Question

The rate of a gaseous reaction is given by the expression $k{\left[A\right]}^2{\left[B\right]}^3$. The volume of the reaction vessel is suddenly reduced to one half of the initial volume. The reaction rate relative to the original rate will be:

• A. $1/24$
• B. $1/32$
• C. $32$
• D. $24$

Answer 1

The correct option is C $32$

Rate $=k{\left[A\right]}^2{\left[B\right]}^3$ ...(i)
When volume is halved, the concentration will become double.
$\therefore$ Rate $=k{\left[2A\right]}^2{\left[2B\right]}^3$
$=32k{\left[A\right]}^2{\left[B\right]}^3=32\times$original rate from (i)