The rate of true neutralization reactions has proved to be immeasurably fast. Eucken’s Lehrbuch der — Physical Chemistry — Thermodynamics Chemistry Question
Ultrafast reactions
The rate of true neutralization reactions has proved to be immeasurably fast.
Eucken’s Lehrbuch der Chemischen Physik, 1949
The main problem with studying ultrafast reactions is mixing the reactants. A smart way to circumvent this problem is the so-called relaxation technique. Neutralization is a good example of an ultrafast reaction:
Here, k1 and k2 are the rate constants for the forward and backward reaction, respectively. The mean enthalpy for this reaction is −49.65 kJ mol−1 in the temperature range 298 – 373 K. The density of water is 1.000 g cm−3.
Water has pH = 7.00 at 298 K. Calculate the apparent equilibrium constant K = [H2O] / [H+][OH-] of the neutralization reaction shown above. Calculate also the entropy change for the reaction.
Model Answer
The equilibrium constant of neutralization is given as
The constant K is related to the free energy change of the reaction:
ΔG° = −RT lnK = −89.8 kJ mol−1
Note that the Gibbs free energy change calculated in this way corresponds to the standard state c0 = 1 mol dm−3 for all species, including the water solvent. The Gibbs free energy change can be expressed via the enthalpy and entropy change for the reaction
ΔG° = ΔH° − T ΔS°
from which
Estimate the pH of boiling water (T = 373 K).
Model Answer
To estimate the pH of boiling water we need to evaluate Kw at 373 K using the van ’t Hoff’s formula (alternatively, we could recalculate the constant K). Note that Δ𝐻° was defined for a reverse reaction, here we have to use ΔH = 49.65 × 103 J mol−1.
The temperature change is given as
After substitution
we get
Kw,T2 = 56.23 × 10−14
which translates into proton concentration at the boiling point of water
[H+]T2 = (Kw2) ½ = (56.23×10−14) ½ = 7.499 × 10−7
or pH
pH = −log[H+]T2 = 6.125
Heavy water undergoes an analogous neutralization reaction, yet it is less dissociated than light water at the given temperature: Kw(D2O) = 1.35 × 10−15 at 298 K.
What is pD of heavy water at 298 K?
Model Answer
pD is analogical to pH, i.e. pD = −log[D+]. The concentration of [D+] cations at 298 K is given as
[D+] = [Kw (D2O)] ½ = (1.35 × 10-15) ½ = 3.67×10−8
and pD is given by
pD = −log[D+] = 7.435
Write the rate law for the change of the concentration of D2O in terms of the concentrations of D+, OD− and D2O.
Model Answer
(d[D2O] / dt) = k1 [D +][OD−] – k2 [D2O]
The composition of the equilibrium system depends on temperature. If we apply an external stimulus, for example a very fast heat pulse on the system, we disturb the equilibrium and observe a subsequent relaxation to the equilibrium composition. We can describe the relaxation with a new quantity x, a deviation from the equilibrium concentrations:
x = [D2O]eq − [D2O] = [OD−] − [OD−]eq = [D+] − [D+]eq
Express the time change dx/dt in terms of x. Give both the exact equation and the equation in which you neglect the small terms of x2.
Model Answer
We start from the rate equation derived in 5.4
(d[D2O] / dt) = k1 [D +][OD−] – k2 [D2O]
All concentrations can be expressed via the quantity x
Expanding the right hand side of the equation, we get
Using the equality of the backward and forward reaction rates at equilibrium
k1[D +]eq[OD−]eq = k2[D2O]eq
and neglecting the (small) quadratic term x2, we can rewrite the equation as
Solving the equation derived in 5.5, we get:
х = х(0) × exp(− t × (k1 [D+]eq + k1[OD−]eq + k2))
where х(0) is the deviation from equilibrium at the moment of perturbation.
For heavy water at 298 K, the relaxation time τ (time at which the deviation from equilibrium drops to 1/e of the initial value) was measured to be 162 µs. Calculate the rate constant for he forward and backward reaction. The density of heavy water is ρ = 1.107 g cm−3 and its molar mass is Mr = 20.03.
Model Answer
The relaxation time is given as
At equilibrium, the backward and forward reaction rates are the same. The concentration of heavy water [D2O]eq is given as
The relaxation time is the given as
Substituting the values of all quantities
we get
k1 = 8.41 × 1010 dm3 mol−1 s−1
We get k2 from the equilibrium constant K
k2 = k1 K = 2.05 ×10−6 s−1
Ultrafast reactions can also be triggered by a pH jump. Using an ultrafast laser pulse, we can induce a pH jump in a system with so-called photoacids. These compounds have dramatically different acid-base properties in the ground and excited electronic states. For example, the pKa of 6-hydroxynaphthalene-2-sulfonate is 9.12 in the ground state and 1.66 in the excited state.
1 cm3 of 5.0×10−3 mol dm−3 6-hydroxynaphthalene-2-sulfonate solution was irradiated by light with the wavelength of 297 nm. The total absorbed energy was 2.228×10−3 J. Calculate the pH before and after irradiation. Neglect the autoprotolysis of water in both cases.
Note that the standard state for a solution is defined as c0 = 1 mol dm−3 and assume that the activity coefficient is γi = 1 for all species. It may be of advantage to use an online cubic equation solver.
Model Answer
The pH before irradiation is calculated from the dissociation constant of the ground state of 6-hydroxynaphthalene-2-sulfonate.
where [A−] is the concentration of 6-oxidonaphthalene-2-sulfonate and [HA] is the concentration of 6-hydroxynaphthalene-2-sulfonate.
The concentration of [H+] is equal to the concentration of [A−] due to electroneutrality and can be denoted as y. The equilibrium concentration of the undissociated acid [HA] is c − y, where c is the analytical concentration of the acid. The equilibrium constant is then given as
Because the amount of dissociated acid is very small, we can neglect y in the denominator
Ka = y2 / c
From which
y = (Ka × c)½ = (7.59×10−10 × 5.0×10−3) ½ = 1.9×10−6
pH = − log(1.9×10−6) = 5.72
During irradiation, 1 cm3 of sample absorbs 2.228×10-3 J of energy. 1 dm3 would thus absorb 2.228 J. The number of absorbed photons corresponds to the number of excited molecules of 6-hydroxynaphthalene-2-sulfonate.
One photon has energy
The number of absorbed photons in 1 dm3 is
The number of moles of excited molecules of 6-hydroxynaphthalene-2-sulfonate is
The pH can again be calculated from the pKa in the excited state; the analytical concentration c* of the excited acid is now 5.5×10−6 mol dm−3.
Let us denote by x the proton concentration [H+] and by y* the concentration of the 6-oxidonaphthalene-2-sulfonate in the excited state [A−]∗ . The electroneutrality condition implies
x = y* + y
The two equilibrium constants are expressed as
where we assumed c – c* − y ≈ c in the denominator of the last equation.
These three equations constitute a system of equations from which we get
x3 + Ka *x2 – (Ka c + Ka* c*) x – KaKa * c = 0
or
x3 + 0.022 x2 – 1.21×10-7 x – 8.35×10-14 = 0
We can solve this equation e.g. with any on-line solver of cubic equations
x = 6.12×10−6 mol dm−3
which corresponds to
pH = −log (6.12 × 10−6) = 5.21
It is possible to avoid solving cubic equations by an iterative solution. In the first step, we assume that y ≈ 0. The equation for Ka* then transforms to
y* can be calculated from the quadratic equation
Next, we update the concentration of the anion in the ground state y from the corresponding equilibrium constant
From which y can be obtained by solving a quadratic equation
This again leads to the quadratic equation
y2 + y × 5.5×10−6 − 7.59×10−10 × 5.0×10−3 = 0
y = 6.2×10−7 mol dm−3
The concentration of [H+] is
x = y* + y = 5.5 ×10−6 + 6×10−7 = 6.1× 10−6 mol dm−3
pH = − log (6.1×10−6) = 5.21
We could now repeat the whole cycle: with the first estimate of x, we would get a new value of y* and continue with these new values of y and x until convergence is reached. At the level of precision in our calculations, the concentration is already converged in the first iteration. Generally, more iterative cycles would be needed.