The rate of a gaseous reaction is given by the expression `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is reduced to one half of the initial volume. What will be the reaction rate as compared to the original rate a?

Options

A. `1//10`
B. `1//8`
C. 8
D. 16

Answer

D

Explanation

`r=k[A][B] =(k n_(A) n_(B))/(V^(2))`
Where `n_(A) and n_(B)` are moles of A and B respectively and V is volume.
`r'=(kn_(A) n_(B))/((V//4)^(2))=16 (kn_(A)n_(B))/(V^(2))=16 r, (r')/(r)=16`