Question

The reaction $\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\stackrel{}{\mathbf{\to }}\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$was studied, and the following data were obtained,

Where,$\mathbf{Rate}\mathbf{=}\mathbf{-}\frac{\mathbf{d}\left[O_2\right]}{\mathbf{dt}}$.

${\left[\mathrm{NO}\right]}_{\mathbf{0}}\left(\mathrm{molecules}/{\mathrm{cm}}^3\right)$	${\left[O_2\right]}_{\mathbf{0}}\left(\mathrm{molecules}/{\mathrm{cm}}^3\right)$	role="math" localid="1663754227568" $\mathbf{Initial}\mathbf{}\mathbf{Rate}\mathbf{\left(}{\mathbf{molecules}}^{\mathbf{-}\mathbf{3}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}$
$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	role="math" localid="1663754491156" $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{16}}$
$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	$\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{17}}$
$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{18}}$	$\mathbf{3}\mathbf{.}\mathbf{13}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{17}}$

What would be the initial rate for an experiment where

${\left[\mathrm{NO}\right]}_{\mathbf{0}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{21}\mathbf{x}{\mathbf{10}}^{\mathbf{18}}\mathbf{molecules}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}\mathbf{and}{\left[O_2\right]}_{\mathbf{0}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{35}\mathbf{x}{\mathbf{10}}^{\mathbf{18}}\mathbf{molecules}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$ ?

Short answer

The initial rate of reaction will be $5.66\mathrm{x}{10}^{18}{\mathrm{molecules}}^{-3}{\mathrm{s}}^{-1}$.

Step-by-step solution

Finding the order of reaction with respect to NO and O2 separately.

The general rate equation for the given reaction will be,

$\mathrm{Rate}=\mathrm{k}{\left[\mathrm{NO}\right]}^{\mathrm{a}}{\left[{\mathrm{O}}_2\right]}^{\mathrm{b}}$

Obtaining the rate equation for the 1^st column, we get

$2\times {10}^{16}=\mathrm{k}{\left(1\times {10}^{18}\right)}^{\mathrm{a}}{\left(1\times {10}^{18}\right)}^{\mathrm{b}}$... eq. ( i )

Same for the 2^nd column,

$1.8\times {10}^{17}=\mathrm{k}{\left(3\times {10}^{18}\right)}^{\mathrm{a}}{\left(1\times {10}^{18}\right)}^{\mathrm{b}}$ ... eq. ( ii )

Dividing eq. ( i ) by eq. ( ii ), we get

$\mathrm{b}=2$

Similarly, for the 3^rd column, we get the rate equation,

$3.13\times {10}^{17}=\mathrm{k}{\left(2.5\times {10}^{18}\right)}^{\mathrm{a}}{\left(2.5\times {10}^{18}\right)}^{\mathrm{b}}$... eq. ( iii )

Dividing eq ( ii ) by eq. ( iii ) and substituting the value of b, we get $\mathrm{a}=1$.

Finding the value of k

Putting the value of a and b in eq. ( i ), we get

$\mathrm{k}=2\mathrm{x}{10}^{-38}$

Finding the required initial rate

According to the problem, the rate equation becomes

$\mathrm{Initial}\mathrm{rate}=\mathrm{k}{\left(7.35\times {10}^{18}\right)}^{\mathrm{a}}{\left(6.21\times {10}^{18}\right)}^{\mathrm{b}}$

Putting the value of a, b, and k, we get,

$\mathrm{Initial}\mathrm{Rate}=5.66\mathrm{x}{10}^{18}$