3- CN$(aq) + NH4+(aq) HCN(aq)+ NH3(aq) A solution is prepared with all reactants

Question

3- CN$(aq) + NH4+(aq) HCN(aq)+ NH3(aq) A solution is prepared with all reactants at 0.45 M and all products at 0.25 M. a) Calculate G° and K (at 298K). b) Calculate Q and G and indicate in what direction the reaction will proceed. c) Determine the concentrations once equilibrium is reached. d) What would be the effect of heating the solution at equilibrium? EXPLAIN your reasoning. -----------------------------------Give detail Show ALL WORK, UNITS and use ICE/reaction table and ML (no millimoles). Please be clear in how you get numbers for everything like the reaction table... I've asked this before and have gotten incomplete answers. SHOW UNITS whats units for Delta G/Q and so on Thank you :)

Explanation / Answer

CN-(aq) + NH4+(aq) HCN(aq)+ NH3(aq)

a.

K = [HCN][NH3] / ([CN-] [NH4+])

     = Kw [HCN][NH3] / ([CN-][H+][HO-] [NH4+])

     = Kw / (Ka,HCN · Kb, NH3 )

      = 1x10-14 / (6.17x10-10 x 1.76x10-5)

       = 0.921      (adimensional value)

dGo = - RTlnK

       = - 0.008314 kJ/molK x 298Kx ln0.921

        = + 0.204 kJ/mol

b.

Q = [HCN][NH3] / ([CN-] [NH4+]

    = 0.25x0.25/(0.45x0.45)

    = 0.308   (adimensional value)

dG = dGo + RTlnQ    and also   

dG = RT ln (Q/K)

      = 0.008314 kJ/molK x 298 x ln (0.308/0.921)

      = -2.71 kJ/mol

The reaction goes from left to right.

c.

CN-(aq) + NH4+(aq) HCN(aq)+ NH3(aq)

0.45              0.45                  0.25              0.25    initial conc.

0.45-x           0.45-x              0.25+x         0.25+x

K= 0.921 = (0.25+x)2/(0.45-x)2

(0.25+x)/(0.45-x) = 0.960

X = x = 0.093

[CN-] = [NH4+] = 0.343 M

[HCN]=[HN3]= 0.357 M

d.

The equilibrium constant will change if the temperature is changed.

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