Question

The rate of reaction between two reactants $A$ and $B$ decreases by a factor of $4$ if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is:

Answer 1

$xA$ + $yB$ $\to$ $Product$
$Rate$ = $k\left[A\right]6x$ . $[B{]}^y$ .........(i)
When conc. of B is doubled,
$Rate/4$ = $k[A{]}^x$ .$[2B{]}^y$...........(ii)
Equation (i)/(ii)
$4$ = $1/2^y$
⇒$2^y$ = {1\2}²
⇒y= $-2$ with respect to $B$