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The rate of a gaseous reaction is given by the expression $k[A][B]$. If the volume of the reaction vessel is suddenly reduced to $ \large\frac{1}{4}$ th of the initial volume , the reaction rate relating to original rate will be

$(A)\;\frac{1}{10} \\ (B)\;\frac{1}{8} \\(C)\;8 \\(D)\;16 $

1 Answer

Since active mass = concentration in moles / litre.
Thus when the volume of the reaction vessel reduces to $\large\frac{1}{4}$th of its initial volume
The molar concentration of each reactant increases four times, Thus,
$r= K[4A][4B]$
$\quad= 16\;K[A][B]$
16 times
Hence D is the correct answer.
answered Apr 4, 2014 by meena.p
 

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