Thermal decomposition of gaseous `X_(2)` to gaseous `X` at `298K` takes place according to the following equation:
`X(g)hArr2X(g)`
The standard reaction Gibbs energy `Delta_(r)G^(@)`, of this reaction is positive. At the start of the reaction, there is one mole of `X_(2)` and no `X`. As the reaction proceeds, the number of moles of `X` formed is given by `beta`. Thus `beta_("equilibrium")` is the number of moles of `X` formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
[Given, `R=0.083L` bar `K^(-1) mol^(-1)`)
The equilibrium constant `K_(p)` for this reaction at `298K`, in terms of `beta_("equilibrium")` is

Options

A. `(8(beta^(2))_("equilibrium"))/(2-beta_("equilibrium"))`
B. `(8(beta^(2))_("equilibrium"))/(4-beta_("equilibrium")^(2))`
C. `(4(beta^(2))_(equilibrium))/(2-beta_("equilibrium"))`
D. `(4(beta^(2))_("equilibrium"))/(4-beta^(2)_("equilibrium"))`

Answer

b

Explanation

`X_(2)(g) to 2X(g)`
`{:(t=0, 1,-1),(t=t_(eq), 1-(beta_(eq))/2, beta_(eq)),(px_(2)) =2(1-beta_(eq)/2)/(1+beta_(eq)/2) , p_(x) =2(beta_(eq)/(1+beta_(eq)/2))`
`K_(p) = (((2beta_(eq))/(1+beta_(eq)/2)^(2))/(2(1-beta_(eq)/2)/(1+beta_(eq)/2))) = (2beta_(eq)^(2))/(1-beta_(eq)^(2)/4) =(8beta_(eq)^(2))/(4-beta_(eq)^(2))`