Question

For the reaction having rate law expression:Rate$=k\left[A{]}^{3/2}\right[B{]}^{-1/2}$If the concentration of both $A$ and $B$ becomes four times, the rate of reaction:

• A. becomes four times
• B. becomes $16$ times
• C. decreases four times
• D. remains same

Answer 1

The correct option is A becomes four times

$Rate=k{\left[A\right]}^{3/2}{\left[B\right]}^{-1/2}$

Now the Concentration of both $A$ and $B$ becomes four times

So rate becomes ${\left[4\right]}^{3/2}{\left[4\right]}^{-1/2}$ times = four times