Question

A(g) -> B(g)+C(g) Find the equilibrium concentrations of A, B, and C for each of the following different values of Kc. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products.

UNITS_18_19_HW Exercise 15.61 A Review Constants 1 Peric Submit Request Answer Consider the following reaction: A(9) B(9) + C(9) Find the equilibrium concentrations of A, B, and C for each of the following different values of Ke. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products. Make any appropriate simplifying assumptions Part B K 1.2x10-2 Express your answer using two significant figures. Enter your answers numerically sep by commas. * AXO * O ? (A).B]. [C] = 0.91,0.095,0.095 Submit Request Answer Part C K 1.0x10-5

Answer 1

Initial conc. of A = 1.0 (a) Kc = 1.20 x 10-2 Let x be the change in the conc. of each species at equilibrium, ICE table is М А B с Initial Change Equillibrium 1.0 - X X.X Kc = ** =1.2x102 = 0.012 1.0-X x? +0.012x – 0.012 = 0 by solving the quadratic equation, we get, x = 0.103709 .:. the equilibrium concentraions are [A]=1.0- x=1.0-0.103709 = 0.896291 = 0.90 M [B]= x=0.103709 = 0.10 M [C]=x=0.103709 = 0.10 M [A]=0.90 M [B]=0.10 M [C]=0.10 M

Part - C :

Initial conc. of A = 1.0 M Kc = 1.0 x 10-5 Let x be the change in the conc. of each species at equilibrium, ICE table is М А B с Initial Change Equillibrium 1.0 - x KC +1.0- x – XX_=1.0x10- x +(1.0x10-)x-(1.0x10–5)=0 by solving the quadratic equation, we get, x=0.003157 .. the equilibrium concentraions are JA=1.0-x=1.0-0.003157 = 0.996843 = 1.0 M [B] = x=0.003157 = 0.0032 M [C] = x=0.003157 =0.0032 M (A)=1.0 M [B] =0.0032 M [C]=0.0032 M