Saved Bookmarks
| 1. |
Thermal decomposition of gaseous X_(2) to gaseousX at 298 K takes place according to the 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 positive . At the start of the reaction , there is one mole of X_(2) andno. As the reaction proceeds, the number of moles of X formed is given by beta." Thus " beta _(equilbrium )is the number ofmoles 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* 0833 " L bar " K^(-1)mol^(-1)). Theequilibrium constantK_(p) for this reaction at 298 K, in terms ofbeta _(equilibrium ) , is |
|
Answer» `(8 beta_("equilibrium ")^(2))/(2 - beta _("equilibrium")` As the number of moles of X formed at equilibrium `= beta _(eq).` ` :. " No. of moles of " X_(2)" reacted at equilibrium " = (beta_(eq))/2 ` `{:(,X_(2)(g) ,hArr,2X(g)),(" INTIAL moles" ,1,,0),("Moles at eqm",1-beta_(eq)/2,,beta_(eq)/2):}` Total no. of moles at equilibrium `= 1- (beta_(eq.))/2 + beta_(eq.) =1+ beta_(eq.)/2` `:. p_(X_(2)) = (1-beta_(eq.)/2)/(1+beta_(eq)/2)xx2, p_(X) = beta_(eq.)/(1+(beta_eq)/2)xx2, p_(X) = beta_(eq.)/(1+(beta_(eq))/2)xx2` (as `P_(total ) = 2 "BAR")` ` :. K_(P) = (p_(X)^(2))/p_(X)^(2) = ((2beta_(eq.))/(1+(beta_(eq.))/2))^(2) xx ((1+(beta_(eq.))/2))/(2(1-(beta_(eq.))/2))` `=(4beta_(eq.)^(2))/(1+beta_(eq.)/2)xx1/(2(1-beta_(eq)/2)) ` ` (4beta_(eq).)/(2(1-beta_(eq)^(2)/4 ))= (2beta_(eq)^(2))/((4-beta _(eq)^(2))/4)=(8beta_(eq.)^(2))/(4-beta_(eq.)^(2))` |
|