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$ I_2 + I^- \rightleftharpoons I^-_3 $ This reaction is set-up in aqueous medium. Initially we have 1 mole of $I_2$ and 0.5 mole of $I^-$ in 1 L flask. After equilibrium is reached, excess of $AgNO_3$ gave 0.25 mole of yellow ppt. Equilibrium constant is

(a) 1.33

(b) 2.66

(c) 2.00

(d) 3.00

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1 Answer

Answer: 1.33
 
The reaction given,
$I_2 + I^- \rightleftharpoons I^-_3$
1M__0.5M_________at t=0
(1-x)M (0.5-x)M x mol _______ at t=t'
 
Since, Moles of AgI = Moles of $I^-$ in solution
 
$\therefore [I^-]_{eqm} = 0.25 M$
0.5 - x = 0.2
$\therefore$ x = 0.25
 
Equilibrium constant of the reaction at equilibrium, $K = \frac{x}{(1-x)(0.5-x)}$
$K = \frac{0.25}{0.75 \times 0.25}$ = 1.33

 

answered Nov 20, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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