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The decomposition of dimethyl ether is a...

The decomposition of dimethyl ether is a fractional order reaction. The rate of reaction is givenby 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 constant?

A

bar `"min"^(-1), "bar"^(2) "min"^(-1)`

B

bar `"min"^(-1), "bar"^(2), "min"^(-1)`

C

`"bar"^(-1//2) "min"^(-1),"bar"^(2),"min"^(-1)`

D

bar `"min"^(-1), "bar"^(1//2) "min"^(-1)`

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The correct Answer is:
To solve the problem, we need to determine the units of the rate of reaction and the rate constant (k) based on the given rate law for the decomposition of dimethyl ether. ### Step 1: Understand the Rate of Reaction The rate of a reaction is defined as the change in concentration (or pressure in this case) over time. The general formula for the rate can be expressed as: \[ \text{Rate} = \frac{\Delta P}{\Delta t} \] where \(\Delta P\) is the change in pressure and \(\Delta t\) is the change in time. ### Step 2: Determine the Units of Rate Given that pressure is measured in bar and time is measured in minutes, we can express the units of rate as: \[ \text{Units of Rate} = \frac{\text{Pressure (bar)}}{\text{Time (minutes)}} = \text{bar} \cdot \text{min}^{-1} \] ### Step 3: Analyze the Rate Law The rate law provided is: \[ \text{Rate} = k \cdot (P)^{\frac{3}{2}} \] where \(P\) is the pressure of dimethyl ether. ### Step 4: Rearranging for the Rate Constant (k) From the rate law, we can express the rate constant \(k\) as: \[ k = \frac{\text{Rate}}{(P)^{\frac{3}{2}}} \] ### Step 5: Substitute the Units into the Equation Now, we substitute the units of rate and pressure into the equation for \(k\): - The units of rate are \(\text{bar} \cdot \text{min}^{-1}\). - The units of pressure are \(\text{bar}\). Thus, we have: \[ k = \frac{\text{bar} \cdot \text{min}^{-1}}{(\text{bar})^{\frac{3}{2}}} \] ### Step 6: Simplifying the Units of k Now, we simplify the units of \(k\): \[ k = \frac{\text{bar} \cdot \text{min}^{-1}}{\text{bar}^{\frac{3}{2}}} = \text{bar}^{1 - \frac{3}{2}} \cdot \text{min}^{-1} = \text{bar}^{-\frac{1}{2}} \cdot \text{min}^{-1} \] ### Final Result The units of the rate of reaction and the rate constant are: - Units of Rate: \(\text{bar} \cdot \text{min}^{-1}\) - Units of Rate Constant (k): \(\text{bar}^{-\frac{1}{2}} \cdot \text{min}^{-1}\)

To solve the problem, we need to determine the units of the rate of reaction and the rate constant (k) based on the given rate law for the decomposition of dimethyl ether. ### Step 1: Understand the Rate of Reaction The rate of a reaction is defined as the change in concentration (or pressure in this case) over time. The general formula for the rate can be expressed as: \[ \text{Rate} = \frac{\Delta P}{\Delta t} \] where \(\Delta P\) is the change in pressure and \(\Delta t\) is the change in time. ...
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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 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 ?

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