Question

I need help with the rate constants for trials 1 through 4. I put as much info as I thought would be necessary. The last question I posted I was told I didn't have enough info, so I hope this is enough.

This is all the info I was given to solve the questions, so this has to be enough to solve them.
Rate Constant, k', for the reaction occurring in Solution 1 (M1.51) NG 0.5 Info Rate Constant, k', for the reaction occurring in Solution 2 (M?•s?) NG 0.5 Info Rate Constant, k', for the reaction occurring in Solution 3 (M1.5-1) NG 0.5 Info Rate Constant, k', for the reaction occurring in Solution 4 (M1.51) NG 0.5 Info

Answer 1

| H₂O KI after I KI Trial R02 mi H₂O2 after mixing ml ml CM7 1:05 0.700 0.101 3.99 0.5 0.5 0.703 0.050 |-00 2:0 0.362 0.098 201 1.05 0.699 4:08 10.102 bia Rall of formation of 02 (Lorr/5 Rate of formation O2 + mol/Ls Rate of decomposition H₂O₂ (moe / 25 0.0000 150 - 1. 18.79 0.274 9.9 E-005 0.0000 256 18.79 0.0685 0.00000,376 18.79 0.0662 0.00000 3636 0.00002424 - 4. 32.18 4:0791 0.000214 0.0014)

Date Rate of Reaction = -1 d[ Hz03] British Rale el Reaction I Trial 1 R = 9.9E-005 - 000099 22 - 6-9503 Trial2 R2 = 0000 256 - 1.28es = 1.212 e 5 Trial 3 R3 00002424 2 Trialu 0.000705 Ryc 0.00141 2

Date: Now Calculation for Rate constant Initial Rate = K (H₂O₂] [I] Initial Rate 4.95e-5 [H027 - I - 0.700 0.101 Trial 2 K= Initial Rate & 495e-5 [12 Oe] [I ] (0.700] [o 10 11 - 10.0007071 Initial Rate - 1.28e-5 [H₂O2) = 0.703 (I-] -0.050 K= 1.28e-5 = 10.000364 / [0.703] [0.050 ] Initial Rate = 1.212 e-s. TH₂O2) = 0.362 1-T - 0.098 K= 1.212es - 10.0003416 [0.862] [0.098) Trial 3 Trially Initial Rate = 0.000 tos [H₂O2) = 0.699 IL - 0.102 K 0.000 705 - 10.699] (0.102) 0.00988