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

Answer

B

Explanation

`{:(,X_(2) (g) ,hArr,2X(g)),("At t=0,",1,,0):}`
At eqm. `(1-(beta_(eq))/(2)) " "beta_(eq)`
`K_(p) =((pX)^(2))/((pX_(2)))`
`P_(X) =((beta_(eq))/(1-(beta_(eq))/(2)+beta_(eq))) P_("total") =((beta_(eq))/((1-beta_(eq))/(2)))P_("total")`
`PX_(2)=((1-(beta_(eq))/(2))/(1+(beta_(eq))/(2)))P_("total")`
`K_(p) =[((beta_(eq))/(1+(beta_(eq))/(2))) P_("total") ]^(2)/(((1-(beta_(eq))/(2))/(1+(beta_(eq))/(1))) P_("total"))=((beta_(eq)^(2))/(1-(beta_(eq)^(2))/(4)))p_("total")`
`=((4beta_(eq)^(2))/(4-beta_(eq)^(2))) P_("total") =((4beta_(eq)^(2))/(4-beta_(eq)^(2))) xx2 =(8beta_(eq)^(2))/(4-beta_(eq)^(2))`