The results given in the below table were obtained during kinetic studies of the following reaction : `2A+BtoC+D`

X and Y in the given table are respectively :

Options

A. 0.4, 0.4
B. 0.3, 0.4
C. 0.4, 0.3
D. 0.3, 0.3

Explanation

`2A+BtoC+D`
Exp. (I) `6xx10^(-3)=K(0.1)^(p)(0.1)^(q)`
(II) `2.4xx10^(-2)=K(0.1)^(p)(0.2)^(q)`
(III) `1.2xx10^(-2)=K(0.2)^(p)(0.1)^(q)`
`(exp(I))/(exp(II))" "(1)/(4)=((1)/(2))^(q)impliesq=2`
`(Exp(I))/(Exp(III))" "(1)/(2)=((1)/(2))^(p)impliesp=1`
exp. (I)+exp (IV)
`(0.6xx10^(-2))/(7.2xx10^(-2))=((0.1)/(x))^(1).[(0.1)/(0.2)]^(2)`
`(1)/(12)=(0.1)/(x)-(1)/(4)`
[x]=0.3
`exp(I)-:exp(V)`
`(2.88xx10^(-1))/(0.6xx10^(-2))=((0.1)/(0.3))^(1)xx((0.1)/(y))^(2)`
`(1)/(48)=(1)/(3)xx(10^(-2))/(y^(2))impliesy^(2)=0.16`
y=0.4