Question

The following data were obtained for the reaction

$\mathbf{2}{\mathbf{ClO}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{2}{\mathbf{OH}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{\to }\mathbf{Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{ClO}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

Where,$\mathbf{Rate}\mathbf{=}\mathbf{-}\frac{\mathbf{d}\left[{\mathrm{ClO}}_2\right]}{\mathbf{dt}}$

${\left[{\mathrm{ClO}}_2\right]}_{\mathbf{0}}\left(\mathrm{mol}/L\right)$	role="math" localid="1663766560289" ${\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}_{\mathbf{0}}\mathbf{(}\mathbf{mol}\mathbf{/}\mathbf{L}\mathbf{)}$	Initial Rate $\left({\mathrm{molL}}^{-1}{s}^{-1}\right)$
0.0500	0.100	$\mathbf{5}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$
0.100	0.100	$\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{1}}$
0.100	0.0500	$\mathbf{1}\mathbf{.}\mathbf{15}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{1}}$

a. Determine the rate law and the value of the rate constant.

b. What would be the initial rate for an experiment with${\left[{\mathrm{ClO}}_2\right]}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{175}\mathbf{mol}\mathbf{/}\mathbf{L}$and${\left[{\mathrm{OH}}^{-}\right]}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0844}\mathbf{mol}\mathbf{/}\mathbf{L}$?

Short answer

1. The rate law is $\mathrm{k}{\left[{\mathrm{ClO}}_2\right]}^2\left[{\mathrm{OH}}^{-}\right]$and the value of rate constant localid="1663767493869" $230\times 100{\mathrm{L}}^2{\mathrm{mol}}^{-2}{\mathrm{s}}^{-2}$
2. The initial rate for an experiment with ${\left[{\mathrm{ClO}}_2\right]}_0=0.175\mathrm{mol}/\mathrm{L}$and ${\left[{\mathrm{OH}}^{-}\right]}_0=0.0844\mathrm{mol}/\mathrm{L}$will be$0.594{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$

Step-by-step solution

Determining the rate law and the value of the rate constant

Rate law of reaction is an expression gives us the relationship between the rate of reaction and the concentration of reactants.

So, the rate law equation will be

$\mathrm{Rate}=\mathrm{k}{\left[{\mathrm{ClO}}_2\right]}^{\mathrm{x}}{\left[{\mathrm{OH}}^{-}\right]}^{\mathrm{y}}$

From question we can determine two equations from first and second row in table

$\begin{array}{rcl}k{\left[0.0500\right]}^x{\left[0.100\right]}^y& =& 5.75\times {10}^{-2}-\text{i}\\ k{\left[0.100\right]}^x{\left[0.100\right]}^y& =& 2.30\times {10}^{-1}-\text{ii}\end{array}$

On solving equations (i) and (ii) we get $x=2$.

Now we can determine two equations from second and third row in table

$k{\left[0.100\right]}^x{\left[0.0500\right]}^y=1.15\times {10}^{-1}-iii$

On solving equations (ii) and (iii) we get $y=1$.

We get$x=2$and $y=1$.

$\mathrm{Rate}=\mathrm{k}{\left[{\mathrm{ClO}}_2\right]}^2{\left[{\mathrm{OH}}^{-}\right]}^1$

For determining the rate constant put the value of concentrations of reactants in rate law equation.

$\begin{array}{rcl}\text{Rate}& =& k{\left[Cl{O}_2\right]}^2{\left[O{H}^{-}\right]}^1\\ 5.75\times {10}^{-2}& =& k\times {\left(0.0500\right)}^2\times 0.100\\ k& =& 230\end{array}$

So, the value of rate constant is$230{\mathrm{mol}}^{-2}{\mathrm{L}}^2{\mathrm{s}}^{-2}$.

Determining the initial rate

Initial values are${\left[{\mathrm{ClO}}_2\right]}_0=0.175\mathrm{mol}/\mathrm{L}$and${\left[{\mathrm{OH}}^{-}\right]}_0=0.0844\mathrm{mol}/\mathrm{L}$.

role="math" localid="1663767424647" $\begin{array}{c}\mathrm{Rate}=\mathrm{k}{\left[{\mathrm{ClO}}_2\right]}^2{\left[{\mathrm{OH}}^{-}\right]}^1\\ =230\times {\left(0.175\right)}^2\times 0.0844\\ =0.594\end{array}$

Rate with initial values will be $0.594{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$.