At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

\(x = 1.5 \) 6. Calculate the equilibrium partial pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 4.5 - 1.5 = 3.0\ \mathrm{atm}\) \(P_{\mathrm{NO}_{2}} = 2 (1.5) = 3.0\ \mathrm{atm}\) #tag_title# Part B: Calculate equilibrium partial pressures when starting with NO2 #tag_content# 1. Let the change in pressure of \(NO_{2}\) be y at equilibrium. 2. Since two moles of \(NO_{2}\) yields one mole of \(N_{2}O_{4}\), the change in pressure of \(N_{2}O_{4}\) is y/2. 3. Write the expressions for the equilibrium pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = \frac{y}{2}\) \(P_{\mathrm{NO}_{2}} = 9.0-y\) 4. Substitute the equilibrium pressure values into the Kp expression: \(0.25 = \frac{(9.0-y)^{2}}{\frac{y}{2}}\) 5. Solve for y: \(y = 6.0\) 6. Calculate the equilibrium partial pressures: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = \frac{6.0}{2} = 3.0\ \mathrm{atm}\) \(P_{\mathrm{NO}_{2}} = 9.0 - 6.0 = 3.0\ \mathrm{atm}\) #tag_title# Part C: Discuss the dependence of the equilibrium position on the initial conditions #tag_content# In both cases, the equilibrium partial pressures are the same: \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 3.0\ \mathrm{atm}\) and \(P_{\mathrm{NO}_{2}} = 3.0\ \mathrm{atm}\). This indicates that, for this reaction, the equilibrium position does not depend on the initial conditions or the direction from which the equilibrium is approached.