InterviewSolution
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InterviewSolution
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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 isA. Decrease in the total pressure will result in formation of more moles of gaseous XB. At the start of the reaction , dissocition of gaseous `X_(2)` takes place spontaneouslyC. `beta_("equilibrium") =0.7`D. `K_(c)lt 7` |
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Answer» 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` |
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