A 250 ml flask and 100 ml flask are separated by a stopcock. At 350 K, in larger flask exerts 0.4 at — Chemical Equilibrium Chemistry Question
Question
A 250 ml flask and 100 ml flask are separated by a stopcock. At 350 K, $NO$ in larger flask exerts 0.4 atm and $O_2$ in smaller has 0.8 atm. Gases are mixed by opening the stopcock. First reaction 2$NO$ + $O_2$ -> 2$NO_2$ is complete while second 2$NO_2$ ⇌ $N_2O_4$ is at equilibrium. If total pressure is 0.3 atm, calculate $K_p$ for second reaction.
💡 Solution & Explanation
Step 1 - Use PV products to represent moles (T constant) Two flasks: V1 = 250 mL with NO at 0.4 atm; V2 = 100 mL with O2 at 0.8 atm; T = 350 K. Since \(T\) is constant, \(PV \propto n\): \[(PV)_{\ce{NO}} = 0.4 \times 250 = 100\text{ atm·mL}\] \[(PV)_{\ce{O2}} = 0.8 \times 100 = 80\text{ atm·mL}\] Total volume after mixing: \(V_\text{total} = 350\text{ mL}\). Step 2 - First reaction goes to completion: 2NO + O2 → 2NO2 NO is limiting (needs 2× O2 but only 100/2 = 50 atm·mL O2 required): \[(PV)_{\ce{NO2}} = 100\text{ atm·mL}, \quad (PV)_{\ce{O2,rem}} = 80 - 50 = 30\text{ atm·mL}\] Total PV after reaction 1 = 130 atm·mL. Step 3 - Second reaction reaches equilibrium: 2NO2 ⇌ N2O4 Let \(y\) atm·mL of \ce{N2O4} form: \[(PV)_{\ce{NO2}} = 100 - 2y, \quad (PV)_{\ce{N2O4}} = y, \quad (PV)_{\ce{O2}} = 30\] Total PV = \(130 - y\). Using \(P_\text{total} = 0.3\text{ atm}\): \[0.3 = \frac{130 - y}{350} \implies 130 - y = 105 \implies y = 25\text{ atm·mL}\] Step 4 - Calculate equilibrium partial pressures \[p_{\ce{NO2}} = \frac{100 - 50}{350} = \frac{50}{350} = \frac{1}{7}\text{ atm}\] \[p_{\ce{N2O4}} = \frac{25}{350} = \frac{1}{14}\text{ atm}\] \[p_{\ce{O2}} = \frac{30}{350} = \frac{3}{35}\text{ atm}\] Check: \(\frac{1}{7} + \frac{1}{14} + \frac{3}{35} = \frac{10+5+6}{70} = \frac{21}{70} = 0.3\text{ atm}\) ✓ Step 5 - Calculate Kp for 2NO2 ⇌ N2O4 \[K_p = \frac{p_{\ce{N2O4}}}{(p_{\ce{NO2}})^2} = \frac{1/14}{(1/7)^2} = \frac{1/14}{1/49} = \frac{49}{14} = \boxed{3.5\text{ atm}^{-1}}\] Note: Some answer keys list 0.87 atm^-1, which results from the error of writing \((2p_{\ce{NO2}})^2\) in the denominator instead of the correct \((p_{\ce{NO2}})^2\). The thermodynamically correct answer is 3.5 atm^-1 (option A). Step 6 - Evaluate all options - **Option (A) 3.5 atm^-1**: Correct. Kp = P_N2O4/P_NO2² = (1/14)/(1/49) = 3.5. - **Option (B) 0.87 atm^-1**: Incorrect (common error using 2P_NO2 in denominator). - **Option (C) 0.07 atm^-1**: Incorrect. Calculation error. - **Option (D) 7.0 atm^-1**: Incorrect. Off by factor of 2.