The decomposition of dimethyl ether leads to formation of CH_4 , H_2…

The correct answer is Option (1) → \(bar^{–1/2} min^{–1}\)

The unit of a rate constant follows from the rate law. For a gaseous reaction the rate is measured as change in pressure per unit time, so here the rate has the unit bar min⁻¹, and the concentration term is a pressure in bar.

    \( \displaystyle r = k\,[\ce{CH3OCH3}]^{3/2} \)

    \( \displaystyle k = \frac{r}{[\ce{CH3OCH3}]^{3/2}} \)

    \( \displaystyle \text{unit of } k = \frac{\mathrm{bar\,min^{-1}}}{\mathrm{bar^{3/2}}} = \mathrm{bar^{1-3/2}\,min^{-1}} = \mathrm{bar^{-1/2}\,min^{-1}} \)

So the rate constant has the unit bar^−1/2 min⁻¹.

Shortcut: For a reaction of order n, the unit of k is (pressure)¹⁻ⁿ time⁻¹. With n = 3/2 the power of bar is 1 − 3/2 = −1/2.

Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Intext Questions – “The decomposition of dimethyl ether leads to the formation of \( \ce{CH4} \), \( \ce{H2} \) and CO and the reaction rate is given by

    \( \text{Rate} = k\,[\ce{CH3OCH3}]^{3/2} \)

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.,

    \( \text{Rate} = k\,(p_{\ce{CH3OCH3}})^{3/2} \)

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

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: