A fristorderreaction is 40%complete in 50 minutes. Calculatethe value of the constant. In what timewill the reaction be 80 %complete ? (ii)K_(sp) " of" Ag_(2)CrO_(4) " is "1.1 xx 10^(-12) . Whatis the solubility ofAg_(2)CrO_(4) in 0.1MK_(2)CrO_(4)

A fristorderreaction is 40%complete in 50 minutes. Calculatethe value

1.	A fristorderreaction is 40%complete in 50 minutes. Calculatethe value of the constant. In what timewill the reaction be 80 %complete ? (ii)K_(sp) " of" Ag_(2)CrO_(4) " is "1.1 xx 10^(-12) . Whatis the solubility ofAg_(2)CrO_(4) in 0.1MK_(2)CrO_(4)
Answer» Solution :(i) For the firstorder reation ` K = (( 2.303)/t) log ( a/(a-x))` When x= ` ( 40/100) a = 0.4 a ` t = 50 m ` thereforek = (( 2.303)/50) log (a/(a-0.4a))` `k = ((2.303)/50) log (1/0.6)` `0.010216min^(-1)` t = ?when x = 0.8a From aboveK = ` 0.010216 min^(-1)` ` thereforet = (2.303/(0.010216)) log (a/(a-0.8a))` ` (2.303/(0.010216)) log (1/0.2) = 157.58 min` The timeat WHICHTHE reaction will be 80%complete is 157.58 min. (ii) ` Ag_(2) CrO_(4)iff_(2x)2Ag^(+) + underset(x)(CrO_(4)^(2-)` x is the solubility of ` Ag_(2)CrO_(4) " in"0.1 M K_(2)CrO_(4)` ` K_(2)CrO_(4) IFF 2K^(+) + CrO_(4)^(2-)` ` [ Ag^(+)]= 2x ` ` [ CrO_(4)^(2-)] = ( x+ 0.1) = 0.1 ` ` K_(SP) = [Ag^(+)]^(2) [ CrO_(4)^(2-)]` `1.1xx 10^(-12) = (2x)^(2) (0.1)` ` 1.1 xx 10^(-12) = 0.4 x^(2)` ` x^(2) = ( 1.1 xx 10^(-12))/(0.4)` ` x = sqrt((1.1 xx 10^(-12))/(0.4))` ` x = sqrt(2.75 xx 10^(-12))` ` x = 1.65 xx 10^(-6) M`