Question

You have two bulbs connected by a valve. One bulb contains 3.5 L of nitrogen at 750 torr, and the other contains 21 L of hydrogen at 720 tor. The valve is opened and the gases are allowed to mix and react according to the equation below at SOC. What is the final total pressure? N2 (g)Hy (g) - NH, (g) Submit response ++ Change seat Send a message to the instructor < Join another session

Answer 1

Piv, No. of moles of N2 = RT, 750X3.5 303x8.314 5 1.05 No. of more of No. of moles of All H₂ = P. RT2 720x 2.) 's 0.606 300 Y 8.314 No. of moles of NH3 formed = 2x. no. of moles of Hz. the is limiting He Reagent} = 2x0.606 = 1.212 moles. P₂ /3 = 1.212 RTB . .. 1.2128 Pa = 1.212 X 323X8.314 5.6 " 1= + 2 h = 581.4 Pressure due T-273 + 50 = 321 K - to NH₂ Remaining wo. of moles of Na = 1.05-0.006 Por Vg 0.444 0.444 X 323X8.314 = i. Py 0.444 x 224 5.6 Final - 212.927 Pressure due to remaining N2. molecules Pressure = P + py - 581.4+ 212.92 = 794.32 tour