Consider the dissociation of a
Consider the dissociation of aqueous HCN at 25°C:HCN(aq) –> H+(aq) + CN–(aq) K =6.2×10–10 .
a) Compute ΔG° at 25°C.
b) If 0.120 mol of HCN is dissolved to make 200. mL of solution,then what are the equilibrium concentrations of all of thespecies?
c) Suppose 0.040 mol of HCN is dissolved in the same solutionwithout appreciably increasing the volume. Compute the derivativedG/dξ for the system before it has a chance tore-establish equilibrium. Use this result to predict thespontaneous direction of reaction. (Justify the answer.)
d) Based on Le Chatelier’s principle, which is the spontaneousdirection of reaction after addition of the HCN? Explainqualitatively.
e) What are the new concentrations of all species afterequilibrium is re-established?
Answer:
a)
dG = -RT*ln(K)
dG = -8.314*298*ln(6.2*10^-10)
dG = 52527.75J/mol
dG = 52.527 kJ/mol
b)
[HCN] = mol/V = 0.12/0.2 =0.6 M
find equilbirium
First, assume the acid:
HCN
to be HA, for simplicity, so it will ionize as follows:
HA <-> H+ + A-
where, H+ is the proton and A- the conjugate base, HA ismolecular acid
Ka = [H+][A-]/[HA]; by definition
initially
[H+] = 0
[A-] = 0
[HA] = M;
the change
initially
[H+] = + x
[A-] = + x
[HA] = – x
in equilbrirum
[H+] = 0 + x
[A-] = 0 + x
[HA] = M – x
substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
x^2 + Kax – M*Ka = 0
if M = 0.6 M; then
x^2 + (6.2*10^-10)x – 0.6*(6.2*10^-10) = 0
solve for x
x =1.928*10^-5
substitute
[H+] = 0 + 1.928*10^-5 = 1.928*10^-5M
[CN-] = 0 + 0.01434 = 1.928*10^-5 M
[HCN] = M – x = 0.6-1.928*10^-5= 0.59998 M
c)
dG/dE –> will favour spotnaneous process, since HCN ismolecular, then H+ and CN- should be produced
d)
First, let us state the Le Chatelier principle which deals withchanges in an equilibrium:
The statement is as follows:
If any equilibrium is disturbed, that is, change in conditionssuch as P,T, concentration, partial pressure, etc.., the systemwill counterbalance such change in order to favour the system’sequilbirium.
- Change in concentration
- Change in pressure
- Change in temperature
In this cas,e if we add more HCN, then H+ and CN- must increasessince HCN molecules is increasesd with direct addition
e)
total [HCN]initially = (0.12 + 0.04)/0.2 = 0.8
repeat step b:
x^2 + Kax – M*Ka = 0
if M = 0.6 M; then
x^2 + (6.2*10^-10)x – 0.8*(6.2*10^-10) = 0
solve for x
x =2.227*10^-5
substitute
[H+] = 0 + 2.227*10^-5 = 2.227*10^-5
[CN-] = 0 + 2.227*10^-5= 2.227*10^-5
[HCN] = M – x = 0.8-2.227*10^-5 0.79997M
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