Question

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 the onscreen virtual nemeric keypad in the place designed 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^(-)`
At `25^(@)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 `.........`.

Answer 1

Correct Answer - (53*85)
For forward reaction , ` ((dx)/(dt))_(f) = k_(f) [PtCl_(4)^(2-)]`
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^(-)]`
Comparing with 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`