Consider the dissociation of aqueous HCN at 25°C: HCN(aq) --> H + (aq) + CN – (a
ID: 587083 • Letter: C
Question
Consider the dissociation of aqueous HCN at 25°C:
HCN(aq) --> H+(aq) + CN–(aq) K = 6.2×10–10 .
a) Compute G° 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 the species?
c) Suppose 0.040 mol of HCN is dissolved in the same solution without appreciably increasing the volume. Compute the derivative dG/d for the system before it has a chance to re-establish equilibrium. Use this result to predict the spontaneous direction of reaction. (Justify the answer.)
d) Based on Le Chatelier's principle, which is the spontaneous direction of reaction after addition of the HCN? Explain qualitatively.
e) What are the new concentrations of all species after equilibrium is re-established?
Explanation / 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 is molecular 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 is molecular, then H+ and CN- should be produced
d)
First, let us state the Le Chatelier principle which deals with changes in an equilibrium:
The statement is as follows:
If any equilibrium is disturbed, that is, change in conditions such as P,T, concentration, partial pressure, etc.., the system will counterbalance such change in order to favour the system's equilbirium.
In this cas,e if we add more HCN, then H+ and CN- must increases since 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