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For the following question , enter the correct numerical value, ( in decimal - notation , truncated / rounded - off to the second decimal place , e.g., 6*25, 7*00 -0*33, 30*27, -127*30) using the mouse and theonscreenvirtual nemeric keypad in the placedesigned to enter the answer. The approach of the following equilibrium was observed kinetically from both direction : PtCl_(4)^(2-) + H_(2)O hArr Pt (H_(2)O) hArr Pt (H_(2)O) Cl_(3)^(-) + Cl^(-) At25^(@)C , it was found that(d[PtCl_(4)^(-)])/dt=(3*90 xx10^(-5)s^(-1) ) [PtCl_(4)^(2-)] - (2*1 xx 10^(-3) mol^(-1)s^(-1)) [Pt(H_(2)O ) Cl_(3)^(-)] The value of the equilibrium constant when fourth Cl^(-) ion is complexed , i.e., for the step involving backward reaction will be .......... |
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Answer» For backward reaction, `((dx)/(dt))_(B) = k_(b) [ Pt (H_(2)O)Cl_(3)^(-)][Cl^(-)]` NET ` (dx)/(dt) = k_(f) [ Pt Cl_(4)^(2-)] -k_(b) [Pt (H_(2)O)Cl_(3)^(-)][Cl^(-)]` Comparingwith the given expression ` k_(f) = 3*90 xx 10^(-5) s^(-1)` and `k_(b) = 2*1 xx 10^(-3) L " mol "^(-1)s^(-1)` ` :. K_("forward ") =1/K_("forward") = 1/(1*86 xx 10^(-2))=53*85` |
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