Question

The decomposition of dimethyl ether leads to the formation of $C{{H}_{4}}$ , ${{H}_{2}}$ and $CO$ and the reaction rate is given by
Rate = $k{{\left[ C{{H}_{3}}OC{{H}_{3}} \right]}^{\dfrac{3}{2}}}$
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 = $k{{\left( {{p}_{C{{H}_{3}}OC{{H}_{3}}}} \right)}^{\dfrac{3}{2}}}$
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Answer 1

The rate value of rate is either change in concentration per unit time or change in partial pressure per unit time. Here, the unit given is of pressure, thus the rate will be defined as change in partial pressure per unit time. Also, the unit of rate constant can be found out by making it the subject of the equation.

Complete answer: As shown here, the units given are bar and minute, thus we have to calculate rate in terms of partial pressure per unit time.
Therefore, Rate = $\dfrac{bar}{\min }$ , which can also be written as
$\Rightarrow$ $rate= bar\times {{\min }^{-1}}$
Now, as said in the question, the rate can also be expressed in terms of partial pressure of dimethyl ether,
$Rate = k{{\left( {{p}_{C{{H}_{3}}OC{{H}_{3}}}} \right)}^{\dfrac{3}{2}}}$
Thus, the unit of rate expressed in terms of partial pressure is bar.
Therefore, $Rate = k{{\left( bar \right)}^{\dfrac{3}{2}}}$ .
From this, we can find the unit of rate constant by making it the subject. Thus, rate constant
k = $\dfrac{bar\times {{\min }^{-1}}}{ba{{r}^{\dfrac{3}{2}}}}$ , ($bar\times {{\min }^{-1}}$ is the unit of rate)
Therefore, $\Rightarrow$ $k =ba{{r}^{-\dfrac{1}{2}}}\times {{\min }^{-1}}$ .

Note:
If the units given in the question are in terms of concentration, the units of rate and rate constant will vary accordingly. In these kinds of questions, the compound for which the rate is given does not matter as we only have to find out the unit of rate and rate constant.