Question

Rease Solve Qu aud QS CHEM 212-Equilibrium 2 1) At 208 K A+2B PI2C+D If (A 1.05 M. [B), 0.65 M, [Cl 0.ID),-0, and [Cl O460 calculate K 2) At 218 °C, K = 1.2 x 10 for the equilibrium NHHS (s) RINH, (g)+ HS(g) a) Write an exprension for the equilibrium constant, K b) Calculate the aquilibrium concentrations of NH, and H,Sif a 5.0g smple of NHHS (s) is placed in a sealod container and decomposes until equilibriumis achieved 3) Consider the system 2 N: (g) +6 H,O () H-1530.4 kJ 4 NH, (g)+30 (g) How will the amount of ammonia at equilibrium be affected by a) removing oxygen gas, b) adding nitrogen gas, c) adding water, d) expanding the contaner at constant temperature 4) At 2300 K, the equilibrium constant for the formation of NO (g) is 1.7 x 10 Na (g)+O(g) R2 NO (g) a) Analysis shows that the concentrations of N and O, are both 0.25 aatm, and that of NO is 0.0042 atm. Isthe system at equilibrium? b) If the system is not at equilibrium, in which direction does the reaction proceed? c) When the system is at equilibrium, what are the equilibrium conoentrations? 5) At acertain temperature, the reaction Xe(g)+2 F (g) HHXF. (g) gives a 50.0 % yidd of XeF. starting with Xe (Pxe 0.20 atm) and F (Pra 0.40 atm) and no products (Bye 0 CaaudeK

Answer 1

N2 + O2 Ž 2 No Given penitul Basen levering the 10.00421 atm Equilibrium constant k= 1.78103 quotient (a) Reaction e = ent Q = (PNo12 (Poz) (bmg) = loi0042)2 0.25) (0.25) Q=2082248104 Q = 0.2822x10-3 I ack than reaction not in equilibrium (b) is proceed in forward direction Ik>al than (die. reaction Right side)

(6) N₂ + O2 F 2NO I 0.25 0125 010042 Eam 0.25-) (0:25-1) (0.0042 +2n) at Equilibrian, k= a = (0.0042 +20) 2 (0125-1) (0125-1) 1.78103 - 10.0042 +2n7² 0.25-1 (0.0042 +22) ? (0.25-1)? 0.17 x 102 - ro.oou & an I I 0. 25-1

V0:17 x 10-2 = 0.0042 +24 0.25-1 (0.0412 ) = 010042 +2n 0.25-2 (0.0412) (0.25-1) = 010042 +24 0.0103 – 010412 x = 0.0042 +2n 0.0061 = 2:0412 n n = 2.988 x 103 no 0.00298 than ut equilibrian PN2 = 0.25-n= 0.25-0.00298 = 0.24702 alm bog = 0125-1 = 0.24702 alim bno = 0.0042+24 - 00042 + 2X 0.00298 – 0101016 atm

0.2 atm atm - Xe & 2 F2 Xefu intial 0.4 Yield a son than Equilibrin (0.2-n) (0.4-29) p Xe fu= 012 650 (+)= a 100 given oth = 0.1 atm than Equilibria constant I ko n (0.2-1) (0.4-2132 (0.2-0.1) (0•4-2X01]2 - Kp = ou = 25, atma 01 X1012)2 0.04 =Oilatu (based Ion poresion total Pressore u cat cet Equilibriun - Equilibriun - 10 2-n) t o u n - 0.6-in P = 0.6-2X0.1 = 0.4 atin (based on kee kp * (P)2 = 25% (0.4)2 = = 4 Concentration)