5 Br– (aq) + BrO3 – (aq) + 6 H+ (aq) 3 Br2(l) + 3 H2O(l) In a study of the kinet

Question

5 Br– (aq) + BrO3 – (aq) + 6 H+ (aq) 3 Br2(l) + 3 H2O(l)

In a study of the kinetics of the reaction represented above, the following data were obtained at 298 K.

Initial [Br-]

(mol L-1

Initial [BrO3-]

(mol L-1)

Initial [H+]

(mol L-1)

Rate of Dissaperance of BrO3-

(mol L-1 s-1)

a.)  From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning. (i) Br- (ii) BrO3- (iii) H+

b.) Write the rate law for the overall reaction.

c.) Determine the value of the specific rate constant for the reaction at 298 K. Include the correct units

Experiment

Initial [Br-]

(mol L-1

Initial [BrO3-]

(mol L-1)

Initial [H+]

(mol L-1)

Rate of Dissaperance of BrO3-

(mol L-1 s-1)

1 0.00100 0.00500 0.100 2.50x10-4 2 0.00200 0.00500 0.100 5.00x10-4 3 0.00100 0.00750 0.100 3.75x10-4 4 0.00100 0.01500 0.200 3.00x10-4

Explanation / Answer

Let us write the rate expression/law as

Rate = k [Br-]x[BrO3-]y[H+]z where x , y and z are the orders with respect to Br-, BrO3- and H+ respectively.

While caclulating the order with respect to one substrate, selct the ones where other concentrations are constant.

Comparing experiment 1 and 2 [BrO3-] and [H+] are constant.

Thus Rate 1 = k[0.001] x = 2.5 x10-4

Rate 2 = k [0.002]x = 5.0 x10-4

Rate2/rate1 = 2 = 2x  

Thus x = 1

Now comparing rate 3 and rate 1 [Br-] and [H+] are constant.

Rate 3 = k[o.oo75]y = 3.75x10-4

rate 1= k[0.005] y   = 2.5 x10-4

Thus rate 3/rate 1 = [ 1.5]y  = 1.5

hence y = 1

comparing rate 4 and rate 1

rate 4 / rate 1 =[3]1 [2]z = 12 or [2]z = 4

Thus z = 2

Hence the rate law = k [Br-] [BrO3-] [H+]2

and to evaluate k value we can substitute the values from any one experiment.

  K = 2.5x10-4 /[0.001][0.005][0.1]

   = 500 L3/mol3.s

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