Question

Thermal decomposition of gaseous `X_(2)` to gaseous X at 298 K takes place according to following equation
`X_(2) (g) hArr 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 proceeds the number of moles of X formed is given by `beta`. Thus `beta_("equilibrium")` is the the number of moles of X formed at equilibrium . The reaction is carried out a constant total pressure of 2 bar. Consider the gases to behave ideally.
`("Given" : R =0.083 L " bar " K^(-1) " mol"^(-1))`
The incorrect statement among the following for this reaction is
A. Decrease in the total pressure will result in formation of more moles of gaseous X
B. At the start of the reaction , dissocition of gaseous `X_(2)` takes place spontaneously
C. `beta_("equilibrium") =0.7`
D. `K_(c)lt 7`

Answer 1

Correct Answer - C
(C) It is incorrect statement among the
(a) If the pressure on the system is decreased , the equilibrium will shift in the direction in which pressure increases i.e.,increases in no. of moles takes place i.e., in forward idrection .
(b) At the start of the reaction `Q lt K` thus the reaction will proceed in the forward direction i.e., reaction is spontaneous.
(C) `" if "b_(eq) =0.7 " then " K_(p) =(8xx(0.7)^(2))/(4-(0.7)^(2)) gt 1`
`DeltaG^(@) =- RT" In " K_(p) SO, DeltaG^(@) =- " ve but ginen " DeltaG^(@) = + " ve so, " K_(p) " should be less than " 1 " Hence " beta_(eq) ne 0.7`
(d)`K_(p) =K_(c) (RT)^(Dn).`
`K_(c) lt K_(p) ne (":." RT gt 1)`
`" If " K_(p) lt1 " then " K_(c) lt 1`