Question

The decomposition of dimethyl ether leads to the formation of $CH_{4},H_{2}$ and $\text{CO}$ and the reaction rate is given by $rate=k\left[C H_{3} O C H_{3}\right]^{\frac{3}{2}}$

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. $rate=k\left(P_{C H_{3} O C H_{3}}\right)^{\frac{3}{2}}$ ,

If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

• A. $\text{bar}^{1 / 2} \, min$
• B. $\text{bar}^{3 / 2} \, min^{- 1}$
• C. $\text{bar}^{- 1 / 2} \, min^{- 1}$
• D. $\text{bar min}^{- 1}$

Answer 1

Answer: (C)

As $rate=k\left[C H_{3} O C H_{3}\right]^{\frac{3}{2}}$

bar/min $=k\left(b a r\right)^{\frac{3}{2}}$

Unit of $k=bar^{- \frac{1}{2}} \, m i n^{- 1}$