A) \[-\frac{d\,[{{H}_{2}}]}{dt}=-\frac{d\,[{{l}_{2}}]}{dt}=2\frac{d\,[Hl]}{dt}\]
B) \[-2\frac{d\,[{{H}_{2}}]}{dt}=-2\frac{d\,[{{l}_{2}}]}{dt}=\frac{d\,[Hl]}{dt}\]
C) \[-\frac{d\,[{{H}_{2}}]}{dt}=-\frac{d\,[{{l}_{2}}]}{dt}=\frac{d\,[Hl]}{dt}\]
D) \[-\frac{d\,[{{H}_{2}}]}{2dt}=-\frac{d\,[{{l}_{2}}]}{2dt}=\frac{d\,[Hl]}{dt}\]
Correct Answer: B
Solution :
When stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or rate of appearance of products is divided by their stoichiometric coefficients. \[{{H}_{2}}+{{I}_{2}}\xrightarrow{{}}2HI\] Rate of reaction \[=\frac{-d[{{H}_{2}}]}{dt}=\frac{-d[{{l}_{2}}]}{dt}=\frac{1}{2}\frac{d[Hl]}{dt}\] or \[=\frac{-2d[{{H}_{2}}]}{dt}=\frac{-2d[{{l}_{2}}]}{dt}=\frac{d[Hl]}{dt}\] For the given equation, rate of reaction is rate \[=\frac{-d[{{H}_{2}}]}{dt}=\frac{-d[{{l}_{2}}]}{dt}=\frac{1}{2}\frac{d[Hl]}{dt}\] or \[=\frac{-2d[{{H}_{2}}]}{dt}=\frac{-2d[{{l}_{2}}]}{dt}=\frac{d[Hl]}{dt}\]
You need to login to perform this action.
You will be redirected in
3 sec
