Question

1. a sample of n₂o₄(g) is placed in an empty cylinder at room temperature. after equilibrium is reached, the total pressure is 1.5 atm and 16% (by moles) of the original n₂o₄ has dissociated into no₂(g). what is kₚ for this reaction at this temperature?

the volume of the cylinder is increased until the total pressure is 1.0 atm (temperature remains constant). calculate the equilibrium partial pressures of the two gases.

Explanation:

Step1: Write the dissociation reaction

The dissociation reaction of $N_2O_4$ is $N_2O_4(g)
ightleftharpoons 2NO_2(g)$. Let the initial amount of $N_2O_4$ be $n$ moles. The amount of $N_2O_4$ that dissociates is $0.16n$ moles, and the amount of $NO_2$ formed is $2\times0.16n = 0.32n$ moles. The amount of $N_2O_4$ remaining at equilibrium is $(1 - 0.16)n=0.84n$ moles.

Step2: Calculate the mole - fraction of each gas

The total number of moles at equilibrium, $n_{total}=0.84n + 0.32n=1.16n$ moles. The mole - fraction of $N_2O_4$, $x_{N_2O_4}=\frac{0.84n}{1.16n}=\frac{0.84}{1.16}$, and the mole - fraction of $NO_2$, $x_{NO_2}=\frac{0.32n}{1.16n}=\frac{0.32}{1.16}$.

Step3: Calculate the partial pressures

The total pressure at equilibrium, $P_{total}=1.5$ atm. The partial pressure of $N_2O_4$, $P_{N_2O_4}=x_{N_2O_4}P_{total}=\frac{0.84}{1.16}\times1.5$ atm $\approx1.086$ atm. The partial pressure of $NO_2$, $P_{NO_2}=x_{NO_2}P_{total}=\frac{0.32}{1.16}\times1.5$ atm $\approx0.414$ atm.

Step4: Calculate $K_p$

The expression for $K_p$ for the reaction $N_2O_4(g)
ightleftharpoons 2NO_2(g)$ is $K_p=\frac{P_{NO_2}^2}{P_{N_2O_4}}$. Substituting the values of $P_{NO_2}$ and $P_{N_2O_4}$, we get $K_p=\frac{(0.414)^2}{1.086}\approx0.158$.

Now, when the total pressure is changed to $P_{total}=1.0$ atm. Let the partial pressure of $N_2O_4$ be $P_1$ and the partial pressure of $NO_2$ be $P_2$. We know that $P_1 + P_2=1.0$ atm and $K_p = 0.158=\frac{P_2^2}{P_1}$. From $P_2 = 1 - P_1$, we substitute into the $K_p$ expression: $0.158=\frac{(1 - P_1)^2}{P_1}$. Expanding, we get $0.158P_1=1 - 2P_1+P_1^2$. Rearranging to a quadratic equation: $P_1^2-2.158P_1 + 1=0$.
Using the quadratic formula $P_1=\frac{2.158\pm\sqrt{(2.158)^2-4\times1\times1}}{2}=\frac{2.158\pm\sqrt{4.657 - 4}}{2}=\frac{2.158\pm\sqrt{0.657}}{2}=\frac{2.158\pm0.81}{2}$. We get two solutions for $P_1$: $P_1=\frac{2.158 + 0.81}{2}=1.484$ (rejected since $P_1+P_2 = 1$) and $P_1=\frac{2.158 - 0.81}{2}=0.674$ atm. Then $P_2=1 - 0.674 = 0.326$ atm.

Answer:

$K_p = 0.158$, $P_{N_2O_4}=0.674$ atm, $P_{NO_2}=0.326$ atm