Hydrogenation of vegetable ghee at 25° C reduces pressure of H 2 from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is -

Question

Hydrogenation of vegetable ghee at 25° C reduces pressure of H 2 from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is - (A) 1.09 × 10-6 (B) 1.09 × 10-5 (C) 1.09 × 10-7 (D) 1.09 × 10-9

Tardigrade Admin · Accepted Answer

The change in molarity =( n / V )=(Δ P / R T ) =(0.8/0.0821 × 298)=0.0327 rate of reaction =(0.0327/50 × 60) =1.09 × 10-5 mol litre -1 sec -1

Q. Hydrogenation of vegetable ghee at $2 5^{\circ} C$ reduces pressure of $H_{2}$ from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is -

$1.09 \times 1 0^{- 6}$

$1.09 \times 1 0^{- 5}$

$1.09 \times 1 0^{- 7}$

$1.09 \times 1 0^{- 9}$

The change in molarity $= \frac{n}{V} = \frac{Δ P}{RT}$

$= \frac{0.8}{0.0821 \times 298} = 0.0327$

rate of reaction $= \frac{0.0327}{50 \times 60}$

$= 1.09 \times 1 0^{- 5} m o l$ litre $^{- 1} sec^{- 1}$

Q. Hydrogenation of vegetable ghee at 25∘C reduces pressure of H2​ from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is -

1.09×10−6

1.09×10−5

1.09×10−7

1.09×10−9