Consider the equilibrium `X_(2) + Y_(2) hArr P`. Find the stoichiometric coefficient of the P using the data given in the following table

Options

A. 1
B. 2
C. 3
D. `0.5`

Answer

B

Explanation

Given, `X_(2) + Y_(2) rarr ?P` Let the coefficient of P be x.
Now, we know that `K= ([P]^(x))/([X_(2)][Y_(2)])`
According to the data given in question, `K_(1)= ((2.52 xx 10^(-2))^(x))/((1.14 xx 10^(-2))xx (0.12 xx 10^(-2)))`
Also, `K_(2)= ((3.08 xx 10^(-2))^(x))/((0.92 xx 10^(-2)) xx (0.22 xx 10^(-2)))`
Now, substitute the values of x given in option one by one.
(a) `K_(1)= ((2.52 xx 10^(-2)))/((1.14 xx 10^(-2)) xx (0.12 xx 10^(-2)))=1842.10`
`K_(2)= ((3.08 xx 10^(-2)))/((0.92 xx 10^(-2)) xx (0.22 xx 10^(-2)))=1521.73`
`because K_(1) ne K_(2)`, thus option (a) is incorrect.
(b) `K_(1)= ((2.52 xx 10^(-2))^(2))/((1.14 xx 10^(-2)) xx(0.12 xx 10^(-2)))=46.42`
`K_(2)= ((3.08 xx 10^(-2))^(2))/((0.92 xx 10^(-2)) xx (0.22 xx 10^(-2)))=46.36`
`because K_(1) ne K_(2)`
`therefore` Option (b) is correct