The conversion of molecules $X$ to $Y$ follows second order kinetics. If concentration of $X$ is increased to three times how will it affect the rate of formation of $Y ?$
The conversion of molecules $X$ to $Y$ follows second order kinetics. If concentration of $X$ is increased to three times how will it affect the rate of formation of $Y ?$
The reaction $x \rightarrow Y$ follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate $=k[X]^{2}(1)$
Let $[X]=a$ $mol$ $L^{-1},$ then equation $(1)$ can be written as:
$Rate_{1}=k \cdot(a)^{2}$
$=k a^{2}$
If the concentration of $X$ is increased to three times, then $[X]=3 \,a$ $mol$ $L^{-1}$
Now, the rate equation will be:
Rate $=k(3 a)^{2}$
$=9\left(k a^{2}\right)$
Hence, the rate of formation will increase by $9$ times.
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