Question

Nitrogen trifluoride decomposes to form nitrogen and fluorine gases according to the following equation:2NF_3(g) N_2(g) + 3F_2(g) When 1.82 mol of NF_3, is placed in a 3.00 L container and allowed to come to equilibrium at 800 K, the mixture is found to contain 0.0261 mol of N_2. What is the value of k_p, at this temperature? 398 times 10^-4 1.91 times 10^-5 1.80 times 10^-3 4.43 times 10^-7 3.87 times 10^-4 A red blood cell placed in pure water will swell because water. the osmotic is greater in the cell than in the water water moves from a higher osmotic pressure to a lower osmotic pressure. The vapor pressure of the water in the cell is greater than the vapor pressure of pure water. The osmotic pressure is the same in the cell and the water. Which of the following equilibria would not be affected by pressure changes at constant temperature? CO(g) + 1/2O_2(g) CO_2(g) CO_2(g) + H_2(g) CO(g) + H_2O(g) 2H_2(g) + O_2(g) 2H_2O(l) CaCO(t) CaO(x) + CO_2(g) 2Hg(l) + O_2(g) 2HgO(s) A solution consisting of 0.182 mol of methylbenzene. C_4H_2CH_3 in 258 g of nitrobenzene. C_6H_3NO_3 forces at 1.1 degree C. Pure nitrobenzene fluencies at 6.0 degree C. What is the freezing-point depression constant of nitrobenzene? 27 degree C/m 14.0 degree C/m 3.5 degree C/m 7.0 degree C/m

Answer 1

1.82 mol 0.607 M 3.00 L 21) Initial concentration of NF, [NE Equilibrium 2NE (g) Ini N2(g) + 3F (g) itial (M 0.607 Change (M):2x Equil (M):0.607 2x 3x From the ICE table 00261 mol o.0087 M 3.00 L x = 0.0087 Equilibrium concentration of NH3, [NH,] 0.607-2x) M(0.607-2(0.0087))M = 0.590M Equilibrium concentration of F. E] 3(x) (3(0.0087)M 0.0261M Equilibrium constant, K.22 NH (0.0087)(0.0261) 0.590 4.44x10- Equilibrium equation: 2NF, (g) N2(g) + 3F, (g) Change in the number of moles of gasses, An -np-4-2 2 Equilibrium constant, K 4.44x107 Temperature, T 800 K Equilibrium constant in terms of parti al pressures, K K (RT)* -(444x10)00.0821x800) -1.91x10-3