Thermal decomposition of gaseous `X_(2)` to gaseous X at 298 K takes place according to the following equations.
`X_(2)(g) + 2X(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 procees, the number of moles of X formed is given by `beta`. Thus, `beta_("equlibrium")` 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=0.083 L bar `K^(-1) mol^(-1)`).
The equilibrium constant `K_(p)` for this reaction at 298 K, in terms of `beta_("equilibrium")`, 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)=1`

Answer

a

Explanation

It is true statement `X_(2)(g) + 2X(g)`
If P is decreased then reaction will move in forward direction according to Le-Chatelier's principle.
(B) It is true statement
At start of reaction Q=0
`therefore DeltaG = DeltaG^(@) + RT ln Q`
If Q=0
Then `DeltaG = -ve`, Reaction is spontaneous
(D) It is true statement
`DeltaG^(@) = +ve`
`-RT ln K_(p) = DeltaG^(@)`
`rArr K_(p) lt 1` Since, `K_(c) lt K_(p) rArr K_( c)` is also less than 1 .