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Order of Reaction

The decomposi...

Question

The decomposition of dimethyl ether leads to the formation of $C H_{4}$ , $H_{2}$ and $C O$ and the reaction rate is given by:

Rate $= k [C H_{3} O C H_{3}]^{\frac{3}{2}}$ . The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = $k (p_{C H_{3} O C H_{3}})^{\frac{3}{2}}$ . If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

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Solution

Unit of rate $= b a r m i n^{- 1}$

Unit of k $= \frac{R a t e}{[(P_{C H_{3} O C H_{3}})]^{\frac{3}{2}}}$

$= \frac{b a r m i n^{- 1}}{(b a r)^{\frac{3}{2}}} = b a r^{- \frac{1}{2}} m i n^{- 1} = \frac{1}{\sqrt{b a r} m i n}$

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Similar questions

C H_{4}, H_{2}

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

R a t e = k (P_{C H_{3} O C H_{3}})^{\frac{3}{2}}

C H_{4}, H_{2}

K [C H_{3} O C H_{3}]^{\frac{3}{2}}

K (p_{C} H_{3} O C H_{3})^{\frac{3}{2}}

R a t e = k (p_{C H_{3} O C H_{3}})^{3 / 2}

C H_{3} O C H_{3} \to C H_{4} + H_{2} + C O

6.72 \times 10^{- 3} m i n^{- 1}

0.4 a t m

C H_{3} O C H_{3 (g)} \to C H_{4 (g)} + C O_{(g)} + H_{2 (g)}

4.78 \times 10^{- 3} m i n^{- 1}

4.5

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