# For the reaction, BrO3+5Br-+6H+→3Br2+3H2O, . Find the value of .Solution Rate of reaction can be expressed as:-15d[Br-]dt=13dBr2dt= (1.5×10-4) mol L-1 s-1. HOW THIS 1/3 CAME ? -dBr-dt=53×1.5×10-4molL-1s-1= 2.5×10-4 mol L-1 s-1 PLZ ANSWER TO MY QUESTION .

Reaction:
BrO3-(aq) + 5Br-(aq) + 6H+(aq) $\to$ 3Br2(aq) +  3H2O(l)

Rate of reaction = Rate of disappearance of reactants = Rate of appearance of products

Now, to get the unique value of the reaction rate (independent of the concentration terms chosen), we divide the rate of the reaction defined with any of the reactants or products by the stoichiometric coefficient of that reactant or product involved in the reaction. Thus, for the above reaction,

Rate of reaction = $-\frac{\mathrm{\Delta }\left[{{\mathrm{BrO}}_{3}}^{-}\right]}{\mathrm{\Delta t}}=-\frac{1}{5}\frac{\mathrm{\Delta }\left[{\mathrm{Br}}^{-}\right]}{\mathrm{\Delta t}}=-\frac{1}{6}\frac{\mathrm{\Delta }\left[{\mathrm{H}}^{+}\right]}{\mathrm{\Delta t}}=\frac{1}{3}\frac{\mathrm{\Delta }\left[{\mathrm{Br}}_{2}\right]}{\mathrm{\Delta t}}=\frac{1}{3}\frac{\mathrm{\Delta }\left[{\mathrm{H}}_{2}\mathrm{O}\right]}{\mathrm{\Delta t}}$

Since concentration of reactants are decreasing therefore negative sign is incorporated.

So, 1/3 came because the stoichiometric coefficient of Br2 is 3.

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