The decomposition of dimethyl ether leads to the formation of CH4,H2 and textCO and the reaction rate is given by rate=k[C H3 O C H3](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(PC H3 O C H3)(3/2) , If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

Question

The decomposition of dimethyl ether leads to the formation of CH4,H2 and textCO and the reaction rate is given by rate=k[C H3 O C H3](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(PC H3 O C H3)(3/2) , If the pressure is measured in bar and time in minutes, then the unit of rate constants is: (A)  textbar1 / 2 min (B)  textbar3 / 2 min- 1 (C)  textbar- 1 / 2 min- 1 (D)  textbar min- 1

Tardigrade Admin · Accepted Answer

As rate=k[C H3 O C H3](3/2) bar/min =k(b a r)(3/2) Unit of k=bar- (1/2) m i n- 1

$bar^{1/2} min$

$bar^{3/2} mi n^{- 1}$

$bar^{- 1/2} mi n^{- 1}$

$bar min^{- 1}$

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

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

Unit of $k = ba r^{- \frac{1}{2}} mi n^{- 1}$

bar1/2min

bar3/2min−1

bar−1/2min−1

bar min−1

As rate=k[CH3​OCH3​]23​ bar/min =k(bar)23​ Unit of k=bar−21​min−1

$bar^{1/2} min$

$bar^{3/2} mi n^{- 1}$

$bar^{- 1/2} mi n^{- 1}$

$bar min^{- 1}$

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

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

Unit of $k = ba r^{- \frac{1}{2}} mi n^{- 1}$