Question

CN^-(aq)+NH₄(aq) <-> HCN(aq) + NH₃(aq)

A solution is prepaired with all reactants at 0.45M and all products at 0.25M

a) Calculate (delta)G^o and K (at 298K)

b) Calculate Q and G and indicates in what direction the reaction will proceed

c) Determine the concentrations once equilbrium is reached

d) What would be the effect of heating the solution at equilbrium.

Answer 1

A) CN-(aq)+NH4(aq) ↔ HCN(aq) + NH3(aq) (neutralization )

Can be divided into two reactions,

At 298K

CN-(aq) +H+ ↔ HCN(aq) ka=6.17*10^-10

Kb=kw/ka=10^-14/6.17*10^-10=0.162*10^-4

Or, Kb=0.162*10^-4 (equilibrium constant )=[HCN]/[CN-]

And

NH4+ (aq) ↔ NH3(aq) + H+   ka=5.6*10^-10)=[NH3]/[NH4+]

So for net rxn ,

CN-(aq)+NH4(aq) ↔ HCN(aq) + NH3(aq)

Keq=kb*ka=(0.162*10^-4)( ka=5.6*10^-10)=[HCN][NH3]/[CN-][NH4+]

Keq=0.907*10^-14

∆Go (at equilibrium)=-RT ln Keq=-(8.314 J/K mol)(298K) ln (0.907*10^-14)

                                                             =-2477.6 J/mol *2.303 log (0.907*10^-14)

                                                              =-5705.91 J/mol (-14.013)

                                                         ∆Go     =79956.96 j/mol=799.57 KJ/mol

B)Q=[HCN][NH3]/[CN-][NH4+]=(0.25M)(0.25M)/(0.45M)(0.45M)=0.0625/0.2025=0.3086

Q=0.3086

∆G=∆Go+RT ln Q=799.57 KJ/mol +(8.314 J/K mol)(298K) ln (0.3086)

                                 =799.57 KJ/mol + 2477.6 J/mol *2.303 log (0.3086)

                                  =799.57 KJ/mol + 5705.91 J/mol (-0.51)= 799.57 KJ/mol -2910.01 J/mol

                                                                                                       =799.57 KJ/mol -2.910 k J/mol

                                                                                                    ∆G    =796.66 kj/mol

Q>Keq so reverse rxn is favoured

C)ICE table

CN-(aq)+NH4(aq) ↔ HCN(aq) + NH3(aq)

	[CN-]	[NH4+]	[HCN]	[NH3]
Initial	0.45	0.45	0.25	0.25
change	-x	-x	+x	+x
equilibrium	0.45-x	0.45-x	0.25+x		0.25+x

Keq=[HCN][NH3]/[CN-][NH4+]

Or, 0.907*10^-14=(0.25-x)^2/(0.45-x)^2

Or, 0.907*10^-14=0.0625 +x^2-0.5x/0.2025 +x^2-0.9x

Ignoring x^2 terms x<<< than 0.45,0.25 .(weak acid and base dissociation is very low)

0.907*10^-14=0.0625 -0.5x/0.2025 -0.9x

0.1837*10^-14-0.8163*10^-14x=0.0625 -0.5x

Or, 0.1837*10^-14-0.0625=-0.5x+0.8163*10^-14x

Or, -0.0625=-0.5x

X=0.125M

Equilibrium conc of all species=0.125M

D)heating the solution at equilibrium would shift equilibrium in forward direction, as the rxn will be forward favoured