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 incorrect statement among the following for this reaction, is

Options

A. Decrease in the total pressure will result in formation of more moles of gaseous X
B. At the start of the reaction, dissociation of gaseous `X_(2)` takes place spontaneously
C. `beta._("equilibrium ")= 0*7`
D. `K_(c) lt1`

Answer

C

Explanation

(a) If the pressure on the system is decreased , equilibrium will shift in the direction in which pressure increases , i.e., in which the number of moles is more , i.e, forward direction. Hence, (a) is correct .
(b) At the start of the reaction , `Q lt V` Hence, reaction will proceed in the forward direction . So (b) is also correct.
(c) If `beta_(eq.) = 0*7 , " then " K = (8 xx (0*7)^(2))/(4 - (0*7)^(2) ) gt 1 `
`Delta G^(@) =-RT "In" K_(p) :. Delta G^(@) "will be" - "ve but we are given that " Delta G^(@) " is " +" ve which can be so if " K_(p) lt 1. " Hence " beta_(eq.) != 0*7 , " i.e, (c) is incorrect . "`
(d) `K_(p) = K_(c) (RT)^(Deltan_(g))= K_(c) (RT) or K_(c) = K_(p)/(RT)`.
Thus, `K_(c) lt K_(p)`
As `K_(p) lt 1, "therefore " K_(c) lt 1 . ` Hence, (d) is also correct.