Question

The rate of reaction ${\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{CO}\left(g\right)\mathbf{\to }\mathbf{NO}\left(g\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$depends only on the concentration of Nitrogen dioxide at temperatures below ${\mathbf{225}}^{\mathbf{o}}\mathbf{C}$. At a temperature below ${\mathbf{225}}^{\mathbf{o}}\mathbf{C}$, the following data were collected.

Time (s)	$\left[{\mathrm{NO}}_2\right]\left(\mathrm{mol}/L\right)$
0	0.500
$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$	0.444
$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$	0.381
$\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$	0.340
$\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$	0.250
$\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$	0.174

Determine the integrated rate law, the different rate law, and the value of the rate constant at this temperature. Calculate$\left[N{O}_2\right]$at$\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{s}$after the start of the reaction.

Short answer

The rate of reaction refers to the speed at which the products are formed from the reactants in a chemical reaction.

• According to collision theory the rate of reaction increases with the increase in the concentration of the statement.
• Time plays a major role in changing the concentration of reactants and products.

Step-by-step solution

Determination of the order of reaction by using data

By plotting the graph between $\frac{1}{\left[N{O}_2\right]}$versus time, a straight line is found so that we can conclude that the reaction is second order in respect to $N{O}_2$.

The rate law $\mathrm{Rate}=\mathrm{K}{\left[{\mathrm{NO}}_2\right]}^2$

Time (s)	localid="1663854662654" role="math" $\frac{1}{\left[N{O}_2\right]}$
0	2
localid="1663854735621" role="math" $1.20\times {10}^3$	2.25
localid="1663854779341" role="math" $3.00\times {10}^3$	2.62
localid="1663854799922" role="math" $4.50\times {10}^3$	2.94
localid="1663854813582" role="math" $9.00\times {10}^3$	4
localid="1663854834459" role="math" $1.80\times {10}^4$	5.74

Integrated rate law.

$\frac{1}{{\left[N{O}_2\right]}_t}-\frac{1}{{\left[N{O}_2\right]}_{\circ }}=Kt$

Which can be written as

$\frac{1}{{\left[N{O}_2\right]}_t}=Kt+\frac{1}{{\left[N{O}_2\right]}_{\circ }}$

Value of rate constant.

The slope of the plot $\frac{1}{\left[N{O}_2\right]}$ versus time gives the value of K.

$\begin{array}{rcl}\mathrm{Slope}& =& \frac{\mathrm{\Delta x}}{\mathrm{\Delta y}}\\ & =& \frac{2-2.25}{0-1.20\times {10}^3}\\ & =& \frac{0.25}{1.2\times {10}^3}\\ & =& 2.08\times {10}^{-4}\end{array}$

Calculation of the as after the start of the reaction

$$\begin{gathered} \frac{1}{{\left[{\mathrm{NO}}_2\right]}_{\mathrm{t}}}=\frac{=2.08\times {10}^{-4}}{\mathrm{mols}}\times 2.70\times {10}^4+\frac{1}{{\left[0.500\right]}_{\circ }} \\ \frac{1}{{\left[{\mathrm{NO}}_2\right]}_{\mathrm{t}}}=7.62,\left[{\mathrm{NO}}_2\right]=0.131\mathrm{M} \end{gathered}$$