The decomposition of dimethyl ether leads to the formation of `(CH)_(4) and (H)_(2)` and `(CO)` and the reaction rate is given by Rate `=(k)[(CH)_(3) (OCH)_(3)]^(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)P (C H)_(3) (OCH)_(3)^(3 / 2)` If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Explanation

Ans: The decomposition reactiuon is `(CH)_(3) (OCH)_(3) rarr` `(CH)_(4)+(H)_(2)+(CO)`
In terms of concentrations. Units of rate=`(mol) (L)^(-1)` `(min)^(-1)`
therefore units of `(R)=(Rate )(CH)_(3)(OCH)_(3)^(3/ 2)=((mol)^(-) (min)^(-1))(mol) (L)^(-1)^(3/2)`

=`(L)^(1 / 2) (mol)^(-1 / 2) (min)^(-1)`

In terms of pressures, units of rate = `bar min ^(-1)`
therefore units of `R=bar min )^(-1)(b a r^(3/2)=bar^(-1/2) min^(-1)`