The rate of a gaseous reaction is given by the expression k[A][B]. If volume of the reaction vessel is suddenly reduced to \(\frac{1}{4}\)th the initial volume, then what will be the reaction rate with respect to the original rate? 

16 times

Concentration of reactant increases with decrease in volume.

Concentration α \(\frac{1}{volume}\)

As rate r = k[A][B] 

New rate r' = k[4A][4B] = 16k[A][B] = 16r