Questions 22-25 refer to the following information. N2O4(g) ⇄ 2 NO2(g) Kp = 3.0 at 70°C colorless br — Equilibrium Chemistry Question

STEPS:

1. Understand the concept of the partial pressure equilibrium constant ($K_p$):
For a gas-phase reaction, the equilibrium constant $K_p$ is formulated using the partial pressures of the gaseous products and reactants, with each species raised to the power of its stoichiometric coefficient from the balanced chemical equation.
2. Write the $K_p$ expression for the given reaction:
Using the balanced equilibrium equation:
$$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\ \text{NO}_2(g)$$
The expression is:
$$K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$$
3. Substitute the known values into the expression:
We are given that $K_p = 3.0$ and the equilibrium partial pressure of dinitrogen tetroxide ($P_{\text{N}_2\text{O}_4}$) is $1.33\text{ atm}$:
$$3.0 = \frac{(P_{\text{NO}_2})^2}{1.33}$$
4. Solve for the squared partial pressure of nitrogen dioxide, $(P_{\text{NO}_2})^2$:
Multiply both sides of the equation by $1.33$:
$$(P_{\text{NO}_2})^2 = 3.0 \times 1.33$$
$$(P_{\text{NO}_2})^2 = 3.99$$
5. Calculate the final partial pressure of $\text{NO}_2$:
Take the square root of both sides to solve for $P_{\text{NO}_2}$:
$$P_{\text{NO}_2} = \sqrt{3.99} \approx \sqrt{4.0} = \mathbf{2.0\text{ atm}}$$
This matches Option B.

*

WHY_OTHERS_WRONG:

* Option A is incorrect (0.44 atm): This error occurs if a student writes the $K_p$ expression upside down (reactant over product) and forgets to square the $\text{NO}_2$ term:
$$K_p = \frac{P_{\text{N}_2\text{O}_4}}{P_{\text{NO}_2}} \implies 3.0 = \frac{1.33}{P_{\text{NO}_2}} \implies P_{\text{NO}_2} = \frac{1.33}{3.0} \approx 0.44\text{ atm}$$
* Option C is incorrect (2.3 atm): This is approximately the value of $\frac{K_p}{P_{\text{N}_2\text{O}_4}}$ (specifically, $\frac{3.0}{1.33} \approx 2.26\text{ atm}$). A student would choose this if they made an algebraic error while rearranging the $K_p$ formula or incorrectly assumed a linear relationship where they divided rather than multiplied.
* Option D is incorrect (4.0 atm): This represents the squared partial pressure, $(P_{\text{NO}_2})^2 = 3.99 \approx 4.0\text{ atm}$. A student would choose this option if they correctly set up the equilibrium equation and performed the multiplication but forgot to take the square root at the very end. It also results from setting up the equation without squaring the product:
$$K_p = \frac{P_{\text{NO}_2}}{P_{\text{N}_2\text{O}_4}} \implies 3.0 = \frac{P_{\text{NO}_2}}{1.33} \implies P_{\text{NO}_2} = 3.0 \times 1.33 \approx 4.0\text{ atm}$$