Chemistry - Chemical Kinetics Question with Solution | TestHub

$r=k[A{]}^x[B{]}^y$.

$2\times 10^{-2}=k[10^{-2}{]}^x[3\times 10^{-3}{]}^y\dots (1)$

$4\times 10^{-2}=k[2\times 10^{-2}{]}^x[3\times 10^{-3}{]}^y\dots (2)$

$8\times 10^{-2}=k[2\times 10^{-2}{]}^x[6\times 10^{-3}{]}^y\dots (3)$

Dividing equation (2) by (1), we get:

$\frac{4\times 10^{-2}}{2\times 10^{-2}}=\frac{k[2\times 10^{-2}{]}^x[3\times 10^{-3}{]}^y}{k[10^{-2}{]}^x[3\times 10^{-3}{]}^y}$

$2=(2{)}^x$

$x=1$.

Similarly, dividing equation (3) by (2), we get:

$\frac{8\times 10^{-2}}{4\times 10^{-2}}=\frac{k[2\times 10^{-2}{]}^x[6\times 10^{-3}{]}^y}{k[2\times 10^{-2}{]}^x[3\times 10^{-3}{]}^y}$

$2=(2{)}^y$

$y=1$.

So, $r=k[A][B]$.

$2\times 10^{-2}=k[10^{-2}][3\times 10^{-3}]$

$k=\frac{2\times 10^{-2}}{10^{-2}\times 3\times 10^{-3}}$

$k=\frac{2}{3\times 10^{-3}}=\frac{2\times 10^3}{3}{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$

To convert this to ${\mathrm{M}}^{-1}{\min }^{-1}$, we multiply by 60:

$k=\frac{2\times 10^3}{3}\times 60{\mathrm{M}}^{-1}{\min }^{-1}$

$k=2\times 10^3\times 20{\mathrm{M}}^{-1}{\min }^{-1}$

$k=4\times 10^4{\mathrm{M}}^{-1}{\min }^{-1}$