# The decomposition of dimethyl ether leads to the formation of

Question:

The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by

Rate = [CH3OCH3]3/2

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

Rate $=k\left(p_{\mathrm{CH}_{3} \mathrm{OCH}_{3}}\right)^{3 / 2}$

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

Solution:

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min−1

Rate $=k\left(p_{\mathrm{CH}_{3} \mathrm{OCH}_{3}}\right)^{3 / 2}$

$\Rightarrow k=\frac{\text { Rate }}{\left(p_{\mathrm{CH}_{3} \mathrm{OCH}_{3}}\right)^{3 / 2}}$

Therefore, unit of rate constants $(k)=\frac{\operatorname{bar} \min ^{-1}}{\operatorname{bar}^{3 / 2}}$

$=b a r^{-1 / 2} \min ^{-1}$