Question

The decomposition of nitrosyl chloride was studied:

$\mathbf{2}\mathbf{NOCl}\left(g\right)\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\left(g\right)$

The following data were obtained, where

$\mathbf{Rate}\mathbf{=}\frac{\mathbf{-}\mathbf{d}\left[\mathrm{NOCl}\right]}{\mathbf{dt}}$

1. What is the rate law?
2. Calculate the rate constant.
3. Calculate the rate constant for the concentrations given in moles per liter.

Short answer

(a) $\mathrm{Rate}=\mathrm{K}{\left[\mathrm{NOCl}\right]}^2 \mathrm{K}=\mathrm{rate} \mathrm{constant}$

(b) $\mathrm{R}=6.6\times {10}^{-29}{\mathrm{cm}}^3{\mathrm{molecules}}^{-1}{\mathrm{s}}^{-1}$

(c)$4.0\times {10}^{-8}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$

Step-by-step solution

Solve for finding the rate law.

(a)

$\mathrm{Rate}={\left[\mathrm{NOCl}\right]}^2$as $\mathrm{Rate}=\frac{-\mathrm{d}\left[\mathrm{NOCl}\right]}{\mathrm{dt}}$

From experiments 2 and 3,

$\begin{array}{c}\frac{2.66\times {10}^4}{6.64\times {10}^3}=\frac{K{\left(2.0\times {10}^{16}\right)}^x}{K{\left(1.0\times {10}^{16}\right)}^x}\\ 4.01={\left(2\right)}^x\\ 2^2=2^x\\ x=2\end{array}$

Therefore, the rate can be given as:$\mathrm{Rate}=\mathrm{K}{\left[\mathrm{NOCl}\right]}^2 \mathrm{K}=\mathrm{rate} \mathrm{constant}$

Now calculate the rate constant.

(b)

We know that$\text{Rate=K}{\left[\mathrm{NOCl}\right]}^2 \mathrm{K}=\mathrm{rate} \mathrm{constant}$

So,

$\begin{array}{c}\mathrm{K}=\frac{\mathrm{Rate}}{{\left[\mathrm{NOCl}\right]}^2}\\ \mathrm{K}=\frac{5.98\times {10}^4}{{\left(3.0\times {10}^{16}\right)}^2}\left(\mathrm{Fromexperiment}1\right)\\ \mathrm{K}=6.6\times {10}^{-24}{\mathrm{cm}}^3{\mathrm{molecules}}^{-1}{\mathrm{s}}^{-1}\end{array}$

Do the unit conversion for K

(c)

$\begin{array}{c}\mathrm{K}=\frac{6.6\times {10}^{-24}{\mathrm{cm}}^3}{\mathrm{molecules}}\times \frac{1\mathrm{L}}{1000{\mathrm{cm}}^3}\times \frac{6.022\times {10}^{23}\mathrm{molecules}}{\mathrm{mol}}\\ =\frac{4.0\times {10}^{-8}\mathrm{L}}{\mathrm{mols}}\\ =4.0\times {10}^{-8}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\end{array}$