The decomposition of dimethyl ether leads to the formation of C H 4 , H 2 and CO and the reaction rate is given by r a t e = k C H 3 O C H 3 3 2 The rate of reaction is followed by increase in pressur

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 $r a t e = k {[C H_{3} O C H_{3}]}^{\frac{3}{2}}$

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. $r a t e = k {(P_{C H_{3} O C H_{3}})}^{\frac{3}{2}}$ ,

If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

Options:

Answer:

C

Solution:

As $r a t e = k {[C H_{3} O C H_{3}]}^{\frac{3}{2}}$

bar/min $= k {(b a r)}^{\frac{3}{2}}$

Unit of $k = b a r^{- \frac{1}{2}} {m i n}^{- 1}$

TestHub · Accepted Answer

As r a t e = k C H 3 O C H 3 3 2 bar/min = k b a r 3 2 Unit of k = b a r - 1 2 m i n - 1

Chemistry - Chemical Kinetics Question with Solution | TestHub

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