Explain the rate expression for a reaction in which stoichiometri…

Question

Explain the rate expression for a reaction in which stoichiometric coefficients of reactants and produch are not same or unit.

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Answer

$\rightarrow$ For reaction, $\ce{n_1A+n_2B \rightarrow n_3C + n_4D}$
$\rightarrow$ Rate of reaction
$-\frac{1}{n_1} \frac{d[A]}{d t}=-\frac{1}{n_2} \frac{d[B]}{d t}=+\frac{1}{n_3} \frac{d[ C ]}{d t}=+\frac{1}{n_4} \frac{d[ D ]}{d t}$
$\rightarrow$ For expressing the rate of such a reaction where stoichiometric coefficients of reactants or product are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of the products is divided by their respective stoichiometric coefficients.
$\rightarrow$ For reaction, $\ce{Hg (l)+ Cl_2(g) \rightarrow HgCl_2(s)}$
where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as Rate of reaction $=-\frac{\Delta[ Hg ]}{\Delta t}=-\frac{\Delta\left[ Cl _2\right]}{\Delta t}=\frac{\Delta\left[ HgCl _2\right]}{\Delta t}$
$\rightarrow$ For reaction, $\ce{2Hl( g ) \rightarrow H_2(g)+ I_2(g)}$ the rate of decomposition of $HI$ is twice the rate of formation of $H$ or $I _2$, to make them equal, the term $\Delta[ HI ]$ is divided by $2 $.
The rate of this reaction is given by Rate of reaction $=-\frac{1}{2} \frac{\Delta[ HI ]}{\Delta t}=\frac{\Delta\left[ H _2\right]}{\Delta t}=\frac{\Delta\left[ I _2\right]}{\Delta t}$
Similarly for the reaction
$ 5 \ce{Br ^{-}( aq )+BrO_3^{-}(aq)+6H ^{+}(aq) \rightarrow 3Br_2(aq)+3H_2O(l) }$
$\text { Rate }=-\frac{1}{5} \frac{\Delta\left[ Br ^{-}\right]}{\Delta t}$
$=-\frac{\Delta\left[ BrO _3^{-}\right]}{\Delta t}$
$=-\frac{1}{6} \frac{\Delta\left[ H ^{+}\right]}{\Delta t}$
$=\frac{1}{3} \frac{\Delta\left[ Br _2\right]}{\Delta t}$
$=\frac{1}{3} \frac{\Delta\left[ H _2 O \right]}{\Delta t} $