COLLEGE OF SAINT MARY     FALL 2008    PHYSICAL CHEMISTRY    

INSTRUCTOR: DR. PETER ILICH        YOUR NAME: _____________________________

EXAM # 2B  -- SOLUTIONS & ANSWERS

Copyright 2008/2010 by Peter Ilich

I love lysine:

STRATEGY & SOLUTION:

(a) There are five characteristic pH values for lysine: (1) at pH < 2.18 all the O- and N- groups are protonated and the dominant form is DICATION,  (2) at pH = 2.18 there is (approximately) a 50 : 50 mixture of dication and ε-cation, (3) at pH = 8.95 there is (approximately) a 50 :50 mixture of ε-cation and the neutral zwitterion, (4) at pH = 10.53 there is (approximately) a 50 : 50 mixture of the zwitterion and anion, and (5) at pH > 10.53 the dominant form is the lysine anion.

(b) This ion would travel toward cathode.

(c)  According to the Hendersohn-Hasselbalch equation:

pKa = pH - lg {[zwitterion] / [ε-cation]}

Assuming that the dominant pKa is the on a-NH3+ group this equation reads:

8.95 ≈ 7 - lg {[x] / [1.37*10^-3  - x]} 

where "x" is the concentration of the zwitterion, the "anionic" form.  From there we get "x" as:

x = 10^-1.95 * c / (1 + 10^-1.95) = 1.52*10^-5

And the conccnetration of the e-cation is given as the difference of the total concentration and the concentration of the zwitterion:

c[ε-cation] = 1.37*10^-3 - 1.52*10^-5 = 1.35*10^-3 mol L^-1

NOTE that in this and similar problems we are adding and subtracting concentrations which, generally, must never be done (concentrations are non-additive, intensive properties).  However, since all the lysine species are in the same "pot", the volume of the solvent is the same (the same "denominator") and the concentrations can be manipulated as the amounts of substance.

Don't drink and drive:

2. (30 pts)  The redox reaction given below,

where the reduction of C4 in the pyrimidine ring of the nicotineamideadenine dinucleotide corresponds to the oxidation of one ethanol molecule to ethanal (acetaldehyde) provides the electrochemical basis for the metabolism of ethanol with the catalytic help of LADH (liver alcohol dehydrogenase) enzyme in our liver.  Given that E^ө (NAD/NADH) at pH = 7 is -0.320 V, and E^ө (CH₃CHO/CH₃CH₂OH) is -0.197 V, calculate the standard Gibbs energy for this reaction and determine whether the reaction is thermodynamically spontaneous or driven.

STRATEGY & SOLUTION:

This is a relatively  simple redox reaction that involves two half reactions.  We first write the reductive half reactions:

LEFT:    CH₃CHO + 2 e + 2 H⁺ = CH₃CH₂OH

RIGHT:    NAD + 2e^- + H⁺ = NADH

And the total reaction standard potential will be:

E^ө(RIGHT) - E^ө(LEFT) = +0.320 - (-0.197) = + 0.517 V

And the standard Gibbs free energy is:

ΔG^ө = -zFE  = 9.97*10⁴ J

So the reaction is spontaneous.

The little N-bomb in your heart:

3. (30 pts)    The less well known plutonium isotope, ²³⁸Pu₉₄ undergoes the following radioactive decay reaction:

²³⁸Pu₉₄ → ²³⁴U₉₂ + [He]⁺²

This is a highly exothermic nuclear reaction producing 0.54 kW/kg plutonium and is used in the so-called RTG (Radioisotope Thermoelectric Generator) power units in interplanetary space probes (e.g. Cassini) as well as in some experimental-type pacemakers.  A newly manufactured ²³⁸Pu₉₄-powered pacemaker is surgically implanted into a patient.  Keeping in mind that the Pu-238 half life is 87.74 yr calculate how much Pu-238 fuel, in grams, is needed in order to maintain a minimum of 100 mW output for 25 years?

SOLUTION:

STRATEGY:

       Use the formula for the first-order reaction:

[A] = [A₀] * exp[ -kt]                            (1)

SOLVING THE PROBLEM:

In the specific case when the concentration (radioactivity, in this case) [A] reaches one-half of the initial value, i.e. [A] = [A₀]/2  we write:

0.5* [A₀] = [A₀] * exp[-k t]

Dividing the equation by [A0] we find the value for k:

0.5 = exp[-k t]     or     ln 0.5 = -k t    from where     k = ln 2 / t                            (2)

The "t" in this case is the so-called "half-life", that is the time when the initial activity (concentration) falls to half values, for ²³⁸Pu₉₄ t_1/2 = 87.74 years, so we get for k:

k = ln 2 / 87.74 = 7.90*10^-3 [y^-1]            the unit is reciprocal year

Now we use this value k for further calculations.   We re-write the equation (1):

[A] = [A₀] * exp[-kt]                            (1)'

What is that we know in this equation?  We know the time, t = 25 [y], we know the rate constant, k = 7.90*10^-3 [y^-1] and we know the energy output, which is [A], after 25 years, [A] = 100 μW = 0.1 mW.  What we do not know is the energy output at the beginning of the pacemaker's life, [A₀].  We calculate this value by inserting the known quantities:

[A₀] = 0.1 [mW] * exp {-7.9*10^-3 [y^-1] * 25 [y]}

[A₀] = 1.21853*10^-1 = 0.12 mW

How do we figure out the initial weight of the ²³⁸Pu₉₄ pellet in the pacemaker device?  By using simple proportion calculation: 1 kg plutonium will output 0.54 kW or 540 W energy, and the value for the plutonium pellet which outputs 0.12 mW should be proportionally much less:

m₀ = 0.12 *10^-3[W] / 540 [W kg^-1] = 2.256*10^-7 [kg] = 0.0023 g ²³⁸Pu₉₄

So a rather small pellet wouldn't you say.

Not published yet:

4. (30 pts)   Xanthine dehydrogenase, XDH [E.C. 1.1.1.204], is a molybdo-pterin oxidoreductive enzyme involved in the processing of nucleic acid metabolites in human body. The reductive rate constant of XDH, k_red, is 67 [s^-1] for xanthine substrate [Leimkühler et al., J. Biol. Chem. 2004, 279, 40437, Table 2].   Electronic structure calculations suggest a drop in the activation Gibbs energy from 76 kcal mol^-1 to 66 kcal mol^-1. with the transition-state-coupled proton release  [Ilich & Hille, J. Inorg. Biochem. 2008, submitted].  Assuming the same, pseudo-first order kinetics, use the Arrhenius activation energy formalism to calculate the change in the reaction rate induced by this activation energy decrease.

SOLUTION:

STRATEGY:

    This is a problem about reaction rates and activation energies.    Two activation energies, at two different reaction conditions, are given 76 kcal mol^-1 and 66 kcal mol^-1.  Given also is one reaction rate constant, the question asks to figure the other reaction constant.  So we will use the Arrhenius equation:

k = A₀ * exp [-Ea/RT]                                (0)

SOLVING THE PROBLEM:

    We have two activation energies and to those will correspond two reaction rate constants.  SO we write two Arrhenius equations:

k₁ = A₀ * exp [-Ea(1)/RT]                        (1)

k₂ = A₀ * exp [-Ea(2)/RT]                        (2)

We solve this as two algebraic linear equations with two unknowns, k₂ and A₀ (we actually do not have to calculate A₀).   If we divide one equations by another A0 will cancel and we will be left with only unknown:

k₂ / k₁ = A₀ * exp [-Ea(2)/RT] / A₀ * exp [-Ea(2)/RT]                 (3)

k₂ / k₁ = exp [-Ea(2)/RT] / A₀ * exp [-Ea(2)/RT]  = exp [(-Ea(2) - (-Ea(1)) / RT]            (4)

Since we do not know k2 we will rewrite this equation as:

k₂ = k₁ * exp [-Ea(2) + Ea(1)] / RT            (5)

So let us insert the numbers we have been given and solve the equation for k₂:

CAVEAT:  Keep in mind that the activation energies are given in kcal mol^-1, that is, in units of 1000 cal per mol while the gas constant, R is given in J mol^-1 so you will have to convert the activation energies to kJ mol^-1 by multiplying by 4.186 and then multiply by 1000 to express in J mol^-1:

k₂ = 67 [s^-1] * exp [(-66 [kcal mol^-1] + 76 [kcal mol^-1] * 4.186 [kJ mol^-1 / kcal mol^-1]) / 8.31 J [mol^-1 K^-1] * 298.15 [K]]

k₂ = 67 [s^-1] * exp [16.89] = 1.46*10⁹ [s^-1]

So the 10 kcal mol^-1 drop in the activation energy is predicted to bring about 22 000 000 - fold increase in the rate constant -- a large reaction rate acceleration.

COLLEGE OF SAINT MARY     FALL 2008    PHYSICAL CHEMISTRY EXAM # 2

YOUR NAME: _____________________________