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Question

Chemistry Question on Chemical Kinetics

The decomposition of dimethyl ether leads to the formation of CH4,H2CH_{4},H_{2} and CO\text{CO} and the reaction rate is given by rate=k[CH3OCH3]32rate=k\left[C H_{3} O C H_{3}\right]^{\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. rate=k(PCH3OCH3)32rate=k\left(P_{C H_{3} O C H_{3}}\right)^{\frac{3}{2}} , If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

A

bar1/2min\text{bar}^{1 / 2} \, min

B

bar3/2min1\text{bar}^{3 / 2} \, min^{- 1}

C

bar1/2min1\text{bar}^{- 1 / 2} \, min^{- 1}

D

bar min1\text{bar min}^{- 1}

Answer

bar1/2min1\text{bar}^{- 1 / 2} \, min^{- 1}

Explanation

Solution

As rate=k[CH3OCH3]32rate=k\left[C H_{3} O C H_{3}\right]^{\frac{3}{2}} bar/min =k(bar)32=k\left(b a r\right)^{\frac{3}{2}} Unit of k=bar12min1k=bar^{- \frac{1}{2}} \, m i n^{- 1}