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The reaction A(g)+2B(g) rarrC(g)+D(g) i...

The reaction `A(g)+2B(g) rarrC(g)+D(g)` is an elementary process In an experiment the initial partial pressure of A and B are `p_(A) =0.60` and `p_(B) =0.80` atm when `p_(c )=0.2` atm the rate of the reaction relative to the initial rate is

A

`1//48`

B

`1//24`

C

`9//16`

D

`1//6`

Text Solution

Verified by Experts

The correct Answer is:
D
`r_(1) = k[A][B]^(2) = k[0.6][0.80]^(2)`. The reaction involved is
`{:(A,+,2B,rarr,C,+,D),(0.6-0.2,,0.8-0.4,,0.2,,0.2),(0.4,,0.4,,0.2,,0.2):}`
So, `r_(2) = k(0.4)(0.4)^(2)`. Comparing the rates, we get `(r_(2))/(r_(1)) = ((0.4)(0.4)^(2))/((0.6)(0.8)^(2)) = (1)/(6)`
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