In acid solution, bromate ion (BrO_3^i) oxidizes bromide ion (Br\") to produce e

Question

In acid solution, bromate ion (BrO_3^i) oxidizes bromide ion (Br") to produce elemental bromine according to the following reaction: 5 Br^-(aq) + BrO_3^- (aq) + 6H^+(aq) rightarrow 3Br_2(aq)+ 3H_2O (aq) The kinetics of this reaction were studied using the method of initial rates. When the concentrations of the three reactants were held constant at the values tabulated in each column, the indicated initial rates of consumption of Br^- were observed: What are the orders of the reaction in each reactant? Which orders are the same as or different from the corresponding stoichiometric coefficients? What is the value of the rate constant k for the reaction? What is the rate constant k_H+ for the consumption of H^+ ion?

Explanation / Answer

rate = k [Br-]x[BrO3-]y[H+]z

from data 1 and 2 , we get

x = 1

from data 2 and 3 , we get

y = 1

from data 1 and 4 , we get

z = 2.

so now the rate is

rate = k [Br-]1[BrO3-]1[H+]2

1)

order with respect to [Br-] = 1

order with respect to [BrO3-] = 1

order with respect to [H+] = 2

2)

order from stchiometric coefficients

[Br-] = 5 , [BrO3-] = 1 , [H+] = 6.

but frpm experimentaly we have

[Br-] = 1 , [BrO3-] = 1 , [H+] = 2

so only [Br-] order is same. remaining are different..

3)

rate = k [Br-]1[BrO3-]1[H+]2

0.79 x 10^-3 = k [0.12]1[0.15]1[0.08]2

rate constant k = 6.86 s-1

4)

rate = k [H]2

0.79 x 10^-3 = k [0.08]^2

k = 0.123

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