Question

The rate law for the reaction, $2{Cl}_{2}O⟶2{Cl}_2+O_2$, at ${200}^{o}C$ is found to be: $Rate=k{\left[{Cl}_{2}O\right]}^2$.
$\left(a\right)$ How would the rate change if $\left[{Cl}_{2}O\right]$ is reduced to one-third of its original value?
$\left(b\right)$ How should the $\left[{Cl}_{2}O\right]$ be changed in order to double the rate?
$\left(c\right)$ How would the rate change if $\left[{Cl}_{2}O\right]$ is raised to threefold of its original value?

Answer 1

$\left(a\right)$ Rate equation for the reaction,
$r=k{\left[{Cl}_{2}O\right]}^2$
Let the new rate be $r^l$; so, $r^l=k{\left[\frac{{Cl}_{2}O}{3}\right]}^2=\frac{1}{9}r$
$\left(b\right)$ In order to have the rate $=2r$, let the concentration of ${Cl}_{2}O$ be $x$.
So. $2r=k{x}^2$ ...(i)
We know that, $r=k{\left[{Cl}_{2}O\right]}^2$ ...(ii)
Dividing equation (i) by equation (ii),
$\frac{2r}{r}=\frac{k{x}^2}{k{\left[{Cl}_{2}O\right]}^2}$
or $2=\frac{x^2}{{\left[{Cl}_{2}O\right]}^2}$
or $x^2=2{\left[{Cl}_{2}O\right]}^2$
or $x=\sqrt{2}\left[{Cl}_{2}O\right]$
$\left(c\right)$ New rate $=k{\left[3{Cl}_{2}O\right]}^2=9k{\left[{Cl}_{2}O\right]}^2=9r$
i.e., nine times of the original rate.