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

Options

A. 1/48
B. 1/24
C. 9/16
D. 1/6

Answer

D

Explanation

`r_(1)=k[A][B]^(2)=k[0.6][0.80]^(2)` the reaction involved is
A + 2B `rarr` C+D
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`