Question

just 12

KEF CAND 372 C) de law is: a) Rate = k[Fe(CN)63-1*[I][Fe(CN)64 1°[12] Rate = k[Fe(CN).-1*[1][Fe(CN). [12] Rate = k[Fe(CN).-)]*[1] d) Rate = k[Fe(CN).-1[15] Rate = k[Fe(CN)63-1[1] [Fe(CN).-) Tabulated below are initial rate data for the reaction: 2Fe(CN).- + 21 2Fe(CN).-- + 12 12. [I do 0.01 0.02 0.02 0.02 0.02 [Fe(CN)64-) 0.01 0.01 0.01 3 [12] 0.01 0.01 0.01 0.01 0.02 Rate (M/s) 1x 10-5 2x 10-5 8 x 10-5 - 8 x 10-5 8 x 10-5 0.02 0.02 Run [Fe(CN)63-10 0.01 0.01 0.02 0.02 0.02 What is the value of k? a) 107M b) 103 M3 s- c) 10 M2s! d) 50 M2 5-1 e) none of these = K (m)² CM Ise the following to answer questions 13-14: onsider the following data concerning the equation: + T-ILLO

Answer 1

General Rate Law Rate = k [ Fe(CN) (3-7? (1-7% from experiment 1 and 2. 1x 10-5 = K [0.017² [0.017 2x105 = K[0.01] ² [0.02] equation ② = 0 2x1075 = ķToolf?. 10.027% 1x165 (0.017%. (0.01] / 2 = (2) 7 (2)' = (2) .: 19 = 1 I - ③ from experiment 3. 8x10-5 = K[0.027² [0.027 equation (3 = ③ 8x101 - K [0.02]* [0.0p]* 2x107 410.017². 10/2] I 4 = (2) * (2)² = (2) * 1x=2

Rate = k [ Fe (IN);?-) ? [ 3.77 from experiment d. 1x10 -5 M = K[0.01 m 7². (0.01 m) K = 1 x 10-5 M/S (0.01)? (0.01) M3 K = 1x10-3 (10-2) 2. (10-2) megt K = 924 1x10-5 10-6 K = 1x10 m2 K = 10 2t |