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Q.

Hydrogenation of vegetable oils at $25^{o} C$ reduce the pressure from 2 atm to 1 atm in 6.84 min. The mean rate of the reaction in mol $L^{- 1}$ per sec was found to be $X \times 10^{- 4}$ here X is ___

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Detailed Solution

$T = 25^{o} = 298 K$

Change in molarity $\frac{n}{v} = \frac{Δ P}{R T} = \frac{1}{0.0821 \times 298} = 0.041$

Rate of reaction $= \frac{0.04}{6.84 \times 60} = 1 \times 10^{- 4} m o l / L s$

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