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The decomposition of dimethyl ether lead...

The decomposition of dimethyl ether leads to the formation of `CH_(4),H_(2),` and `CO` and the reaction rate is given by
Rate `=k[CH_(3)OCH_(3)]^(3//2)`
The rate of reaction is followed by increase in the pressure in a closed vessel , so the rate can also be expressed in terms of the partial pressure of dimethyl either, `i.e., `
Rate `=k[p_(CH_(3)OCH_(3))]^(3//2)`
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?

Text Solution

Verified by Experts

If pressure is measured in bar and time in minutes , then
Unit of rate = bar `"min"^(-1)`
Rate = `k(P_(CH_(3) OCH_(3)))^((3)/(2))`
`implies k = ("Rate")/((P_(CH_(3)OCH_(3)))^((3)/(2)))`
Therefore , unit of rate constants `(k) = ("bar min"^(-1))/("bar"^((3)/(2)))`
= `"bar"^((-1)/(2)) "min"^(-1)`
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The decomposition of dimethyl ether leads to the formation of CH_(4) , H_(2) and CO and the reaction rate is given by the expression: rate = k[CH_(3)COOH_(3)]l^(3//2) The rate of reaction is followed by increase in pressure in a close vessel and the rate can also be expressed in terms of partial pressure of dimethyl ether: rate = k[CH_(3)OCH_(3)]^(3//2) The rate of reaction is followed by increase in a close vessel and the rate can also be expressed in terms of partial pressure of dimethyl ether: rate = k[pCH_(3)OCH_(3)]^(3//2) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant?

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