The equilibrium constant for the reaction AgBr (s) --> <--Ag + (aq) + Br - (aq)

Question

The equilibrium constant for the reaction AgBr (s) --> <--Ag+ (aq) + Br- (aq) is the solubility productconstant, Ksp = 7.7 x 10^-13 at 25 °C. If [Ag+] = 1.0 x 10^-2 M and [Br-] = 1.0 x 10^-3 M, which of thefollowing statements is true?

a) the reaction is at equilibrium
b) <>G = 0
c) <>G = <>G °
d) <>G < 0 and a precipitate forms
e) <>G > 0 and a precipitate forms

please explain and show me step by step how to solve this problem,including the solution. Thanks in advance for helping :) !!!

Explanation / Answer

We Know that :    The given equation is :           AgBr(s) <-----> Ag+ (aq) + Br-(aq)           Ksp = 7.7 x 10^-13   [ Ag+ ] = 1.0 x 10^-2 M ; [ Br- ] = 1.0 x 10^-3 M         We Know thatif ionic product is greater than Ksp than the precipitationoccurs.        In order to calculatethe Greaction =   - RT lnK                                                                 = - 2.303 x 8.314 J / mol- K x 298 K x log ( 7.7 x 10-13  )                                                                 =   69117.846J / mol         G > 0 and precipitate forms.   option eis correct.

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