \frac{1.21}{0.28} \approx 4.3214 \text{ atm}$ Since temperature is constant, $K_p$ remains the same for the new equilibrium.

2. Determine instantaneous pressures after volume change: When the volume of the container is doubled at constant temperature, the pressure of each gas is halved (according to Boyle's Law, $P \propto 1/V$). Instantaneous pressures immediately after doubling the volume (before equilibrium shifts): $P'_{\text{N}_2\text{O}_4} = \frac{0.28}{2} = 0.14 \text{ atm}$ $P'_{\text{NO}_2} = \frac{1.1}{2} = 0.55 \text{ atm}$

3. Calculate the reaction quotient ($Q_p$) and predict the shift: Calculate $Q_p$ using the instantaneous pressures: $Q_p = \frac{(P'_{\text{NO}_2})^2}{P'_{\text{N}_2\text{O}_4}} = \frac{(0.55)^2}{0.14} = \frac{0.3025}{0.14} \approx 2.1607 \text{ atm}$ Compare $Q_p$ with $K_p$: Since $Q_p (2.1607) < K_p (4.3214)$, the reaction will shift in the forward direction (towards products) to reach equilibrium.

4. Set up an ICE (Initial, Change, Equilibrium) table for pressures: Let $x$ be the change in partial pressure of N$_2$O$_4$ at equilibrium.

Species	Initial Pressure (atm)	Change (atm)	Equilibrium Pressure (atm)
N$_2$O$_4$	0.14	$-x$	$0.14 - x$
NO$_2$	0.55	$+2x$	$0.55 + 2x$

5. Solve for $x$ using the $K_p$ expression: $K_p = \frac{(P_{\text{NO}_2, \text{eq}})^2}{P_{\text{N}_2\text{O}_4, \text{eq}}}$ $4.3214 = \frac{(0.55 + 2x)^2}{(0.14 - x)}$ Rearrange and solve the quadratic equation: $4.3214 (0.14 - x) = (0.55 + 2x)^2$ $0.6050 - 4.3214x = 0.3025 + 2.2x + 4x^2$ $4x^2 + (2.2 + 4.3214)x + (0.3025 - 0.6050) = 0$ $4x^2 + 6.5214x - 0.3025 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-6.5214 \pm \sqrt{(6.5214)^2 - 4(4)(-0.3025)}}{2(4)}$ $x = \frac{-6.5214 \pm \sqrt{42.5287 + 4.84}}{8}$ $x = \frac{-6.5214 \pm \sqrt{47.3687}}{8}$ $x = \frac{-6.5214 \pm 6.8825}{8}$ Two possible values for $x$: $x_1 = \frac{-6.5214 + 6.8825}{8} = \frac{0.3611}{8} = 0.0451375$ $x_2 = \frac{-6.5214 - 6.8825}{8} = \frac{-13.4039}{8} = -1.6754875$ Since the reaction shifts forward, $x$ must be positive. Also, $0.14 - x$ must be positive, which means $x < 0.14$. Therefore, $x = 0.0451375$ is the valid solution.

6. Calculate the new equilibrium pressures: $P_{\text{N}_2\text{O}_4, \text{eq}} = 0.14 - x = 0.14 - 0.0451375 = 0.0948625 \text{ atm}$ $P_{\text{NO}_2, \text{eq}} = 0.55 + 2x = 0.55 + 2(0.0451375) = 0.55 + 0.090275 = 0.640275 \text{ atm}$

Rounding to two significant figures: $P_{\text{N}_2\text{O}_4, \text{eq}} \approx 0.095 \text{ atm}$ $P_{\text{NO}_2, \text{eq}} \approx 0.64 \text{ atm}$

The new equilibrium pressure of N$_2$O$_4$ is approximately 0.095 atm and that of NO$_2$ is approximately 0.64 atm.

Explanation of the solution:

1. Calculate $K_p$ from initial equilibrium pressures.
2. Halve initial partial pressures due to volume doubling.
3. Calculate $Q_p$ with new pressures; since $Q_p < K_p$, equilibrium shifts forward.
4. Set up an ICE table with changes in pressure ($+2x$ for NO$_2$, $-x$ for N$_2$O$_4$).
5. Substitute equilibrium expressions into $K_p$ and solve the resulting quadratic equation for $x$.
6. Calculate new equilibrium pressures using the valid $x$ value.