The following data were obtained for the reaction between iodide ion and dibromo

Question

The following data were obtained for the reaction between iodide ion and dibromoethane (C2H,Br2) in methanol 2 4 Initial Rate of lemperature Experiment[C H Brl, (M)I1, (M) 0.102 0.102 0.204 Formation of l (M/s)Of Reaction (C) 0.127 0.254 0.127 6.45x10 1.29x10 1.29x10 20 20 20 2 3 Use the initial rate method and the data above to determine the rate law and rate constant for this reaction Select one o a. Rate-4.98 x 10 [C.H.Br,TI O b. Rate 4.98 x 10 [C2H,Br,111 O C. Rate 4.98 x 10 IC,H,Br.II] d. Rate = 200 [CHBr2][1] O

Explanation / Answer

Answer

c) Rate = 4.98×10-3[C2H4Br2][I-]

Explanation

C2H4Br2 + 3I-  - - - - - > C2H4 + 2Br-   + I3-

Rate = k [C2H4Br2]m[I-]n

By comparing experiment 1 and 2, we know that after two fold increase in concentration of C2H4Br2, the rate is increased to two fold

rate = k (2×[C2H4Br2]) m [I-]n

if concentration of C2H4Br2 increased by two fold

rate = k× 2m [C2H4Br2]m [I-]n

So, 2m = 2 and m = 1

By comparing experiment 1and 3, we know that by increasing concentration of I- by two fold, rate of reaction increased by two fold

rate = k [C2H4Br2]m [I-]n

if concentration of I- is increased by two fold

rate = k [C2H4Br2]m (2[I-])n

rate = k[C2H4Br2]m × (2)n [I-]n

2n = 2

So, n = 1

Therefore,

rate = k [C2H4Br2] [I-]

substitute experment 1 concentrations and rate value in the above rate law

6.45×10-5 M/s = k × 0.127M × 0.102M

k = 6.45×10-5M/s / 0.127M ×0.102M

k = 4.98×10-3 M-1s-1

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