1.

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) " andno. " 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

Answer»

Decrease in the total pressure will result in formation of more moles of GASEOUS X
At the START of thereaction , dissocition of gaseous `X_(2)` takes place spontaneously
`beta_("equilibrium") =0.7`
`K_(c)lt 7`

Solution :(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`


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