OC 583- ISOTOPE BIOGEOCHEMISTRY (Spring 2009)
TOPIC 1: INTRODUCTION AND BACKGROUND
References:
1. Chaps 1 and 2 of Robert Criss’s book “Principles of Stable Isotope Distribution”
Oxford Press, 1999. (on reserve in Ocean-Fish Library)
2. Chaps 1-3 of Zachary Sharp’s book “Stable Isotope Geochemisry” Prentice-Hall Press,
2007. (on reserve in Ocean-Fish Library)
I. BACKGROUND
Class Information and Goals (Figs. 1-3)
a. Learn how to use isotopic measurements to improve our understanding of
biogeochemical cycles.
b. Large Range of Applications of Isotopes
-common isotopes: carbon, nitrogen, hydrogen, oxygen and sulfur isotopes
-fundamental cycles: water, CO2, nutrients, food webs, climate, trace gases, etc.
-many scales of interest: global, regional and local scales
-many environments: marine, freshwater, atmospheric, and terrestrial systems
-both modern and paleo environments (modern applications are stressed in class)
c. Try to use isotopic measurements to be quantitative when ever possible
-Need to assess whether isotopic measurements are worth the effort.
-What are the uncertainties of your conclusions based on isotopic information?
d. Isotope Biogeochemistry is a field that continues to grow by finding innovative uses for
isotopic measurements
-isotopic measurements have been shown, over and over again, to be an extremely
useful tool for a diversity of fields within the earth sciences
e. Instruments used for measurements are now easier to use with less experience
-continuous flow techniques have reduced sample size by 1000x
-laser based instruments provide for in situ field measurements
f. New applications for existing isotopic methods and development of new isotopic
methods are still on going despite 50 years of biogeochemical isotopic measurements
1
g. Field is dominated by experimental rather than theoretical results (often to the chagrin
of physical chemists) (Figs. 4 and 5)
- thus the course material is dominated by experimental results and applications
- as a result oceanographers, ecologists, fisheries scientists, atmospheric chemists,
etc. can use isotopic methods to help solve problems without having to understand
the quantum mechanical explanations of isotope effects
-however, they have a responsibility to know limitations of isotopic
methods
A. General Characteristics of Isotopes
1. Isotopes are atoms of a single element that have different atomic weights
a. Thus isotopes have the same number of protons but a different number of
neutrons
b.
-
12
C, 13C, 14C or 16O, 17O, 18O
there are ~300 stable isotopes (versus ~1200 radioactive isotopes)
2. Symmetry rules --at low atomic number (<20) the most abundant isotope often has an
equal number of neutrons and protons (e.g. 12C6, 14N7, 16O8, 32S16) (Fig. 6)
-but not always: (e.g. 7Li3, 9Be4)
3. Since isotopes have the same number of protons and electrons this means isotopes have
same chemical behavior, i.e., all isotopic species enter into same chemical reactions and
form the same bonds
-in this sense, the rare or less abundant isotopic species is an excellent tracer of
the abundant isotopic species
-for example, 13CO2 undergoes photosynthesis just like 12CO2
4. However, the different number of neutrons of an isotope results in slightly different
physical properties
-the substitution of an atom with different mass (13C for 12C) affects molecular
properties that are related to nuclear motion (especially the vibrational energy state which
depends on mass of the bonding atoms)
-e.g. heat capacity, vapor pressure, diffusion rates,
2
-Importantly, reaction rates are slightly different between isotopic species because
of their slightly different energy states
5. Isotopologues are molecules of the same elemental composition but different molecule
weight
-e.g., for CO2:
12
C16O16O (44), 13C16O16O (45), 12C18O16O (46), 13C18O16O (47)
- Isotopomers are isotopologues that have the same molecular weight
-e.g., for CO2: 13C16O16O (45) and 12C17O16O (45)
6. Isotopic Abundances
a. Elements of interest: e.g., C, N, O and H (Fig. 6)
-12C, 13C,
-16O, 17O, 18O
-14N, 15N
-2H, 1H
b. Typically, the heavier isotope is much less abundant than the lighter isotope (Fig. 6)
-for C, N and O: heavier isotope comprises <1% of total (light) isotope
-for hydrogen , 2H (deuterium) comprises only 0.015% of hydrogen
B. Isotopic Notation (Fig. 7)
1. The amount of rare isotope is always expressed as a ratio of the rare to the
abundant isotope
-e.g., 13C/12C, 18O/16O, etc.
- del () notation is used to express the isotope ratio and is represented in parts per
thousand or per mil (‰), e.g., 13C, 15N, etc.
-the measured isotope ratio is expressed as a deviation from the isotope ratio of a
standard using del () notation
 (13C /12C )

e.g.,  13C  13 12 sample  1 * 1000, with 13C in per mil (‰) units
 ( C / C ) stnadard 
-Similar notation is used for reporting all isotope ratios (e.g., 15N, 18O, etc.)
-each element usually has its own specific standard
3
B. Standards and their assigned isotope ratios: (Fig. 7)
1.
-Carbon: PDB (a carbonate): 13C/12C = 0.0112372
-Oxygen: PDB (a carbonate): 18O/16O = 0.0020671
-Oxygen: SMOW (seawater): 18O/16O = 0.0020052
-Oxygen: AIR (molecular O2): 18O/16O = 0.002005308
-Nitrogen: AIR (molecular N2): 15N/14N = 0.0036765
-Sulfur: CANYON DIABLO (meteorite): 34S/32S = 0.0450045
-Hydrogen: sea water (SMOW): D/H = 0.00015576
2. How do they measure the absolute ratio of the standards?
-Gravimetrically, but good only to about ±0.2‰
e.g., Baertschi, in Earth Planet. Sci. Lett. 31: 341-344.
-However, the absolute isotope ratio of the standard isn’t that important because
in most situations we are concerned with knowing the isotopic composition of various
compounds relative to each other.
-thus in practice, the absolute ratio of the rare to abundant isotopes is rarely used
3. Important: The  value of any standard = 0 ‰.
-Thus positive or negative del values imply higher or lower isotope ratios
relative to the isotope ratio of the standard
4. The choice of material used for standard is (somewhat) arbitrary
5. A primary role of standards is to report isotopic measurements on a consistent scale, so
that isotope measurements made in different labs and at different times are comparable.
-thus there are gases available for isotope labs that have been calibrated to the
primary standards and can be used to calibrate “working” gas standards used daily
-this situation has benefited isotope biogeochemistry substantially and does not
occur as frequently as you might expect in earth sciences (e.g., there isn’t a universally
accepted certified standard for nutrient measurements)
-thus 13C/12C measurements reported versus PDB from the 1950s can be
compared to 13C/12C reported versus PDB today (assuming the 13C/12C measurement was
done correctly)
4
6. Many of the original stocks of standards like PDB and SMOW (made in the 1950s) are
no longer available so replacement standards like VPDB and VSMOW (V for Vienna
where IAEA is located), which were calibrated to PDB and SMOW, respectively, are
used in their place.
-Often you will see 13C/12C or 18O/16O values in papers reported versus either
PDB or VPDB and SMOW or VSMOW.
Class Problem- a. Calculate the 13C/12C of atmospheric CO2 and plant cellulose assuming
their 13C values are –8 and –28 ‰ relative to PDB, respectively. b. Let’s say you choose
a new standard called PDQ (my personal favorite), which has a 13C/12C = 0.01100000, to
base your 13C values. Calculate the 13C of atmospheric CO2 and cellulose relative to
PDQ. c. Does the 13C difference between atmospheric CO2 and cellulose change with
the new standard? d. This example points out an important characteristic of del (δ)
values. δ values are not exactly (linearly) proportional to isotope ratios (but they are
close).
C. Range in isotopic composition of natural materials
(Figs. 8 and 9)
1. General observed range in del (values:
Carbon (vs PDB):
13C = -40 to 0 ‰
Oxygen (vs SMOW):
18O = -30 to +30 ‰
Nitrogen (vs AIR):
15N = -20 to +20 ‰
Sulfur (vs CD):
34S = -40 to +40 ‰
Hydrogen (vs SMOW):
D = -300 to +120 ‰
2. Range of 13C, 15N, 18O and 34S are similar at about 40 to 80 ‰, whereas for D it
is 400 ‰.
a. This indicates much greater isotopic fractionation effects for H versus
other elements. Why might this be expected?
5
3. Why are some of the  ranges entirely negative (e.g., 13C) while others are more
symmetric about zero?
D. Precision of isotope ratio measurements
1. Internal Precision of mass spectrometer instrument
a. Typically about ±0.003 to 0.01 ‰
b. How would you determine this?
c. Internal precision indicates the precision limit of the measurement.
2. External precision (or measurement reproducibility)
a. Typically about ±0.02 to 0.2 ‰ for 13C, 15N, 18O and 34S
b. Typically about ±2-5 ‰ for D
c. How would you determine this?
d. Note that the precision is worse for D. Why? (Hint: see Fig. 5)
e. External precision (reproducibility) is generally more useful because it
indicates the overall uncertainty of the isotope ratio measurement
II. ISOTOPIC FRACTIONATION EFFECTS
A. Background
1. Differences in chemical and physical properties of isotopologues (e.g., reaction rates,
diffusion rates, equilibrium distributions), which result from their slight differences in
molecular mass, cause isotope fractionation effects.
Important: If there were no isotope fractionation effects, then isotope measurements
would not be useful. Why?
2. Differences in physiochemical properties (e.g. heat capacity, density, vapor pressure) of
isotopologues (i.e., same molecules with different molecular weights) arise as a result of
quantum mechanical effects. (Fig. 10)
-energy of a molecule restricted to discrete energy levels derived from rotational,
translational and vibrational energies
6
3. Isotopologues have different vibrational energy states because the vibrational energy
state of the molecule depends on the vibrational frequency (v) which, in turn, depends
inversely on mass. Using an analogy to a harmonic oscillator
v=
k
1
*
,

2
where μ is the reduced mass and k is a force or spring constant. The reduced mass is
calculated as follows:
μ=
m1* m 2
m1  m 2
4. For two isotopologues, the ratio of the vibrational frequencies would be:
v1
u
 2
v2
u1
5.This may be easiest thought about with a diatomic molecule like H2 that has two stable
isotopologues H-H and H-D.
-for H-H:
μ1= (1*1)/(1+1) = 1/2
-for D-H:
μ2 = (1*2)/(1+2) = 2/3
-thus
v1
u
 2 =
v2
u1
( 2 / 3)
(1 / 2)
= 1.154
6. The measured H-H to H-D vibrational frequency ratio
vH  H
4401.213cm1
=
= 1.1542
vD  H
3813.15cm1
-for this simplest molecule, the measured ratio of vibrational frequencies equals
that predicted by the harmonic oscillator assumption
-for more complicated molecules, the harmonic oscillator assumption is less good
7. Different isotopologues of molecular hydrogen (e.g., H-H, D-H and D-D), because of
their different masses, have different vibrational energies
7
-the heavier isotopologue has a lower vibrational frequency because v varies
inversely with mass
8. The Zero Point Energy (ZPE) is the energy a molecule has in its ground energy state
(Eo), its inherent energy, at absolute zero (Fig. 11)
-the ZPE depends on the vibrational frequency of the molecule
-thus ZPE = ½*h*v, where h = Planck's constant and v = vibrational frequency of
atoms in the molecular bond
-thus the heavier isotopologue has a lower zero-point energy than the lighter
isotopomer
9. Important: Since the lighter isotopologue has a higher ZPE, it takes less energy to
break the bonds of the lighter isotopologue relative to the heavier isotopologue
-this assumes vibrational energies only affect overall bond strengths
-generally, one can ignore mass dependencies of rotational and translational
contributions to zero point energies since typically they are much smaller than vibrational
energy effects
10. The different zero point energies of the isotopologues cause fractionation effects
during chemical reactions via both equilibrium and kinetic processes:
a. equilibrium processes: the different isotopes of an element (e.g., H or D; 12C or
13
C, 16O, 17O or 18O) do not partition equally between different molecular species or
between different phases of the same molecular species at equilibrium
b. kinetic processes: the isotopologues react at different rates in non-reversible
reactions because of their different ZPEs, with the lighter isotopologue typically reacting
more often than the heavier isotopologue because of its higher ZPE
11. Fractionation effects result from the different molecular diffusion rates of
isotopologues.
-the velocity of a molecule depends on its kinetic energy and inversely on its mass
8
-since KE = ½ mv2, where m= mass and v=velocity. Thus v =
(2 * KE)
m
-since the diffusion rate of the molecule is related to its mean velocity (v), the
molecular diffusion rate depends inversely on mass
-e.g., molecular H-H diffuses slightly faster than molecular D-H.
-different diffusion rates between isotopologues yield isotopic fractionation
effects.
12. Calculate the isotope fractionation effect due to molecular diffusion of H-D vs HH molecules assuming diffusion rate depends on velocity
vH  H
mH  D
3


 1.2247
vH  D
mH  H
2
-in a per mil notation sense (as discussed below), this would mean that there is
about a 224 ‰ fractionation effect during the diffusion of H2 gas
B. Equilibrium Isotopic Fractionations
a. Background and Nomenclature
1. An equilibrium isotope effect describes a situation where the distribution of isotopes is
not equal between different compounds or different phases of the same compound under
equilibrium conditions.
2. Generally the isotopic equilibrium reaction is written as a chemical equilibrium
condition as follows:
AX + BX* = AX* + BX
-where AX and BX are two compounds and * denotes the rare (usually heavier)
isotope of element X
3. At equilibrium, the equilibrium constant (Keq) describes the ratio of the abundances of
each molecule as follows:
Keq =
( AX *)( BX )
,
( AX )( BX *)
-dividing numerator and denominator by (AX)(BX) yields:
9
Keq =
(X * / X )A
( X * / X )B
,
-which represents the isotope ratio of compound A divided by the isotope ratio of
compound B
4. Although values for Keq are usually very close to a value of 1.0000, they do not equal
exactly 1.0000 because the rare isotope (usually heavier) does not distribute itself exactly
equally between the compounds
5. The isotopic fractionation effect (or isotopic fractionation factor) that results from this
equilibrium process is expressed as , where  = Keq
-so α =
(X * / X )A
in the above example
( X * / X )B
- thus, by measuring the isotopic composition of compound A and B under
equilibrium conditions, you can empirically determine the equilibrium isotope
fractionation effect for the reaction between compounds A and B
6. If  equaled exactly one, then the distribution of isotopes between the reacting
compounds would be exactly equal
7. The deviation of  from 1.0000 represents the magnitude of the equilibrium isotope
effect
- is typically a number close to 1.000, for example, 1.0102 or 0.9997
- can also be expressed using epsilon (ε), in parts per thousand or per mil (‰), as
the deviation from 1.000, where
 (‰) = (-1)*1000
8. Thus an isotope fractionation effect can be expressed as a numeric value close to
1.0000 (as ) or as a value in per mil units (as )
-for example, if  = 0.9983, then  = -1.7 ‰
-this implies that there is a “-1.7 per mil isotope effect”
-thus there usually should be five or possibly six significant figures for values
of α (e.g. 1.0102 or 1.01025) if ε values are determined to tenths or hundredths
of a per mil
10
9. The equilibrium reaction between gaseous water and hydrogen sulfide serves to
illustrate an equilibrium isotope fractionation effect (Fig. 12)
H2S + HDO = HDS + H2O
Keq =
( HDS )( H 2 O )
and  = Keq
( H 2 S )( HDO )
rearrange to yield:
=
( HDS ) /( H 2 S )
( HDO ) /( H 2 O)
=
(D / H ) H 2S
= 0.45 (measured)
( D / H ) H 2O
-thus by measuring the D/H of H2S and H2O gases under equilibrium conditions,
one empirically determines the magnitude of α for this reaction
-The measured value of α for this equilibrium reaction is 0.45, which means that
the D/H for water is more than twice the D/H for hydrogen sulfide.
-what determines the equilibrium distribution of isotopes between species?
b. Zero Point Energy
1. The zero point energy (ZPE) of a molecule represents the vibrational energy of the
ground energy state and depends on the vibrational frequency of the molecule
-thus ZPE = ½ hv, as discussed above, where v = vibrational frequency and h =
Planck’s constant
2. The zero point energies (ZPE) of two different isotopomers differ because their
vibrational frequencies differ
-in the above example, H2O has a higher ZPE than HDO and H2S has a higher
ZPE than HDS (Fig. 13)
-Note that the difference between the ZPEs for H2O and HDO is greater than the
difference between the ZPEs for H2S and HDS, because the bond strength between O and
H is greater than between S and H
3. Thus the substitution of D for H in H2O will cause a bigger energy decrease
than the substitution of D for H in H2S (Fig. 14)
b. The energy change in the reaction is
E = ZPEHDS + ZPEH2O – ZPEH2S – ZPEHDO
11
-thus a greater energy change will occur if the D preferentially substitutes for H2O
rather than for H2S
4. At equilibrium: ln (Keq) = -E/RT, where R= gas constant, T= °K and Keq = 
-thus, ln (Keq)*RT = -E = -ZPEHDS - ZPEH2O + ZPEH2S + ZPEHDO
-thus, -E = (ZPEH2S – ZPEHDS) - (ZPEH2O – ZPEHDO)
-since the energy change between HDO and HHO is greater than between
HDS and HHS, then (ZPEH2O – ZPEHDO) > (ZPEH2S – ZPEHDS) and -ΔE <0.
-thus when –ΔE<0, then ln (Keq)*RT < 0 and Keq <1
-since Keq = , which in turn equals (D/H)H2S / (D/H)H2O, then when Keq
<0, (D/H)H2S should be < (D/H)H2O
-this explains why the  value measured for this equilibrium reaction is <1
(measured at 0.45)
5. Important: Generally, the heavier (usually rare) isotope substitutes
preferentially into the molecule or compound with the stronger bond energies (Fig. 15)
-in the above example, the H-O bond is stronger than the H-S bond so D
preferentially substitutes into H2O relative to H2S
6. Another example:
15
NH4 + 14NO3 = 14NH4 + 15NO3
Keq =  =
[15NO3 ] *[ 14 NH 4 ]
[15NH 4 ] *[ 14 NO3 ]
Rearrange to:
=
(15N / 14N ) NO 3
= 1.041 (measured) (Fig. 15)
(15N / 14N ) NH 4
i.e., the 15N/14N for nitrate is higher than for ammonium, when the two
species are at equilibrium, because the N-O bond is stronger than the N-H bond.
7. Experimentally, this means that if we started out with 400 15N atoms and
100,000 14N atoms, at equilibrium there would not be 200 15N atoms as 15NO3 and 200 as
12
15
NH4. Instead, there would be 204 as 15NO3 and 196 as 15NH4 (assuming equal
concentrations of NO3 and NH4)
-this distribution yields a Keq value =  = (15N/14N)NO3 / (15N/14N)NH4
= (204/50,000) / (196/50,000) = 1.041,
-another way of stating this is to say that there is a 41 ‰ enrichment in the 15N of
NO3 relative to the 15N of NH4 at equilibrium
8. There is are also equilibrium isotopic fractionations between different phases of the
same molecule, for reasons analogous to the discussion presented above for equilibrium
gas mixtures (Fig. 16)
-there is a slight difference in the energy change resulting from the substitution of
the heavy isotope into one phase of a species relative to another phase
9. For example, H2O16(l) + H2O18(g) ↔ H2O18 (l) + H2O16(g),
where l =liquid phase and g =gas phase
-thus Keq =  = (18O/16O )H2O(l) / (18O/16O )H2O(g)
-the measured value of  = 1.0098 at 20º C (Fig. 17)
-this means that 18O of liquid water is 9.8 ‰ higher (or enriched) relative
to the 18O of water vapor with which it is at equilibrium at 20º C
10. This liquid-gas phase equilibrium fractionation for water is very important when
considering the global distribution of the 18O/16O in precipitation, as we will see.
11. Empirical determinations and theoretical calculations of equilibrium isotopic
fractionation effects (Richet et al. Ann Rev Earth and Planet Science 5: 65-110, 1977)
-For some simple reactions, equilibrium isotope effects have been calculated
based on a quantum mechanical-based estimate of free energies changes. These
theoretically calculated isotopic effects for simple (usually gaseous) molecules can
compare well with empirical determinations (Fig. 18)
13
-however, theoretically calculated isotope fractionation effects for liquids and
solid phases usually don't compare as well to empirical measurements because of nonideal interactions between distant molecules, e.g. lattice vibrations, intermolecular bonds
-Most often, one has to empirically determine the equilibrium fractionation effects
between compounds because the theoretical calculations for more complex molecules are
too inaccurate.
12. Generally, as temperature increases then equilibrium fractionation effects decrease
(Fig. 19)
-
as temperature increases, the molecules are excited to higher energy levels
(i.e., higher vibrational energy states)
-
at higher energy levels, the difference in the energy states of the isotopologues
is relatively less important and thus the preference for isotopic substitution
between species (or phases) decreases
-since ln (Keq) = -E/RT,
-at the limit, as T approaches infinity, ln (Keq)  0 and Keq  1
- thus   1.0000 and there is no isotopic fractionation effect (ε = 0 ‰)
C. Kinetic Isotope Fractionation Effects
a. Background
1. Different Reaction rates
-one type of kinetic isotope fractionation effect occurs because the rates of
reaction differ for isotopologues of the same molecule
-the ratio of the reaction rates for the isotopologues equals the fractionation effect
2. Kinetic isotope fractionations are often referred to as KIEs (kinetic isotope effects)
3. Isotopic measurements during a non-reversible chemical reaction almost always show
a preferential increase or enrichment of the lighter isotope in the reaction products
a. That is, the isotopic composition of the reaction product is depleted in the
heavier isotope compared to the isotopic composition of the reactants
14
4. This light isotope enrichment occurs because the molecular bonds of the lighter isotope
are easier to break than the bonds of the heavier isotope (Fig. 11)
a. That is, the reaction rate (K) is proportional to e (-Eb/RT), where Eb is the barrier
energy, R is the gas constant and T is temperature. Eb is equal to the depth of the
potential energy well minus the zero point energy (see Fig. 11).
5.Thus the reaction rates for rare (heavy) and abundant (light) isotopomers are:
-thus *K =Aoe(-*Eb/RT) and K = Aoe(-Eb/RT), where Ao is a constant and the *
represents the heavy isotope
-because *Eb > Eb, the heavy isotopologue reacts more slowly, i.e. *K < K
6.Generally, as temperature increases, the magnitude of the KIEs decrease because the
difference in Eb between isotopomers decreases at higher energy states reached as
temperature increases.
-i.e., as temperature approaches infinity, *K and K approach Ao, thus *K/K  1
and thus α = *K/K  1.000 and there is no kinetic isotope effect (KIE)
b. Example of Kinetic Isotope Effect
1. Methane gas reacting with OH radicals (an important reaction in the
atmosphere)
12
CH4 + •OH  12•CH3 + H2O (reaction rate constant referred to as k12)
13
CH4 + •OH  13•CH3 + H2O
(reaction rate constant referred to as k13)
- the methyl radical (•) reacts with OH to form formaldehyde (H2CO)
2. The kinetic isotope fractionation effect or KIE for this reaction is the ratio of
the reaction rate constants, e.g. k12/k13 or k13/k12, which equals .
-NOTE:  can represent either k12/k13 or k13/k12. Unfortunately, there is no
consistently accepted convention, which can lead to ambiguity.
3. The  = k13/k12 for this reaction has been measured empirically at 0.9946 and
theoretically calculated (ab initio quantum mechanical calculations) at 0.990
15
4. Thus 12CH4 reacts slightly faster than 13CH4 , as expected, because the Eb for the
12
C-H bond is slightly less than the Eb for 13C-H bond.
c. How can we experimentally determine the magnitude of the KIE?
1. Let’s derive an expression for the KIE (α) using the time rate of change of the product
Δ[13CH3]/Δt = [13CH4]*[OH]*K13
Δ[12CH3]/Δt = [12CH4]*[OH]*K12, where t = time and [ ] represents
concentration
-dividing one reaction by the other yields:
[13CH3] /[12CH3] = [13CH4] / [12CH4]*, where  = k13/k12
(13C/12C)CH3 / (13C/12C)CH4 = 
-assuming there was zero concentration of CH3 initially, then
 = (13C/12C)CH3 / (13C/12C)CH4
-Generally,  equals the isotope ratio of the product divided by the isotope ratio
of the reactant in a non-reversible reaction
-thus one way to estimate α is to measure the 13C/12C of the CH4 reacting and the
13
C/12C of the CH3 being produced at the same time.
2. NOTE: Typically,  = Rproducts / Rreactants, where R is the isotope ratio (e.g., 13C/12C,
15
N/14N, 18O/16O)
3. Alternatively, we can estimate α by measuring the change in the isotopic composition
of the reactant (CH4 in this case)
Δ[13CH4]/Δt = -[13CH4]*[OH]*K13
Δ[12CH4]/Δt = -[12CH4]*[OH]*K12, where t = time and [ ] represents
concentration
-dividing one reaction by the other yields:
[13CH4] /[12CH4] = [13CH4] / [12CH4]*, where  = k13/k12
(13C/12C)CH4 / (13C/12C)CH4 = 
16
-
thus by measuring the change in the 13C/12C of the methane over time you can
derive an estimate of α, without having to measure 13C/12C of the both the
reactant and the product
- often it is easier to measure the 13C/12C of the only one species (CH4)
than two species (CH4 and •CH3), especially if the second species is
reactive as it is in this situation
d. KIEs during molecular diffusion
1. The molecular diffusion rate of a molecule can be viewed as analogous to reaction rate,
in the sense that the diffusion of a molecule depends on overcoming some barrier to
motion (e.g. movement of atoms through the diffusion medium like gas, liquid or solid)
like a reacting molecule has to overcome a energy barrier to react
-molecular diffusion depends on the properties of the molecule, where as
turbulent diffusion (i.e., mixing) depends on the properties of the medium in which the
diffusion occurs
-no isotopic fractionation effects occur during turbulent diffusion or mixing
-isotopic fractionation effects occur only during molecular diffusion
2. Molecular Diffusion of a pure gas:
a. the molecular diffusion rate (D) of pure gases is inversely proportional to their mass
(m) assuming the diffusion “velocity” depends on available kinetic energy (KE),
-i.e., KE = 1/2 mv2 or v =
-thus:
2 KE / m , where m = mass and v = velocity
v1
m2

v2
m1
b. the ratio of molecular velocities or diffusivities for the two isotopologues of molecular
oxygen, O18-O16 (mass 34) and O16-O16 (mass 32), assuming the kinetic energy applied to
both is the same, would be:
-thus
v16O 18O
m16O 16O
32


 0.970
v16O 16O
m18O 16O
34
17
-the KIE is the ratio of the diffusivities and in this case α = 0.970
-this α represents the KIE of O2 diffusing in pure O2 (no other gases present)
-experimentally, if you could design an experimental set up where there was no
turbulent mixing and could measure the 18O/16O of O2 gas versus distance from an O2
source, the 18O/16O of O2 would decrease as you moved away from the O2 source.
c. theoretical predictions of the isotope fractionation for pure gases, generally, are in good
agreement with empirical determinations
d. However, most geochemical applications involve the molecular diffusion of specific
gases (CO2, O2, N2, etc.) in air
3. Molecular Diffusion of a Gas in a gas mixture:
-the molecular diffusion rate (D) of a gas (e.g., O2) in a gas mixture (e.g. air)
depends on the mass of the medium,
-in this case, the reduced mass (μ) of the mixture is used, rather than the mass of
the pure gas, to determine the KIE
- where μ =
m1 * m2 ,
m1  m2
- where m1 represents the molecular weight of the pure gas of interest and m2 is
the molecular weight of the gaseous medium (e.g. air)
- thus
D1
2

D2
1
-the ratio of the diffusion rates of O18-O16 and O16-O16 in air (assume air has a
mass =28, because it’s mostly N2) would be:
-thus D34/D32 =
=
 32/  34
[(32 * 28) /(32  28)] / [(34 * 28) /(34  28)]
= (32/34) * (34  28)/(32  28)
-thus  = D34/D32 =
0.97254 = 0.9862
-this yields ε = -13.8 ‰ (compared to –30 ‰ effect of O2 diffusion in pure O2)
18
-in essence, taking into account the mass of the diffusive medium reduces the
mass difference between the two isotopologues and reduces the KIE
4. Think of an example of where isotopic fractionation caused by the molecular diffusion
of a gas in air could be important.
5.Molecular Diffusion of Gas in Liquids
-main interest is in the diffusion of a gas in water (like CO2 or O2 gas diffusing at
the air-water interface in the ocean (or rivers, lakes) or CO2 diffusing across a cell
membrane)
-unfortunately, there is no useful theoretical basis to predict  for gases diffusing
in liquids (in contrast to diffusion in a gas phase presented above)
-thus KIE for molecular diffusion of gas in liquids are determined empirically
-Typically, the KIEs for diffusion of gases in water are small compared to KIEs
for gases diffusing in a gas phase
- examples (from Knox et al., J. Geophysical Research 97: 20,335-20343, 1992)
-for CO2 in water,  = D13CO2 / D12CO2 = 0.9991 ( = -0.9 ‰)
-for O2 in water,  = D34O2 / D32O2 = 0.9977 ( = -2.3 ‰)
-for N2 in water,  = D29N2/D28N2 = 0.9988 ( = -1.2 ‰)
-for H2 in water,  = D3HD/D2HH = 0.982 ( = -18 ‰)
-in all cases, these empirical results indicate that molecular diffusion of the heavy
isotopologue is slower than the light isotopologue
D. Mass Dependent and Mass Independent Isotope Fractionation Effects
1. Almost all isotope fractionation effects in nature are mass dependent, that is,
the magnitude of the fractionation effect depends on the magnitude of the
mass difference between the isotopes
a. For example, the isotope fractionation effect for 17O/16O is considered
to be the square root (0.5 power) of the isotope fractionation effect of
18
O/16O
19
b. The mathematical expression describing this effect is as follows.
i. assume there are three isotopes of an element: m1, m2 and m3
ii. an estimate of the relative magnitude of the fractionation
effects between the three isotopes is calculated as:
1. m2/m1 = (m3/m1)z, where z =
m3 (m2  m1)
*
m2 (m3  m1)
c. The magnitude of  for 17O/16O compared to the  for 18O/16O is
calculated as follows:
i. m1=16, m2=17 and m3=18.
ii. z = 18/17 * (17-16) / (18-16) = 0.5294
iii. 17/16 = (18/16)0.5294
d. This means that the fractionation effect for 17O is about 0.529 the
fractionation effect for 18O, as follows:
i. 17/16 = (18/16)0.5294
ii. ln (17/16) = 0.5294*ln(18/16)
iii. ln (X)  X-1 when X is close to 1 (e.g., ln (1.040) = 0.0392)
iv. thus 17/16 – 1  0.529* (18/16 – 1)
v. since ε (‰) = (– 1)*1000
vi. 17/16 (‰)  0.529* 18/16 (‰)
e. In practice, this means that if you plotted the measured 17O vs 18O
for an oxygen containing substance the slope of the line would be
about 0.529
1. for example, the measured 17O vs 18O slope for
rainwater is 0.528±0.001, which agrees well with
theoretical prediction
2. however, the measured 17O vs 18O for O2 consumed
during respiration has a slope of 0.518±0.001, which is
significantly different from the expected slope
20
3. Thus one has to empirically determine the exact
magnitude of the mass dependent fractionation effect
Important: Mass dependent fractionation effects almost always occur in nature,
>99%(?) of the time.
2. However, occasionally mass independent fractionations are observed
3. A classic example is observed in the oxygen isotopic composition of silicate minerals
in certain meteorites (Fig. 20)
-The oxygen isotopic composition of earth and moon -rocks fall along the
same line with a slope of ~0.52 for 17O vs 18O (which is evidence for
their common origin)
-However, Murchison or CAI meteorites, which are thought to have
formed by condensation reactions during the early or primitive stages of
the solar system (molten droplets in contact with nebular gases) have a
17O vs 18O slope = 1
4. A notable example in the modern earth involves oxygen isotopic composition
molecular O2, O3 and CO2 in the stratosphere. (Luz et al. Nature 400: 547-549, 1999)
-
Based on lab experiments it is thought that during the production of ozone in
the stratosphere, the ozone produced has a mass independent composition of
17
O and 18O compared to the isotopic composition of the molecular oxygen
from which is formed (Fig. 21)
-
Subsequently, the photolysis of ozone to produce O2 and an excited O atom
[O(1D)] in the stratosphere and the reaction of O(1D) with CO2, there is a
transfer of the mass independent 17O enrichment from the O3 to CO2
O2 + O  O*3
O*3 + hv  O2 + O*(1D)
O*(1D) + CO2  CO*3  CO*2 + O
21
-where the * indicates the mass independently 17O enriched oxygen
- the net effect is to enrich stratospheric CO2 and deplete stratospheric O2 in 17O
relative to that expected via mass dependent fractionation (Fig. 21)
-because of stratosphere-troposphere mixing, the isotopic anomaly on
stratospheric CO2 and O2 appears in the tropospheric O2 and CO2
-since only a very small portion of the tropospheric O2 pool comes from
the stratosphere, the tropospheric 17O/18O anomaly on O2 is small (~0.1 ‰ level)
5. In general, the reasons for mass independent effects are not completely understood,
despite 20 years or so of research. However, the implication is that whatever reaction
mechanism controls the isotopic composition of the reactants and products in these
situations (e.g., UV photodissociation, spark discharge, thermal decomposition, etc.), it
bypasses the normal thermodynamic (i.e., mass dependent) controls on the distribution of
isotopes.
E. Nomenclature Used to Represent Isotope Fractionation Effects (Review)
a. Epsilon () uses per mil (‰) notation and  uses decimal values that are close to a
value of 1. Both symbols are used to represent isotopic fractionation effects.
-remember:  (‰ ) = ( - 1 )*1000
b. For equilibrium fractionations
-the fractionation factor  equals the ratio of the isotope ratio of the compounds in
equilibrium, in the following example between dissolved CO2 gas and carbonate ion
-e.g.,
13
CO2 + 12CO3=  12CO2 +
13
CO3=
 = (13C/12C)CO3= / (13C/12C)CO2
 = 1.0093 at 25° C (measured, see Fig. 15)
thus  = 9.3 ‰ in this case
c. For kinetic fractionations
1. The isotope fractionation effect equals the ratio of the reaction rate constants (or
diffusion rates) of light and heavy isotopomers
22
2. For example, the photosynthesis reaction
a.
12
CO2 + H2O 
12
CH2O + O2 , where k12 represent rate constant
b.
13
CO2 + H2O 
13
CH2O + O2, where k13 represents rate constant
3.  = K13 / K12  0.980 (measured experimentally) and thus  = -20 ‰
d. Remember: You cannot assume a consistent nomenclature for isotopic fractionation
factors is used, so you need to confirm the  and ε notation definition used in a paper
otherwise you risk having the fractionation effects inverted (i.e., K12/K13 rather than
K13/K12). Thus verify the following:
-
Are they reporting  as Rproduct / Rreactant or Rreactant / Rproduct ?
-
Are they reporting  as Klight / Kheavy or Kheavy / Klight ?
-
Are they reporting  as Dlight / Dheavy or Dheavy / Dlight ?
III. CLOSED SYSTEM ISOTOPE FRACTIONATION EFFECTS
1. Mass and Isotope balances
a. If the isotopic composition of the material removed from a closed reservoir
differs from the isotopic composition of the reservoir itself, then the isotopic
composition of the reservoir will change over time in addition to the total
mass of the reservoir
a. A closed reservoir is one that is not exchanging with the surrounding
environment.
b. An illustration of such a closed system with isotope effects would be a
terrarium in which all the carbon present initially is in the form of CO2 gas and in which a
plant begins to grow from seed and convert CO2 to organic carbon.
c. In this situation, a carbon mass balance can be expressed as follows:
Ct = Co – Cp ,
(1)
where C= carbon in reservoir (as CO2), o=initial, t=time, p=carbon in plant at time
t. That is, the final amount of CO2 in the terrarium equals the initial amount minus the net
amount of carbon photosynthesized by the plant. Net = gross photosynthesis – respiration.
23
d. Here C represents the total carbon [12C (98.9%) + 13C (1.1%)] measured by a
non-mass discriminating analytical technique (e.g. gas chromatography). Since 12C makes
up 99% of the total carbon it is usually a reasonable approximation to assume that 12C is
equivalent to total carbon, i.e., 12C  13C + 12C or 12CO2  13CO2 + 12CO2.
e. The mass balance for the rare isotope is expressed as follows:
Ct*Rt = Co*Ro - Cp*Rp ,
(2)
-where R = isotope ratio (e.g. 13C/12C)
-thus C*R equals the total mass of rare isotope (not it’s ratio)
-i.e., C*R = [12CO2] * (13C/12C)CO2 = 13CO2
f. We can rewrite this rare isotope mass balance using  notation:
-that is: Ct(13Ct/1000+1)Rstd = Co(13Co/1000+1)Rstd – Cp(13Cp/1000+1)Rstd
(3)
-Remember: 13C (‰) = [Rsample/Rstandard – 1]*1000
-Note that the Rstd terms cancel in equation (3), as long as all of the 13C values
were reported relative to the same standard (e.g., PDB).
-If we multiply eq. (3) by 1000, after canceling Rstd, then
Ct(13Ct+1000) = Co(13Co+1000) – Cp(13Cp+1000)
Ct 13Ct +1000Ct = Co 13Co +1000Co – Cp 13Cp -1000Cp
-since Ct = Co – Cp (using mass balance in eq. 1), then
1000Ct = 1000Co - 1000Cp
-Thus, Ct(13Ct) = Co(13Co) - Cp(13Cp)
(4)
-this mass isotope balance is equivalent to the one described in eq. (2)
which is expressing using 13C/12C instead of 13C
2. A quantitative example of a mass and isotope balance
a. Determine the isotope fractionation factor for net carbon uptake during photosynthesis
using a closed system terrarium experiment described above.
-We have a terrarium with photosynthesizing plants for which we measure the
initial and final CO2 concentrations and 13C:
-initial conditions: CO2 concentration = 400 ppm and 13C of CO2 = -8.0 ‰
-final conditions:
CO2 concentration = 365 ppm and 13C of CO2 = -6.0 ‰
24
b. What is 13C isotope fractionation effect during CO2 fixation by photosynthesis based
on the above experimental results?
1. the isotope fractionation effect equals the ratio of the reaction rates for 12CO2
and 13CO2 during photosynthesis, i.e.,  = k12CO2 / k13CO2
2. the fractionation effect also equals the ratio of the 13C/12C of the carbon in the
plant (i.e., photosynthetic product) over the 13C/12C of the CO2 (i.e., photosynthetic
reactant), i.e.,  = (13C/12C)plant / (13C/12C)CO2
3. the most direct method to estimate  would be to measure the 13C of both the
CO2 and the plant material over the same time interval
c. However, with the measurements in hand we can estimate the kinetic isotope
fractionation effect (KIE or ) during photosynthesis, if we assume that the 13C of the
CO2 lost from the terrarium equals the 13C of the plant material. Is this legit?
-if we rearrange the above mass and isotope balance equations
(1)
Ct = Co – Cl,
(2)
Ct*Rt = Co*Ro - Cl*Rp,
where R = (13C/1000 + 1)*Rstd and l = lost.
Solving for Rp yields:
Rp = (RtCt - RoCo)/(Ct-Co);
-converting the measured 13C values (-8 and –6 ‰) to R values and accounting
for the change in pCO2 in the terrarium (from 400 to 365 ppm) yields
Rp = (0.994*365 - 0.992*400) / (365-400);
-NOTE: the Rstd’s in each “R” term cancel
-thus Rp = 0.9711
-converting to 13C = (0.9711-1)*1000 = -28.86 ‰
-Conclusion: based on the mass and isotope balance, we estimate the 13C of the
plant carbon to be –28.86 ‰
d. We can also do the calculation in terms of mass and 13C values (rather than R values)
Ct = Co – Cp
25
Rt*13Ct = 13Co*Co - 13C p*Cp
Combining:
13Cp = (13C t*Ct - 13C o*Co)/(Ct - Co)
13Cp = (365*(-6 ‰ ) - 400*(-8 ‰ ))/(365-400)
13Cp = -28.86 ‰
Conclusion: In this case we can use either 13C/12C or 13C values and get exactly
the same result, as we expected based on equations (2) and (4), described above.
e. In this experiment, 13Cp actually represents the 13C of the CO2 lost from the terrarium
-If we assume the missing CO2 is in the plant (i.e., there are no other CO2 sinks),
then the 13C of the missing CO2 (-28.86 ‰) equals the 13C of the organic carbon fixed
by the plant.
f. If this assumption is valid, then the KIE during photosynthesis () is equal to the ratio
of (13C/12C)product / (13C/12C)reactant
- In terms of our measured values, this implies that:
 = (13C/12C)plant / (13C/12C)CO2
-one source of uncertainty is the choice for the 13C of the CO2 (reactant) which
changes from –8 to –6 ‰ during the experiment.
-If we choose a mean values of –7 ‰ for the 13C of the CO2 then,
=
(-28.86/1000  1) * Rstd
= 0.97114/0.993 = 0.97800
(-7/1000  1) * Rstd
- in terms of ε, the value for the KIE is ( - 1) *1000
ε = (0.9780 – 1)* 1000 = -22.0 ‰
g. Experimental Conclusions:
1. the kinetic fractionation factor during net carbon fixation by plants is 0.978 or ~-22
‰, i.e., the organic carbon accumulating in the plant is ~22‰ more depleted in 13C
than the CO2 substrate used during photosynthesis
26
2. What about respiration effects? Does the net accumulation of carbon in the plant
represent net or gross photosynthesis?
h. Recommendation: Always work through the calculations in terms of isotope ratios
(R), rather than  values, and calculate the fractionation effect as an  value and then
convert to  at the end of the calculation
-The reason is that adding (or subtracting) ε values from  values is
approximately, but not exactly, equal to multiplying R values by  values. It is more
exact to work in terms of R and , and convert your answer to  and ε at the end.
-In the above example, an ε value = -22 ‰ was calculated based on  = (13C/12C)
product /
(13C/12C)reactant. In comparison, if we simply subtracted the 13C values of the
reactant and products we would calculate an ε of –21.86 ‰ (i.e., -28.86 ‰ - -7 ‰). This
is close, but not exactly the same.
- The bigger the difference in  values the less accurate they are. For example, 
= 0.700/ 0.990 = 0.7071 which yields an ε = -293 ‰. Subtracting the  values yields –
290 ‰ and a 3 ‰ difference. This is a particular concern when calculating deuterium
fractionation effects.
i. Important: In almost all mass and isotope balance calculations, the Rstd cancels in all
the “R” terms, so that for the actual calculations the “R” values can have numerical
values close to 1.000, i.e., (13C/1000 + 1), rather than values close to 0.011, i.e.,
(13C/1000 + 1)*0.0112372, where 0.0112372 equals the 13C/12C of the PDB standard.
This makes the calculations easier because you don’t need so many significant figures.
3. Rayleigh Distillation
a. Background
1. To use the mass balance approach, in the above example, one has to estimate how the
13
C/12C of the CO2 is changing over the time of the experiment (we assumed it was a
linear change) to determine the KIE during photosynthesis.
27
2. This assumption will yield inaccuracies if a significant portion of the reacting pool is
consumed because the change in the isotopic composition of the pool is not linear, as we
shall see.
3. Generally, then a mass and isotope balance approach (as used above) should only be
used when a small proportion of the initial material reacts (say <20%)
4. Generally, it is preferable to account explicitly for the isotopic change in the reacting
species as the reaction proceeds.
5.This is what the Rayleigh Distillation approach does, it accounts for the observation
that the isotopic composition of the reacting compounds does not change linearly but the
change accelerates as the fraction of starting material decreases.
b. An example of mass and isotope balance using the Rayleigh Distillation Method
1. Calculate the step-wise change in the 13C of the CO2 pool after three steps of
removing a constant amount (an increasing fraction) of CO2 from the pool. Assume the
initial 13C = -20 ‰ and the KIE or  is -10 ‰. We’ll use del values and ε values for this
illustration. The values 1, 0.2, 0.8, etc. below represent concentrations in units like
moles/liter or ppm.
2. Step 1:
1*(-20‰ ) - 0.2*(-20-10‰) = 0.8 at –17.5‰ ….. 13C = +2.5 ‰
Step 2:
0.8*(-17.5‰) - 0.2*(-17.5-10 ‰) = 0.6 at –14.17 ‰….. 13C =+3.33‰
Step 3:
0.6*(-14.17‰) - 0.2*(-14.17-10 ‰) = 0.4 at -9.17 ‰….. 13C =+5.00‰
-thus the 13C change per step magnifies greatly as CO2 pool shrinks in size
-What would a plot of 13C vs concentration look like in this case?
c. Deriving the Rayleigh Fractionation Relationship
28
1. The Rayleigh distillation describes the isotopic changes of a closed reservoir
undergoing continuous reduction in size.
2. We can derive the Rayleigh Fractionation relationship between reservoir size and
isotopic composition using the same example that we used above, i.e., the effect of
photosynthesis on the 13C/12C of CO2 in a closed system like a terrarium.
3. We start with two differential equations that describe the time rate of change of 13CO2
and 12CO2 as CO2 is photosynthetically consumed, assuming photosynthesis is a first
order process the rate of which depends on the concentration of the substrate and a
reaction rate constant (k):
d(13CO2 )/dt = -k13 * 13CO2
(4)
d(12CO2 )/dt = -k12 * 12CO2
(5)
a. The KIE during photosynthesis () is represented by the ratio of the reaction
rate constants for 13CO2 and 12CO2 , thus  = k13/k12
b. Dividing equation (4) by equation (5) yields:
d(13CO2 )/d(12CO2 ) = k13/k12 * 13CO2 /12CO2
rearranging this expression yields:
d(13CO2 )/13CO2 =  * d(12CO2 )/12CO2
- Integrating between the initial time (o) and final time (t) yields,
ln 13CO2t - ln 13CO2o =  *(ln 12CO2t -ln 12CO2o)
-i.e., the integral of d(13CO2 )/13CO2 = ln 13CO2
-this expression can be rewritten as:
ln (13CO2t/13CO2o) =  * ln (12CO2t/12CO2o)
-this expression is equivalent to:
13
CO2t/13CO2o = (12CO2t/12CO2o)
-dividing this expression by 12CO2t/12CO2o yields:

-1)
(13CO2t/12CO2t)/(13CO2o/12CO2o) =(12CO2t/12CO2o)(
-rearranging this expression yields:

(13C/12C)t / (13C/12C)o =(12CO2t/12CO2o)(
-1)
-this expression is equivalent to:
29

-1)
Rt/Ro = (Ct/Co)(
,
where R = 13C/12C and C = CO2.
-Summary: The expression for the Rayleigh Distillation relationship is:
Rt Ct

Ro Co
( 1)
,
where C = bulk concentration, R = isotope ratio, and  = KIE or isotope
fractionation factor (i.e., k*/k), where * represents the rare isotope and ‘k’ is the reaction
rate constant.
4. The Rayleigh distillation relationship describes the rate of change in the isotopic
composition of a reservoir as the size of the reservoir decreases as a result of a process
that isotopically fractionates during consumption of the reactant
-sometimes the Rayleigh Distillation is expressed as
Rt
 f ( 1) , where f = Ct/Co
Ro
(i.e., the fraction of initial concentration left at time t)
-use this relationship to relate isotope change in the reservoir as a function of the
changing reservoir size
a. Rearranging the Rayleigh equation yields an expression for determining  from
experimental measurements of the change in R and C
ln(Rt/Ro) = ( -1)* ln(f)
 = ln (Rt/Ro)/ ln (f) + 1
b. The Rayleigh expression and  can also be expressed in terms of  values:

-1)
(1+13Ct/1000)/ (1+13Co/1000) = f(
 = ln[( 13Ct +1000)/(13Co+1000)] / ln(f) + 1
c. If we repeat the calculation of  for the terrarium example used above for the
mass and isotope balance, we obtain the following:
 = ln[( -6‰ +1000)/(-8‰+1000)] / ln(365ppm/400ppm) + 1
 = ln(.994/.992)/ln(365ppm/400ppm) + 1
 = .9780
30
-thus we find that the  for net photosynthesis is –22 ‰ similar to the value of –
22 ‰ calculated using the mass and isotope balance approach (as described above)
-in this example, the Rayleigh approach and the mass/isotope balance approach
yield the same value of α.
-this agreement occurs because the change in the reservoir amount was only 10%
of the initial value (400 to 365 ppm). If the change in reservoir amount was significantly
greater, say 50%, then the mass/isotope balance approach would be less accurate than the
Rayleigh Distillation approach.
d. If we measured the concentration of CO2 and its 13C several times during the
experiment (rather than at the beginning and end only), a plot of ln(Rt) vs ln (f) would
yield a slope = ( - 1) and an intercept = ln(Ro).
IV. MEASUREMENT TECHNIQUES (not usually covered in class)
A. Theory behind isotope ratio measurement
1. Separation of molecules of different masses (isotopologues) is based on the
different pathways that ions of different masses have when they move through
a magnetic field
2. The velocity (v) if ions depends on the ionization state (e) and voltage potential
(V) applied to the ions, i.e., kinetic energy = acceleration energy
e*V = 1/2mv2
v=
2eV / m
where v=ion velocity, e= ion charge, V=acceleration potential and m=ion mass
3. As ion passes thru magnetic field, the ion attains a curved path where the
magnetic field force (MF) is balanced by centrifugal force (CF)
MF = H*e*v and CF = m*v2/r,
where H = magnetic field strength, r = radius of curvature
H*e*v = m*v2/r
r = m*v/(H*e) and after substituting for v yields
r = 1/H* 2 *V * m / e
31
4. In Summary:
a. The radius of curvature of ions is proportional to m/e at constant H and V
b. Thus ions of different mass and charge will have different curvature of
radius and separate from each other
c. Heavier ions have larger radius of curvature
-this is why D/H measurements on H2 gas need a separate detector
with a smaller curvature of radius than the other gases like CO2, O2
and N2 (Fig. 22)
B. How do we measure isotope ratios in practice?
1. Mass spec has four basic components (See Fig. 22)
a. gas inlet device
b. ion source/focus
c. mass separator
d. ion detector
2. Main components of mass spec are under high vacuum e.g. 10-8 Torr
-where 1 Torr = 1mm Hg = 1.3 mbar
-high vacuum yields a long mean free path (500m) for ions, i.e. there is
little chance that they will collide other ions within the instrument
3. Inlet system
a. gas sample and standard flasks
f. adjustable pressure reservoir (bellows)
g. capillary tubes that connect gas reservoir to mass spec
i.
~0.1 mm ID and ~0.5 to 1 meter long
ii. admit small amounts of gas under a pressure of ~100 Torr
d. the small ID of the capillaries is further crimped at the mass spec end.
Thus the capillary acts as a flow restriction and allows a significant
pressure difference to exist between the between the sample (and standard)
inlet and inside the mass spectrometer (Fig. 22)
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e. The significant pressure gradient across the length of the capillary
produces viscous flow, which is non-fractionating, and prevents molecular
flow, which would be fractionating.
f. If the pressure gradient across the capillary drops too low (usually because
the sample size is too small), it becomes insufficient to prevent molecular
diffusion fractionation effects that would cause the isotopic composition of
gas admitted to the mass spec to differ from the isotopic composition of
the gas in the sample reservoir
i. This situation typically occurs at gas pressures < 12 torr (mm Hg)
or sample sizes of less than around 0.1 micromole of gas
1. Remember: There are ~22.4 microliters (μL) of gas per
μmole of gas
2. Why would more sample pressure be needed for measuring
the isotopic composition of H2 compared to CO2?
ii. Sample consumption is typically about 0.1 nanomole per sec
(0.002 μL/s or 0.1 uL/min) for a static inlet system
g. Switching Valves
i. On the mass spec side of the capillaries is a switching or changeover valve (Fig. 22)
ii. It does two things: it switches between reference gas and sample
gas and at the same time switches between capillary and vacuum
(waste) pump that provides for constant gas flow through capillary
regardless of whether sample or standard is being measured
4. Ion source
a. pressures inside of ~10-3 to 10-4 torr
b. uses a bombardment of energic electrons (~100 eV) emitted from a hot
(2000°C) tungsten wire or filament to produce positive charged gas
ions (mostly +1 charge)
c. ionization efficiency is about 1 ion per 1000 molecules
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d. positive ions are accelerated using a positive 5-10 kV voltage potential
applied at the ion source housing
e. ions are essentially monoenergetic with a velocity of v= 2neV / m
5. Mass analyzer
a.
The flight tube is curved (e.g., typically 60° for C,N,O,S and 90°
for H) and placed within a magnetic field
b.
Ions accelerated in the source housing travel through the flight
tube
c.
ions of different masses have different curvature of radius in
tube and are thus separated from each other as they move down
flight tube
6. Ion detector
a. Faraday cups or collectors are positioned at the correct radius of
curvature to intercept appropriate ion (Fig. 22)
b. the ions that don’t have the correct mass will not enter the appropriate
Faraday cup
c. when high energy ions collide with collector plate, electrons are
emitted
d. number of electron emitted, i.e. current (10-8 to 10-11 amps), is
proportional to number of ions impacting plate
e. current is amplified (source of instability) and resistors (107 to 1011
ohms) are used to generate enough voltage (1-10V) to easily detect
f. voltage is proportional to number of ions impacting Faraday cup
which, in turn, is proportional to the number of molecules of a certain
mass
g. Voltage from each collector is simultaneously integrated over time and
recorded on computer
h. computer provides statistical filtering of data, electronic and mass
background corrections
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C. Mass versus elemental ratios
1. Remember: Mass spectrometers measure the isotope ratio of a sample gas relative to
the (known) isotope ratio of a standard gas. Mass specs do not measure absolute isotope
ratios.
b. The mass spec switches back and forth between sample and standard gases
every minute or so, typically for ~10 times during each analysis
c. The lab’s “working” gas standard has had its isotope ratio calibrated relative to
an accepted isotope reference gas like VPDB, VSMOW, etc.
d. Thus one rarely measures actual VPDB or VSMOW standards in the lab.
e. Typically, one gets an NBS (National Bureau of Standards) or NIST standard
which in turn has been calibrated to primary standard like VPDB or VSMOW
f. All mass spec measured isotope ratios are reported relative to a standard, never
as an absolute value.
2. Measuring the mass ratio of 45/44 in the mass spec is not the same as measuring
13
C/12C of CO2
- there are two CO2 molecules with mass 45: 13C-16O-16O and 12C-16O-17O
-in this case we want to convert from the measured 45/44 to the 13C/12C
-mass specs have software routines to correct for the C12O16O17 contribution to 45
mass based on measured mass 46/44 (12C-18O-16O / 12C-16O-16O)
-it is assumed that the 18O/17O ratio is constant at ~5 (0.20%/0.04%, see Fig. 6)
-then one corrects the measured 45/44 for the contribution by 12C-16O-17O using
the measured 46/44
D. Continuous Flow Isotope Ratio Mass Spectrometry (CF-IRMS)
1. The continuous flow technique uses a carrier gas to flow the specific gas (e.g.,
CO2, N2, O2) into the mass spectrometer source, rather than let the gas enter via a
capillary (Fig. 23)
2. The advantage is two fold:
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a. Much smaller sample sizes are needed (1000x less)
i. nano- and pico- molar sample sizes instead of micro molar
b. Using CF-IRMS, mixtures of gases or volatile compounds can be
separated in-line and thus eliminate sample preparation steps
3. Procedure:
- mixtures of gas phase compounds of interest (e.g. lipids, amino acids,
gases) are separated on a GC
-the compounds of interest are carried through either a combustion oven
where compounds or gases are combusted to CO2, N2, and H2O or a pyrolysis tube
where the compounds are converted to H2 and CO
- H2O is separated or adsorbed from the combustion gases
4. The continuous flow method admits a slug of sample gas (CO2 , N2, O2, or H2 )
into the mass spec’s source (along with a carrier gas) which is then ionized
-the flow rate into mass spec is typically about 0.5ml/min
-the gas makes up about 1% to 20% of carrier gas depending on sensitivity
of the mass spec
-this approach eliminates the minimum pressure required by the dual-inlet
capillary system
-the mass spec measures the integrated area of the signals for each of the
isotopologues (e.g., 44, 45 and 46 for CO2) simultaneously
-in this manner the entire gas sample is utilized for the isotope ratio
measurement (rather than only a small portion of the gas sample in the traditional static
dual inlet mode)
-thus GC-IRMS allows for isotope ratio measurement on much smaller
samples than dual-inlet method
5. Because of short ionization times (seconds rather than minutes), the
instrumental precision isn't quite as good (±0.1 ‰) as a dual-inlet method but it is often
good enough
-standards are run sequentially, rather than alternately as in dual-inlet mode
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-continuous flow is a very useful technique because individual compounds and
very small volumes can be measured for isotopic composition
-allows one to measure isotope ratios on individual organic compounds, trace
gases, etc.
10. Method can be used for 15N on N2 and D of H2 by having steps to reduce N2O
to N2 and using pyrolysis, rather than combustion, to produce H2 from organic
material
E. Liquid Chromatography- Mass Spectrometry
1. Recent methodological development merges liquid chromatography (LC) with
continuous flow gas ratio mass spectrometry to measure the 13C/12C.
2. It uses a series of steps:
1. HPLC separation of compounds (amino acids, lipids, carbohydrates, drugs)
2. aqueous oxidation of compound of interest to CO2
3. CO2 separation using He carrier gas
4. H2O removal
5. CO2 admission to mass spec via continuous flow inlet.
3. The quoted external 13C precision is around ±0.3 ‰ and sample size limits of 20
nanomoles of C.
F. Laser-based instrumental methods (Fig. 24)
1. A new method for isotope ratio measurement is based on adsorption of light emitted by
lasers at the specific wavelengths that correspond to the slightly different adsorption
peaks for each of the isotopologues of a molecule.
2. The advantage of this technique is the small size (benchtop), cheap cost of the laser
instrument compared to traditional IRMSs ($30K vs $300K) and capability to perform
continuous isotopic measurements under in-situ conditions
-these laser instruments can be taken into the field.
3. The precision is better than ±0.1 ‰ with measurements occurring every second or so,
which can yield very high precision with increased integration times.
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4. Currently, these instruments are available for isotope measurements on only CO2 and
H2O gases.
5. They have a tremendous potential for biogeochemical applications.
V. QUESTIONS
1.What thermodynamic property of isotopologues is the primary cause the equilibrium
and kinetic fractionation effects?
2. Why is the D (‰) range ~10x the range in 13C, 15N and 18O?
3. What two general rules do kinetic and equilibrium fractionation effects typically
follow?
4. Why are most isotope fractionation effects empirically, rather than theoretically,
obtained?
5. Why are there KIEs resulting from molecular diffusion but not turbulent diffusion?
6. What is the relationship that describes the magnitude of the isotope effect and mass
difference between isotopes?
6. Under what conditions does the mass balance calculation approach yield a similar
fractionation effect as the Rayleigh approach?
7. What would be the isotopic composition of the plant in the terrarium experiment,
described in the notes, if all the CO2 were photosynthetically fixed or organic carbon?
8. Think of a real situation where you could use the Rayleigh relationship to calculate ?
9. What condition that often applies in real situations prevents the application of either
mass balance or Rayleigh distillation approaches?
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