Question

please solve the problem(in 20 mins)

X E M The answers must be with correct significant figures. 1009 07 10 For the reaction M(g) + N(g) = P(s) K = 400 at 40.0 C. If 2.00 mol each of M(g),N(g), and P(s) are placed in a 4.00-L vessel. (10%) (a) P(s) at equilibrium = mol(with correct significant figures) (b) N(g) at equilibrium = mol (with correct significant figures) E- F-16 • 9 9H - ERI 11:03 提交測驗

Answer 1

EP K = 5 M(9) + N(%) = ply) - 400 out 90°c - [M] EN Now, initially (m) = 2 = 0.5M) IN]- 0.5(M) [ts, volume of vessel (P) = 0.5 (1) =ALI it .. at 20 M(g) of Ni(9) 0.5 0.5 P (3) 0.5 0.5t 0 -2 0.5-2 at equile brum 0.5fx 400 i K 10.5-x) (0.5-x 2 r 0.5ta 400 2 = 3 0.25-x+ x²-x + 0.25 = 10254103 + 25x102 x 2 - 1.00250 & 0-2487 = 0 1,0025 F (1.0025) - (4 70 2987) 8 775 1.0025 1 0.10003 = 0.5512 and 0.9512 here, a=0.5512 is not possible because initial concentration of Mon N was 0.5 (M) = 0.9572 (M) at equilibrium, (P) = (0.5+ 0.4512) 50.950 and, [N] = (0.5-0.9512) = 0.0988 (4)