Question

$2NO_2 \rightleftharpoons N_2O_4; K_p = 6.8 \text{ atm}^{-1}$ $NO + NO_2 \rightleftharpoons N_2O_3$

In an experiment when NO and $NO_2$ are mixed in the ratio of 1:2, the final total pressure was 5.05 atm and the partial pressure of $N_2O_4$ was 1.7 atm. Calculate

a. the equilibrium partial pressure of NO.

b. $K_p$ for $NO + NO_2 \rightleftharpoons N_2O_3$.

Answer 1

a. 1.05 atm, b. 24/7 atm$^{-1}$

Answer 2

The problem involves two simultaneous equilibrium reactions and requires calculating equilibrium partial pressure and an equilibrium constant.

Given reactions and data:

1. $2NO_2 (g) \rightleftharpoons N_2O_4 (g)$; $K_p = 6.8 \text{ atm}^{-1}$
2. $NO (g) + NO_2 (g) \rightleftharpoons N_2O_3 (g)$

Initial condition: NO and $NO_2$ are mixed in the ratio of 1:2. Final total pressure ($P_{total}$) = 5.05 atm. Partial pressure of $N_2O_4$ ($P_{N_2O_4}$) = 1.7 atm at equilibrium.

Let's denote the initial partial pressures of NO and $NO_2$ as $P_{NO,0}$ and $P_{NO_2,0}$ respectively. Given $P_{NO,0} : P_{NO_2,0} = 1:2$, so $P_{NO_2,0} = 2 P_{NO,0}$.

Step 1: Calculate the equilibrium partial pressure of $NO_2$ from the first reaction. For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g)$, the equilibrium constant $K_p$ is given by: $K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}$ Given $K_p = 6.8 \text{ atm}^{-1}$ and $P_{N_2O_4} = 1.7 \text{ atm}$. $6.8 = \frac{1.7}{(P_{NO_2})^2}$ $(P_{NO_2})^2 = \frac{1.7}{6.8} = \frac{1}{4} = 0.25$ $P_{NO_2} = \sqrt{0.25} = 0.5 \text{ atm}$ (This is the equilibrium partial pressure of $NO_2$).

Step 2: Set up expressions for equilibrium partial pressures based on initial pressures and changes. Let $x$ be the partial pressure of $N_2O_4$ formed at equilibrium, so $x = P_{N_2O_4} = 1.7 \text{ atm}$. This means $2x = 3.4 \text{ atm}$ of $NO_2$ was consumed in the first reaction. Let $y$ be the partial pressure of $N_2O_3$ formed at equilibrium, so $y = P_{N_2O_3}$. This means $y$ atm of NO and $y$ atm of $NO_2$ were consumed in the second reaction.

The equilibrium partial pressures are: $P_{NO,eq} = P_{NO,0} - y$ $P_{NO_2,eq} = P_{NO_2,0} - 2x - y = P_{NO_2,0} - 2(1.7) - y = P_{NO_2,0} - 3.4 - y$ $P_{N_2O_3,eq} = y$ $P_{N_2O_4,eq} = x = 1.7 \text{ atm}$

We know $P_{NO_2,eq} = 0.5 \text{ atm}$ from Step 1. So, $P_{NO_2,0} - 3.4 - y = 0.5 \implies P_{NO_2,0} - y = 3.9 \text{ atm}$.

Step 3: Use the total pressure to find the initial partial pressure of NO. The total pressure at equilibrium is the sum of the partial pressures of all gases: $P_{total} = P_{NO,eq} + P_{NO_2,eq} + P_{N_2O_3,eq} + P_{N_2O_4,eq}$ $5.05 = (P_{NO,0} - y) + 0.5 + y + 1.7$ Notice that the 'y' terms cancel out. $5.05 = P_{NO,0} + 0.5 + 1.7$ $5.05 = P_{NO,0} + 2.2$ $P_{NO,0} = 5.05 - 2.2 = 2.85 \text{ atm}$.

Step 4: Calculate 'y' and the equilibrium partial pressure of NO (Part a). We have $P_{NO_2,0} = 2 P_{NO,0} = 2(2.85) = 5.7 \text{ atm}$. Substitute $P_{NO_2,0}$ into the equation from Step 2: $P_{NO_2,0} - y = 3.9$ $5.7 - y = 3.9$ $y = 5.7 - 3.9 = 1.8 \text{ atm}$.

This means $P_{N_2O_3,eq} = 1.8 \text{ atm}$.

a. The equilibrium partial pressure of NO ($P_{NO,eq}$): $P_{NO,eq} = P_{NO,0} - y = 2.85 - 1.8 = 1.05 \text{ atm}$.

Step 5: Calculate $K_p$ for the second reaction (Part b). For the reaction $NO (g) + NO_2 (g) \rightleftharpoons N_2O_3 (g)$, the equilibrium constant $K_p$ is: $K_p = \frac{P_{N_2O_3,eq}}{(P_{NO,eq})(P_{NO_2,eq})}$ Substitute the equilibrium partial pressures: $P_{N_2O_3,eq} = 1.8 \text{ atm}$ $P_{NO,eq} = 1.05 \text{ atm}$ $P_{NO_2,eq} = 0.5 \text{ atm}$

$K_p = \frac{1.8}{(1.05)(0.5)} = \frac{1.8}{0.525}$ $K_p = \frac{1800}{525}$ (Multiply numerator and denominator by 1000 to remove decimals) Divide by 25: $K_p = \frac{72}{21}$ Divide by 3: $K_p = \frac{24}{7}$ $K_p \approx 3.4286 \text{ atm}^{-1}$.

The final answer is $\boxed{a. 1.05 \text{ atm}, b. 24/7 \text{ atm}^{-1}}$

Explanation of the solution:

1. From the given $K_p$ for $2NO_2 \rightleftharpoons N_2O_4$ and the equilibrium partial pressure of $N_2O_4$, calculate the equilibrium partial pressure of $NO_2$.
2. Define initial partial pressures based on the given ratio.
3. Express equilibrium partial pressures of all species in terms of initial partial pressures and extents of reaction ($x$ for $N_2O_4$ formation, $y$ for $N_2O_3$ formation).
4. Use the given total equilibrium pressure to set up an equation. This equation simplifies to directly yield the initial partial pressure of NO.
5. Substitute the initial partial pressure of NO and the calculated equilibrium partial pressure of $NO_2$ into the expressions to find the extent of reaction for $N_2O_3$ formation ($y$). This gives the equilibrium partial pressure of $N_2O_3$.
6. Calculate the equilibrium partial pressure of NO using its initial partial pressure and $y$.
7. Finally, use the equilibrium partial pressures of $NO$, $NO_2$, and $N_2O_3$ to calculate $K_p$ for the second reaction.

Answer: a. The equilibrium partial pressure of NO is 1.05 atm. b. $K_p$ for $NO + NO_2 \rightleftharpoons N_2O_3$ is 24/7 atm$^{-1}$ (approximately 3.43 atm$^{-1}$).