question_answer

The rate of a gaseous reaction is equal to\[k[A][B]\]. The volume of the reaction vessel containing these gases is reduced by one-fourth of the initial volume. The rate of the reaction would be

A) \[\frac{1}{16}\]                                    

B) \[\frac{16}{1}\]

C) \[\frac{1}{8}\]                                      

D) \[\frac{8}{1}\] Rate (r) = k [A] [B] = kab When volume is reduced by one fourth the concentration becomes 4 times. Hence, \[r'=k(4a)(4b)=16\,kab\] \[\therefore \] \[r'16\,r\]