Hydrogenation of vegetable ghee at $25^{o}$ C reduces the pressure of $H_{2}$ from 2 atm to 1.2 atm in 50 minute. Calculate the rate of reaction in terms of change:
(a) In pressure per minute.
(b) In molarity per second.

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Solution

$P_{i} (H_{2}) = 2 a t m$
$P_{f} (H_{2}) = 1.2 a t m$
$Δ t$ =50 minutes
$a . I n] p r e s s u r e p e r m i n u t e R a t e o f f o r m a t i o n = \frac{d P}{d t}$
= $- \frac{(1.2 - 2)}{50}$
= $\frac{0.8}{50} a t m m i n^{- 1}$
Rate of reaction= $0.016 a t m m i n^{- 1}$ .
b. In molarity per second
PV=nRT
$∴ \frac{P}{R T} \frac{n}{V} = M o l a r i t y$
$∴ M_{1} = \frac{P_{1}}{R T}$
= $\frac{2 a t m}{0.0821 \times 298}$
= $0.08 M$
$∴ M_{1} = \frac{P_{2}}{R T}$
= $\frac{1.2 a t m}{0.0821 \times 298}$
= $0.05 M$
$R a t e o f R e a c t i o n = \frac{- d [M]}{d t}$
= $\frac{0.05 - 0.08}{50 \times 60} m o l a r i t y s e c^{- 1}$
= $1 \times 10^{- 5} M s e c^{- 1}$

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