Question

Write the ICE table, K_t and K_ in terms of x based on the results from the table the following reactions: 3O_2(g) 2O_2(g) (Assume only [Y] M O_2 at the start) K_c = K_c = b. PCl_2(s) P^2+(aq) + 2Cl^-(aq) (Assume only PCl_2 at the start) K_c = K_c = c. N_2(g) + 3H_2(g) 2NH_3(g) (Assume only [Y] M NH_3 at the start) K_c = K_c = CH_3CO_2K(aq) + H_2O(l) H_3O^+(aq) + CH_3CO^-_2(aq) (Assume only [Y] M acetic acid at the start) K_c = K_c =

Answer 1

3O_2 (g) rightleftharpoons 20_3 (g) k_f = [o_2]^3 k_b = [0_3]^2 k_c = k_b/k_f = [0_3]^2/[0_2]^3 30_2 (g) rightleftharpoons 20_3 (g) 1 mole a at t = 0 1 - x mol 2/3 times mole k_f = [2/3 x]^2/[1 - x]^3 pcl_2 = p^2- (aq) + 2cl^- (aq) k_c = [p^2] [cl^-]^2/[pcl_2] pcl_2 rightleftharpoons p^2 (aq) + 2 cl^-1 (aq) 1 mol 0 0 1- x x 2x k_d = [x] [2x]^2/(1 - x) \r\n N_2 (g) + 3 H_2 (g) rightleftharpoons 2 NH_3 k_c = [NH_3]^2/[N_2][H_2]^ N_2 (g) + 3 H_2 (g) rightleftharpoons 2 NH_3 1 mol 3 mol x 0 mol at t = 0 1 - x 3 (-x) 2x k_d = [2x]^2/(1 - x) (3 [1 - x]^3 CH_2 CO_2 H (aq) + H_2O (l) rightleftharpoons H_3 O^+ (aq) + CH_2 CO_2^- k_c = [H_3 o^+] [CH_2 CO_2]/(CH_2 CO_2 H)[H_2 o] 1 mol 1 mol Rightarrow 0 0 at t = 0 (1- x) (1 - x) Rightarrow x x at t = t k_c = [x][x]/[1 - x][1 - x] = x^2/(1 - x)^2