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

Answer

B

Explanation

`underset(1)(X_(2))(g)hArr2XC(g)`
`-(beta_(e))/(2)" "beta_(e)`
Total number of moles at equlibrium
`implies1-(beta_(e))/(2)+beta_(e)implies1+(beta_(e))/(2)`
`K_(P)=((P_(x))^(2))/(P_(x_(2)))`
`=((((beta_(e)xx2)/(1+(beta_(e))/(2)))^(2))/((1-(beta_(e))/(2))xx2))/(1+(beta_(e))/(2))=(2beta_(e)^(2))/(1-(beta_(e^(2)))/(4))impliesK_(p)=(8beta_(e)^(2))/(4-beta_(e)^(2))`