9.4 The response of equilibrium to temperature (continued)
In the last lecture, we studied how the chemical equilibrium responds to the
variation of pressure and temperature. At the end, we derived the van’t Hoff equation:
d ln K
dT
=
∆r H θ
RT 2
(9.4.1)
For the exothermic reaction, ∆rHө is negative, therefore d(lnK)/dT is negative,
which means that the increase of temperature will have a negative impact on the chemical
equilibrium. This is in accord with the analysis based on Le Chatelier’s principle.
There is another expression of the van’t Hoff equation. We can try to find a way
to eliminate the T2 from the equation. Such a task can be accomplished by the following
algebra: We calculate the derivative of 1/T with respect to T
d(1/T)/dT = -1/T2
d ln K
1
−T 2d ( )
T
=
∆r H θ
RT 2
Rearrange the above equation, we get
d ln K
1
d( )
T
∆r H θ
= −
R
(9.4.2)
With this type of differential equation, we can easily get an analytical solution by
integrating it over certain temperature interval:
Please keep it in mind that the standard enthalpy of formation is also a
function of temperature!!!
Therefore, when we integrate the above equation from T1 to T2, we should assume
that ∆rHө is constant within that interval.
∫
K2
K1
so
T2
d ln( K ) =
∫−
T1
ln(K2) – ln(K1) =
∆H 1
d( )
R T
− ∆r H θ 1
(
R
T2
−
1
)
T1
(9.4.3);
From equation 9.4.3, it is obvious that if we know the standard reaction enthalpy
and the chemical equilibrium constant at one temperature, we would be able to calculate
the chemical equilibrium under another temperature.
Example 9.4a, The Haber reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
At 298 K degree, equilibrium constant K = 6.1x105. The standard enthalpy of
formation for NH3 equals -46.1 kJ mol-1. What is the equilibrium constant at 500K?
Answer:
First, calculate the standard reaction enthalpy ∆rHө
∆rHө = 2*∆fHө(NH3) - 3* ∆fHө(H2) - ∆fHө(N2)
= 2*(-46.1) – 3*0 - 1*0
= - 92.2 kJ mol-1
then ln(K2) – ln(6.1*105) =
−1
*(-92.2*1000 J mol-1) (1/500 – 1/298)
8.3145
ln(K2) = -1.71
K2 = 0.18
Example 9.4b, The equilibrium constant of the reaction
2SO2(g) + O2(g) ↔ 2SO3(g) (sulfur dioxide to sulfur trioxide)
is 4.0* 1024 at 300K
2.5*1010 at 500K
3.0*104
at 700K
Estimate the standard reaction enthalpy at 500 K
Answer: First, we assume that the standard reaction enthalpy does not change
within the above temperature interval.
Then, solvingthe equation 9.4.3,
ln( K 2) − (ln( K1) − ∆rH θ
=
1
1
R
−
T 2 T1
Plug in the above numbers, we get ∆rHө= -200kJ mol-1
Here, we actually only need two data in order to calculate the standard reaction
enthalpy. Another approach to resolve the above question is via graphic method (hint:
∆rH θ
plot ln(K) as a function of (1/T), the slope is .
R
9.5 The response of equilibrium to pH
Definition of pH: An indication of hydronium ion concentration
pH = - logα(H3O+)
+
where, H3O is hydronium, a representation of the state of proton in aqueous solution
9.5a Acid-base equilibrium in water
Three equilibrium constants need to be known in order to characterize the
influence of pH on equilibria:
1) acidity constant Ka, which is defined as
HA(aq) + H2O (l) ↔ H3O+ (aq) + A (aq)
Ka
=
a H O + a A−
(9.5.1)
3
a HA
Quite often, we report Ka with pKa, which is defined as –log(Ka).
2) Basicity constant Kb
B(aq)
+ H2O (l) ↔ HB+ (aq) + OH (aq)
a +a −
K b = HB OH
(9.5.2)
aB
we can also use pKb to represent Kb, which is pKb = -logKb
3) Autoprotolysis constant Kw
H3O+ (aq) + OH (aq)
2H2O(l) ↔
Kw = α(H3O+)*α(OH )
( 9.5.3)
-
α(OH-)
because pH = -logα(H O ), and now introduce pOH = - log α(OH )
pKw = -log(Kw) = -log(α(H3O+)*α(OH )) = -logα(H3O+) - log
3
+
we get
pKw = pH + pOH
(9.5.4)
9.5b The pH of acids and bases
Assuming the solutions are ideal solution, the activities in the equation 9.5.1 can
be replaced by the concentration of [HA], [H3O+] and [A-], respectively . Because of the
1:1 stoichiometric relationship, we have [H3O+] = [A-]. Another assumption is that the
concentration of HA or B are unchanged from its nominal value due to the small proton
transfer. After all, the equation 9.5.1 can be simplified to
Ka =
[ H 3O + ] 2
[ HA]
[H3O+] = (Ka* [HA])1/2
or
pH = -log[H3O+] = - (1/2)logKa
- (1/2)log[HA]
= (1/2)pKa – (1/2)log[HA]
(9.5.5)
The pH value for bases can be derived from the similar way. Based on 1:1 stoichiometric
relationship, we have [HB+] = [OH-],
[OH − ] 2
Thus Kb =
[ B]
[OH] = (Kb*[B])1/2
pOH = -log [OH]
= -(1/2)logKb - (1/2)log[B]
(pKb = -logKb !!!)
= (1/2)pKb – (1/2)log[B]
Because pH = pKw – pOH
So,
pH = pKw – (1/2)pKb + (1/2)log[B]
(9.5.6)
9.5c Acid-base titration
Our aim here is to develop a quantitative description of the pH value at any stage
of the titration.
Case 1. Using strong base to titrate weak acid.
While the exact solution of the pH value can be obtained, for simplicity we here
are only interested in a simplified situation, where a number of approximation are
invoked.
First step: examine the source of H3O+:
Source 1: HA(aq) + H2O (l) ↔ H3O+ (aq) + A (aq)
↔
H3O+ (aq) + OH (aq)
Source 2: 2H2O(l)
Approximation 1: Because it is a weak acid, the amount of A- from source is very small
comparing with the amount of HA.
Approximation 2: The amount of hydronium ion coming from source 2 is negligible,
Though HA is a weak acid.
Assuming the initial concentration of [HA] = A0, and volume VA
From equation 9.5.4, we know the initial pH value of the weak acid can be
calculated as
pH = (1/2)pKa – (1/2)log[A0]
After certain time interval, volume VB of titrant is added, the total volume become
V = VA + VB; After the addition of strong base, the following reaction occur
-
HA (aq) + OH (aq)
↔
-
A (aq) + H2O(l)
Note that the concentration of [A-] comes from the salt which is formed due to the
acid and base reaction. The contribution from source 1 is negligible. We represent it with
[A ] = S.
Recall equation 9.5.1
Ka
=
a H O + a A−
3
a HA
=
[ H 3 O + ][ A − ]
[ HA]
After adding VB amount of base, the concentration of HA becomes A’
We get pH = pKa – log(A’/S) (this expression is called Henderson-Hasselbalch
equation )
In a general form, the above equation can be expressed as
pH = pKa – log ([acid]/[base])