For the reaction H_2 + I_2 rightleftharpoons | vedclass

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(C) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by: Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d

Vedclass · Accepted Answer

$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$

Vedclass · Answer

(C) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by: Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$. For the given reaction $H_2 + I_2 \rightleftharpoons 2HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively. Therefore,the rate of reaction is: Rate $= -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$. Rearranging this,we get $-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.

$t/s$	$0$	$50$	$100$	$150$	$200$	$300$	$400$	$700$	$800$
$[C_{4}H_{9}Cl]/mol\ L^{-1}$	$0.100$	$0.0905$	$0.0820$	$0.0741$	$0.0671$	$0.0549$	$0.0439$	$0.0210$	$0.017$

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$,write the correct relationship between the rates of change of concentration of reactants and products.

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