In dynamic-R/RTM: Reaction Transport Modelling: Course Tutorials and Exercises
Introduction
Evaporation and precipitation of water
It is recommended that you first become familiar with the model describing water evaporation and precipitation. You can do this by typing the following command in the R-console:
RTMexercise("evaporation")
The key equations governing that model are summarized here.
- The change in the amount of water molecules in the liquid and gas phase is described by the following differential equation:
\begin{equation}
\label{eq1a}
\frac{dN_{liq}}{dt} = - \frac{dN_{vap}}{dt} = - k_f\cdot A_{liq}
-
k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}},
\end{equation}
where $k_f$ (in $mol~m^{-2}~s^{-1}$) and $k_b$ (in $m~s^{-1}$) are rate constants describing the forward (evaporation) and backward (precipitation) reaction, respectively, $V_{gas}$ (in $m^3$) is the volume of the gas phase (containing water vapor), and $A_{liq}$ (in $m^2$) is the area of the liquid water in contact with the gas phase.
-
The rate constants can be quantified using the following formulas:
\begin{equation}
\label{eq2a}
\begin{array}{l}
\displaystyle
k_f = \frac{P_{vap}^}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v), \[5mm]
\displaystyle
k_b = \frac{R\cdot T}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v),
\end{array}
\end{equation}
where $\Delta \bar{H}{vap}$ is the molar enthalpy of vaporization of water ($\approx 40.66\,kJ\,mol^{-1}$), $P{vap}^$ is the equilibrium vapor pressure above the liquid water, $v$ is the wind velocity (in $m~s^{-1}$), and $c_1 = 0.0888~m~s^{-1}$ and $c_2=0.0783$ are empirical constants (Carrier, 1918).
For a specific temperature, $T$, the equilibrium vapor pressure can be calculated from the Clausius-Clapeyron equation according to
\begin{equation}
\label{eq3a}
P_{vap}^(T) = P_{vap}^(T_{boil}) \cdot \exp\left[
-\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T} - \frac{1}{T{boil}}\right)\right],
\end{equation}
where $T_{boil} = 373.15\,K$ is the boiling temperature of water at a pressure of $1\,Atm$ and $P_{vap}^*(T_{boil}) = 1\,Atm = 101325\,Pa$.
Isotope ratio, atom fraction, $\delta$-notation
Before we expand the above equations towards water isotopologues $H_2{}^{16}O$ and $H_2{}^{18}O$, we introduce quantities that are commonly used in stable isotope research. For a given sample, $s$, we denote the number of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the sample by ${}^{16}N_s$ and ${}^{18}N_s$, respectively, and define the ${}^{18}O/{}^{16}O$ isotope ratio according to
\begin{equation}\label{eq1}
R_s = \frac{{}^{18}N_{s}}{{}^{16}N_{s}}. \qquad
\end{equation}
When performing mass balance calculations, a more convenient quantity is the so-called ${}^{18}O$ atom fraction, which is defined according to
\begin{equation}\label{eq1x}
x_s = \frac{{}^{18}N_{s}}{{}^{18}N_{s} + {}^{16}N_{s}}.
\end{equation}
In this definition, we assumed that the contribution of the $H_2{}^{17}O$ to all stable isotopologues of water is minor and therefore negligible. We will make this assumption throughout this document.
Equations \ref{eq1} and \ref{eq1x} imply the following relationships between $x_s$ and $R_s$:
\begin{equation}\label{eq2}
x_s = \frac{R_s}{1+R_s}, \qquad R_s = \frac{x_s}{1-x_s}.
\end{equation}
They can be verified by direct substitution.
Since absolute isotope ratios or atom fractions are difficult to determine analytically, they are typically expressed as a relative difference between two samples. These differences are typically expressed in the $\delta$ notation. Specifically, the relative difference between the isotope ratios in sample b versus sample a is defined according to
\begin{equation}\label{eq3}
\delta_a (b) = \frac{R_b}{R_a} - 1,
\end{equation}
where $R_a$ and $R_b$ are the corresponding isotope ratios. Most commonly, the $\delta$ value calculated by Eq. \ref{eq3} is multiplied by 1000 and reported in permil (\textperthousand).
Kinetic isotope fractionation
Following the introduction to the basic terminology and symbols used in stable isotope research, we are now ready to formulate the model of water precipitation and evaporation separately for the $H_2{}^{16}O$ and $H_2{}^{18}O$ isotopologues. We make the following assumptions:
-
the rate of evaporation of $^{18}O$-water is proportional to the probability that the water molecule in the liquid phase contains the isotope $^{18}O$. This probability is equal to the ${}^{18}O$ atom fraction in the liquid phase, $x_{liq}$, which is calculated according to Eqs. \ref{eq1} and \ref{eq2} based on the number of $H_2^{18}O$ and $H_2^{16}O$ in the liquid phase, i.e., $^{18}N_{liq}$ and $^{16}N_{liq}$.
-
the rate of precipitation of $^{18}O$-water is proportional to the probability that the water molecule in the gas phase contains the isotope $^{18}O$. This probability is equal to the ${}^{18}O$ atom fraction in the water vapor, $x_{vap}$, which is calculated according to Eqs. \ref{eq1} and \ref{eq2} based on the number of $H_2^{18}O$ and $H_2^{16}O$ in the vapor, i.e., $^{18}N_{vap}$ and $^{16}N_{vap}$.
-
the rate of evaporation of $^{16}O$-water is proportional to the probability that the water molecule in the liquid phase contains the isotope $^{16}O$. This probability is equal to $1-x_{liq}$.
-
the rate of precipitation of $^{16}O$-water is proportional to the probability that the water molecule in the gas phase contains the isotope $^{16}O$. This probability is equal to $1-x_{vap}$.
Based on these assumptions, the changes in the amounts of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the gas phase are described by the following differential equations (compare with Eq. \ref{eq1a}):
\begin{equation}\label{eq4}
\begin{array}{rcl}
\displaystyle
\frac{{}^{18}N_{vap}}{dt} & = &
\displaystyle
{}^{18}k_f \cdot x_{liq} \cdot A_{liq} -
{}^{18}k_b\cdot x_{vap} \cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}}, \[5mm]
\displaystyle
\frac{{}^{16}N_{vap}}{dt} & = &
\displaystyle
{}^{16}k_f \cdot (1-x_{liq}) \cdot A_{liq} - {}^{16}k_b\cdot (1-x_{vap}) \cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}},
\end{array}
\end{equation}
where ${}^{16}k_f$, ${}^{16}k_b$, ${}^{18}k_f$ and ${}^{18}k_b$ are rate constants and $N_{vap}={}^{18}N_{vap} + {}^{16}N_{vap}$ is the total amount of $H_2O$ molecules in the gas phase. We emphasize that for both the forward and backward reaction, the corresponding rate constants differ between the $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules, i.e., ${}^{16}k_f\neq {}^{18}k_f$ and ${}^{16}k_b\neq {}^{18}k_b$. As we will see later (see section "Rate constants" below), this difference will ultimately lead to the so-called equilibrium fractionation effect, i.e., the fact that in a thermodynamic equilibrium, the $^{18}O/^{16}O$ isotope ratio in the water vapor differs from the isotope ratio in the liquid water.
Kinetic versus equilibrium fractionation
In an equilibrium, the time derivatives are equal to zero. Using an asterisk () to denote equilibrium values, this condition yields the following relationships:
\begin{equation} \label{eq6}
\begin{array}{rcl}
\displaystyle
\frac{{}^{18}N_{vap}}{dt} = 0 & \rightarrow &
\displaystyle
{}^{18}k_f \cdot x_{liq}^ = {}^{18}k_b\cdot x_{vap}^ \cdot \left( \frac{N_{vap}}{V_{gas}}\right)^, \[5mm]
\displaystyle
\frac{{}^{16}N_{vap}}{dt} = 0 & \rightarrow &
\displaystyle
{}^{16}k_f \cdot (1-x_{liq}^) = {}^{16}k_b\cdot (1-x_{vap}^) \cdot \left( \frac{N_{vap}}{V_{gas}}\right)^*.
\end{array}
\end{equation}
By dividing the first equation in Eq. \ref{eq6} with the second equation, we obtain
\begin{equation} \label{eq8}
\frac{{}^{18}k_f}{{}^{16}k_f}\cdot \frac{x_{liq}^}{1-x_{liq}^} = \frac{{}^{18}k_b}{{}^{16}k_b} \cdot \frac{x_{vap}^}{1-x_{vap}^}.
\end{equation}
This relationship can be written in a simpler form if we consider the following definitions. First, we define the kinetic fractionation factor, $\alpha_{liq}(vap)$, according to
\begin{equation} \label{eq9}
\alpha_{liq} (vap) \equiv \frac{\displaystyle
\left(\frac{{}^{18}k_b}{{}^{16}k_b}\right)}{\displaystyle \left(\frac{{}^{18}k_f}{{}^{16}k_f}\right)}.
\end{equation}
Additionally, we define the relative difference in the isotopic composition of water vapor versus liquid water according to (see Eq. \ref{eq3})
\begin{equation} \label{eq10}
\delta_{liq} (vap) = \frac{R_{vap}}{R_{liq}}-1
\end{equation}
and the corresponding equilibrium value according to
\begin{equation} \label{eq10eq}
\delta^{liq} (vap) = \frac{R{vap}^}{R_{liq}^*}-1.
\end{equation}
By combining Eqs. \ref{eq9}--\ref{eq10eq} with Eq. \ref{eq2}, the relationship in Eq. \ref{eq8} can be rewritten according to:
\begin{equation} \label{eq11}
\delta^{liq}(vap) = \frac{1-\alpha{liq}(vap)}{\alpha_{liq}(vap)} \quad \mathrm{or} \quad \alpha_{liq}(vap)=\frac{1}{1+\delta^{liq}(vap)} =
\frac{R^_{liq}}{R^{vap}}.
\end{equation}
This equation shows that there is an intimate relationship between the kinetic and equilibrium isotope fractionation effects. On the one hand, $\alpha_{liq}(vap)$ is a measure of the differences between the rate constants of the forward (evaporation) and backward (precipitation) reactions for the lighter and heavier water isotopologues (Eq. \ref{eq9}). On the other hand, $\delta^_{liq}(vap)$ is a measure of the difference between the isotopic composition (i.e., the ${}^{18}O/{}^{16}O$ isotope ratio) of the water vapor and liquid water in a thermodynamic equilibrium (Eq. \ref{eq10eq}). Equation \ref{eq11} therefore provides an important insight, which underpins the use of stable oxygen isotopes in the study of the hydrological cycle on Earth: if the rate constants of evaporation and precipitation of water differ between the $H_2{}^{16}O$ and $H_2{}^{18}O$ isotopologues such that the kinetic fractionation factor $\alpha_{liq}(vap)$ (Eq. \ref{eq9}) is different from 1, the isotopic composition of the water vapor will differ from the isotopic composition of the liquid water. In a thermodynamic equilibrium, this latter difference will be equal to $\delta^{liq}(vap)$, which can be calculated from $\alpha{liq}(vap)$ according to Eq. \ref{eq11}. In practice, this conclusion is applied in reverse: by determining the equilibrium fractionation value $\delta^*{liq}(vap)$ experimentally, the kinetic fractionation factor $\alpha{liq}(vap)$ can be calculated from the second formula in Eq. \ref{eq11} (see next section).
Rate constants
Here we synthesize the results from the previous sections to show how to calculate the values of the rate constants ${}^{16}k_f$, ${}^{16}k_b$, ${}^{18}k_f$ and ${}^{18}k_b$ based on other empirical values.
Empirically, it was determined that in a thermodynamic equilibrium, water vapor is by about 10\textperthousand\ depleted in ${}^{18}O$ relative to liquid water, i.e., $\delta^_{liq}(vap) \approx -10$ \textperthousand. (Note that although $\delta^{liq}(vap)$ depends on temperature, we neglect this dependency in this reader.) Combining this value with Eq. \ref{eq11}, we obtain $\alpha{liq}(vap)=1.0101$. We emphasize that this value is greater than 1 if $\delta^*_{liq}(vap)$ is negative.
Next, we combine the expressions for $k_f$ and $k_b$ in Eq. \ref{eq2a} with the definition of $\alpha_{liq}(vap)$ (Eq. \ref{eq9}). Assuming that the empirical coefficients $c_1$ and $c_2$ are the same for $H_2{}^{16}O$ and $H_2{}^{18}O$, we obtain
\begin{equation} \label{eq12}
\alpha_{liq}(vap)=\frac{{}^{16}P_{vap}^}{{}^{18}P_{vap}^}.
\end{equation}
Because we showed that $\alpha_{liq}(vap)>1$, this relationship shows that ${}^{16}P_{vap}^ > {}^{18}P_{vap}^$, i.e., the equilibrium vapor pressure for the isotopically lighter water is greater than that for the heavier water.
Next, we combine this result with the Clausius-Clapeyron equation (Eq. \ref{eq3a}). Assuming that the boiling temperature ($T_{boil}$) is the same for $H_2{}^{16}O$ and $H_2{}^{18}O$, we obtain
$$
\Delta^{18} \bar{H}{vap} = \Delta^{16} \bar{H}{vap} + R\cdot \frac{T_{boil}\cdot T}{T_{boil}-T} \cdot \ln\alpha_{liq}(vap).
$$
This relationship allows us to calculate $\Delta{}^{18} \bar{H}{vap}$ based on the value of $\Delta{}^{16} \bar{H}{vap}$. Specifically, it shows that $\Delta{}^{18} \bar{H}{vap} > \Delta{}^{16} \bar{H}{vap}$ (because $\alpha_{liq}(vap)>1$). In practical calculations, we assume that the molar enthalpy of vaporization of pure $H_2^{16}O$ is equal to the value for the natural water (i.e., $\Delta{}^{16} \bar{H}_{vap} = 40.66~kJ~mol^{-1}$), which is a good approximation given that natural water only contains about 0.2% of $H_2{}^{18}O$.
Finally, we use Eq. \ref{eq2a} with the above results to calculate the rate constants for water evaporation (${}^{16}k_f$ and ${}^{18}k_f$) and precipitation (${}^{16}k_b$ and ${}^{18}k_b$).
Tasks
Task 1 --- equilibrium isotope fractionation
Consider a gas-tight container partially filled with liquid water and air (Figure 1). Using differential equations in Eq. \ref{eq4}, implement a model that can predict the amount of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the water vapor and liquid water as a function of time. Specifically, demonstrate that, in an equilibrium, the ${}^{18}O/{}^{16}O$ isotope ratio in the liquid water differs from the isotope ratio in the water vapor. Run the model assuming that the air is initially dry or fully saturated with water vapor (i.e., the initial relative air humidity is 0 or 100%, respectively).
{width=30%}
Assumptions
-
Temperature of the system and the total pressure of the gas phase are constant ($T=298~K$, $P_{tot} = 1~Atm = 101325~Pa$). You can assume that this is accomplished by the walls of the gas-tight container being flexible (to keep the pressure constant) and heat being absorbed or supplied by the surrounding, as needed (to keep the temperature constant).
-
The area of the water surface in contact with the gas phase is $A_{liq} = 1~m^2$ (constant), the initial volume of the liquid water is $V_{liq} = 1~L$, and the initial ${}^{18}O$ atom fraction of the liquid water is 0.002. For the simulation with the initial air humidity of 100%, assume the initial ${}^{18}O$ atom fraction of the water vapor of 0.002.
-
The initial total volume of the gas-tight container is $V_{tot}=V_{gas}+V_{liq} = 1~m^3$.
-
Both air and water vapor behave as an ideal gas, i.e., their volume, partial pressure, and temperature are related through the ideal gas law ($P\cdot V = n\cdot R\cdot T$).
-
The liquid and gas phases are well mixed (i.e., homogeneous). Thus, you can ignore transport processes and base your model entirely on mass balances (Eqs. \ref{eq4}) and the ideal gas law.
-
Dissolution of air components such as $N_2$ or $O_2$ in water can be ignored.
-
The ratio between the molar mass and density of water is the same for both $H_2{}^{16}O$ and $H_2{}^{18}O$.
-
In a thermodynamic equilibrium, water vapor is by about 10\textperthousand\ depleted in ${}^{18}O$ relative to liquid water, i.e., $\delta^*_{liq}(vap) \approx -10$\textperthousand.
Task 2 --- isotope exchange between two water reservoirs
Consider an experimental setup where two open containers are enclosed in a gas-tight container filled with air (Figure 2). Both containers are partially filled with water, and the air is initially fully saturated with water vapor. Although there is no direct mixing of liquid water between water reservoirs 1 and 2, there will be an isotope exchange between them due to evaporation and precipitation (i.e., the two liquid reservoirs are "connected" through the water vapor reservoir). Your task is to model the dynamics of this exchange, i.e., predict the isotopic composition of the water vapor and the two liquid water reservoirs as a function of time. How long does it take for an isotopic equilibrium among all water reservoirs to be reached? This task was inspired by the stable isotope labeling experiment used to study the dynamics of polyphosphate in cable bacteria (Geerlings et al. 2022, https://doi.org/10.3389/fmicb.2022.883807).
{width=50%}
Assumptions
In addition to the assumptions listed for Task 1, assume that
-
for both liquid water reservoirs, the area of the water surface in contact with the gas phase is $0.5~m^2$ (constant), the initial volume is $0.5~L$.
-
the initial ${}^{18}O$ atom fraction is 0.002 for the first liquid water reservoir and 0.004 for the second liquid water reservoir.
-
the initial ${}^{18}O$ atom fraction is 0.002 for the water vapor.
-
the initial total volume of the gas-tight container is $V_{tot}=V_{gas}+V_{liq,1} +V_{liq,2} = 1~m^3$.
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Introduction
Evaporation and precipitation of water
It is recommended that you first become familiar with the model describing water evaporation and precipitation. You can do this by typing the following command in the R-console:
RTMexercise("evaporation")
The key equations governing that model are summarized here.
- The change in the amount of water molecules in the liquid and gas phase is described by the following differential equation: \begin{equation} \label{eq1a} \frac{dN_{liq}}{dt} = - \frac{dN_{vap}}{dt} = - k_f\cdot A_{liq}
-
k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}}, \end{equation} where $k_f$ (in $mol~m^{-2}~s^{-1}$) and $k_b$ (in $m~s^{-1}$) are rate constants describing the forward (evaporation) and backward (precipitation) reaction, respectively, $V_{gas}$ (in $m^3$) is the volume of the gas phase (containing water vapor), and $A_{liq}$ (in $m^2$) is the area of the liquid water in contact with the gas phase.
-
The rate constants can be quantified using the following formulas: \begin{equation} \label{eq2a} \begin{array}{l} \displaystyle k_f = \frac{P_{vap}^}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v), \[5mm] \displaystyle k_b = \frac{R\cdot T}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v), \end{array} \end{equation} where $\Delta \bar{H}{vap}$ is the molar enthalpy of vaporization of water ($\approx 40.66\,kJ\,mol^{-1}$), $P{vap}^$ is the equilibrium vapor pressure above the liquid water, $v$ is the wind velocity (in $m~s^{-1}$), and $c_1 = 0.0888~m~s^{-1}$ and $c_2=0.0783$ are empirical constants (Carrier, 1918). For a specific temperature, $T$, the equilibrium vapor pressure can be calculated from the Clausius-Clapeyron equation according to \begin{equation} \label{eq3a} P_{vap}^(T) = P_{vap}^(T_{boil}) \cdot \exp\left[ -\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T} - \frac{1}{T{boil}}\right)\right], \end{equation} where $T_{boil} = 373.15\,K$ is the boiling temperature of water at a pressure of $1\,Atm$ and $P_{vap}^*(T_{boil}) = 1\,Atm = 101325\,Pa$.
Isotope ratio, atom fraction, $\delta$-notation
Before we expand the above equations towards water isotopologues $H_2{}^{16}O$ and $H_2{}^{18}O$, we introduce quantities that are commonly used in stable isotope research. For a given sample, $s$, we denote the number of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the sample by ${}^{16}N_s$ and ${}^{18}N_s$, respectively, and define the ${}^{18}O/{}^{16}O$ isotope ratio according to \begin{equation}\label{eq1} R_s = \frac{{}^{18}N_{s}}{{}^{16}N_{s}}. \qquad \end{equation} When performing mass balance calculations, a more convenient quantity is the so-called ${}^{18}O$ atom fraction, which is defined according to \begin{equation}\label{eq1x} x_s = \frac{{}^{18}N_{s}}{{}^{18}N_{s} + {}^{16}N_{s}}. \end{equation} In this definition, we assumed that the contribution of the $H_2{}^{17}O$ to all stable isotopologues of water is minor and therefore negligible. We will make this assumption throughout this document.
Equations \ref{eq1} and \ref{eq1x} imply the following relationships between $x_s$ and $R_s$: \begin{equation}\label{eq2} x_s = \frac{R_s}{1+R_s}, \qquad R_s = \frac{x_s}{1-x_s}. \end{equation} They can be verified by direct substitution.
Since absolute isotope ratios or atom fractions are difficult to determine analytically, they are typically expressed as a relative difference between two samples. These differences are typically expressed in the $\delta$ notation. Specifically, the relative difference between the isotope ratios in sample b versus sample a is defined according to \begin{equation}\label{eq3} \delta_a (b) = \frac{R_b}{R_a} - 1, \end{equation} where $R_a$ and $R_b$ are the corresponding isotope ratios. Most commonly, the $\delta$ value calculated by Eq. \ref{eq3} is multiplied by 1000 and reported in permil (\textperthousand).
Kinetic isotope fractionation
Following the introduction to the basic terminology and symbols used in stable isotope research, we are now ready to formulate the model of water precipitation and evaporation separately for the $H_2{}^{16}O$ and $H_2{}^{18}O$ isotopologues. We make the following assumptions:
-
the rate of evaporation of $^{18}O$-water is proportional to the probability that the water molecule in the liquid phase contains the isotope $^{18}O$. This probability is equal to the ${}^{18}O$ atom fraction in the liquid phase, $x_{liq}$, which is calculated according to Eqs. \ref{eq1} and \ref{eq2} based on the number of $H_2^{18}O$ and $H_2^{16}O$ in the liquid phase, i.e., $^{18}N_{liq}$ and $^{16}N_{liq}$.
-
the rate of precipitation of $^{18}O$-water is proportional to the probability that the water molecule in the gas phase contains the isotope $^{18}O$. This probability is equal to the ${}^{18}O$ atom fraction in the water vapor, $x_{vap}$, which is calculated according to Eqs. \ref{eq1} and \ref{eq2} based on the number of $H_2^{18}O$ and $H_2^{16}O$ in the vapor, i.e., $^{18}N_{vap}$ and $^{16}N_{vap}$.
-
the rate of evaporation of $^{16}O$-water is proportional to the probability that the water molecule in the liquid phase contains the isotope $^{16}O$. This probability is equal to $1-x_{liq}$.
-
the rate of precipitation of $^{16}O$-water is proportional to the probability that the water molecule in the gas phase contains the isotope $^{16}O$. This probability is equal to $1-x_{vap}$.
Based on these assumptions, the changes in the amounts of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the gas phase are described by the following differential equations (compare with Eq. \ref{eq1a}): \begin{equation}\label{eq4} \begin{array}{rcl} \displaystyle \frac{{}^{18}N_{vap}}{dt} & = & \displaystyle {}^{18}k_f \cdot x_{liq} \cdot A_{liq} - {}^{18}k_b\cdot x_{vap} \cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}}, \[5mm] \displaystyle \frac{{}^{16}N_{vap}}{dt} & = & \displaystyle {}^{16}k_f \cdot (1-x_{liq}) \cdot A_{liq} - {}^{16}k_b\cdot (1-x_{vap}) \cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}}, \end{array} \end{equation} where ${}^{16}k_f$, ${}^{16}k_b$, ${}^{18}k_f$ and ${}^{18}k_b$ are rate constants and $N_{vap}={}^{18}N_{vap} + {}^{16}N_{vap}$ is the total amount of $H_2O$ molecules in the gas phase. We emphasize that for both the forward and backward reaction, the corresponding rate constants differ between the $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules, i.e., ${}^{16}k_f\neq {}^{18}k_f$ and ${}^{16}k_b\neq {}^{18}k_b$. As we will see later (see section "Rate constants" below), this difference will ultimately lead to the so-called equilibrium fractionation effect, i.e., the fact that in a thermodynamic equilibrium, the $^{18}O/^{16}O$ isotope ratio in the water vapor differs from the isotope ratio in the liquid water.
Kinetic versus equilibrium fractionation
In an equilibrium, the time derivatives are equal to zero. Using an asterisk () to denote equilibrium values, this condition yields the following relationships: \begin{equation} \label{eq6} \begin{array}{rcl} \displaystyle \frac{{}^{18}N_{vap}}{dt} = 0 & \rightarrow & \displaystyle {}^{18}k_f \cdot x_{liq}^ = {}^{18}k_b\cdot x_{vap}^ \cdot \left( \frac{N_{vap}}{V_{gas}}\right)^, \[5mm] \displaystyle \frac{{}^{16}N_{vap}}{dt} = 0 & \rightarrow & \displaystyle {}^{16}k_f \cdot (1-x_{liq}^) = {}^{16}k_b\cdot (1-x_{vap}^) \cdot \left( \frac{N_{vap}}{V_{gas}}\right)^*. \end{array} \end{equation}
By dividing the first equation in Eq. \ref{eq6} with the second equation, we obtain \begin{equation} \label{eq8} \frac{{}^{18}k_f}{{}^{16}k_f}\cdot \frac{x_{liq}^}{1-x_{liq}^} = \frac{{}^{18}k_b}{{}^{16}k_b} \cdot \frac{x_{vap}^}{1-x_{vap}^}. \end{equation} This relationship can be written in a simpler form if we consider the following definitions. First, we define the kinetic fractionation factor, $\alpha_{liq}(vap)$, according to \begin{equation} \label{eq9} \alpha_{liq} (vap) \equiv \frac{\displaystyle \left(\frac{{}^{18}k_b}{{}^{16}k_b}\right)}{\displaystyle \left(\frac{{}^{18}k_f}{{}^{16}k_f}\right)}. \end{equation} Additionally, we define the relative difference in the isotopic composition of water vapor versus liquid water according to (see Eq. \ref{eq3}) \begin{equation} \label{eq10} \delta_{liq} (vap) = \frac{R_{vap}}{R_{liq}}-1 \end{equation} and the corresponding equilibrium value according to \begin{equation} \label{eq10eq} \delta^{liq} (vap) = \frac{R{vap}^}{R_{liq}^*}-1. \end{equation}
By combining Eqs. \ref{eq9}--\ref{eq10eq} with Eq. \ref{eq2}, the relationship in Eq. \ref{eq8} can be rewritten according to: \begin{equation} \label{eq11} \delta^{liq}(vap) = \frac{1-\alpha{liq}(vap)}{\alpha_{liq}(vap)} \quad \mathrm{or} \quad \alpha_{liq}(vap)=\frac{1}{1+\delta^{liq}(vap)} = \frac{R^_{liq}}{R^{vap}}. \end{equation} This equation shows that there is an intimate relationship between the kinetic and equilibrium isotope fractionation effects. On the one hand, $\alpha_{liq}(vap)$ is a measure of the differences between the rate constants of the forward (evaporation) and backward (precipitation) reactions for the lighter and heavier water isotopologues (Eq. \ref{eq9}). On the other hand, $\delta^_{liq}(vap)$ is a measure of the difference between the isotopic composition (i.e., the ${}^{18}O/{}^{16}O$ isotope ratio) of the water vapor and liquid water in a thermodynamic equilibrium (Eq. \ref{eq10eq}). Equation \ref{eq11} therefore provides an important insight, which underpins the use of stable oxygen isotopes in the study of the hydrological cycle on Earth: if the rate constants of evaporation and precipitation of water differ between the $H_2{}^{16}O$ and $H_2{}^{18}O$ isotopologues such that the kinetic fractionation factor $\alpha_{liq}(vap)$ (Eq. \ref{eq9}) is different from 1, the isotopic composition of the water vapor will differ from the isotopic composition of the liquid water. In a thermodynamic equilibrium, this latter difference will be equal to $\delta^{liq}(vap)$, which can be calculated from $\alpha{liq}(vap)$ according to Eq. \ref{eq11}. In practice, this conclusion is applied in reverse: by determining the equilibrium fractionation value $\delta^*{liq}(vap)$ experimentally, the kinetic fractionation factor $\alpha{liq}(vap)$ can be calculated from the second formula in Eq. \ref{eq11} (see next section).
Rate constants
Here we synthesize the results from the previous sections to show how to calculate the values of the rate constants ${}^{16}k_f$, ${}^{16}k_b$, ${}^{18}k_f$ and ${}^{18}k_b$ based on other empirical values.
Empirically, it was determined that in a thermodynamic equilibrium, water vapor is by about 10\textperthousand\ depleted in ${}^{18}O$ relative to liquid water, i.e., $\delta^_{liq}(vap) \approx -10$ \textperthousand. (Note that although $\delta^{liq}(vap)$ depends on temperature, we neglect this dependency in this reader.) Combining this value with Eq. \ref{eq11}, we obtain $\alpha{liq}(vap)=1.0101$. We emphasize that this value is greater than 1 if $\delta^*_{liq}(vap)$ is negative.
Next, we combine the expressions for $k_f$ and $k_b$ in Eq. \ref{eq2a} with the definition of $\alpha_{liq}(vap)$ (Eq. \ref{eq9}). Assuming that the empirical coefficients $c_1$ and $c_2$ are the same for $H_2{}^{16}O$ and $H_2{}^{18}O$, we obtain \begin{equation} \label{eq12} \alpha_{liq}(vap)=\frac{{}^{16}P_{vap}^}{{}^{18}P_{vap}^}. \end{equation} Because we showed that $\alpha_{liq}(vap)>1$, this relationship shows that ${}^{16}P_{vap}^ > {}^{18}P_{vap}^$, i.e., the equilibrium vapor pressure for the isotopically lighter water is greater than that for the heavier water.
Next, we combine this result with the Clausius-Clapeyron equation (Eq. \ref{eq3a}). Assuming that the boiling temperature ($T_{boil}$) is the same for $H_2{}^{16}O$ and $H_2{}^{18}O$, we obtain $$ \Delta^{18} \bar{H}{vap} = \Delta^{16} \bar{H}{vap} + R\cdot \frac{T_{boil}\cdot T}{T_{boil}-T} \cdot \ln\alpha_{liq}(vap). $$ This relationship allows us to calculate $\Delta{}^{18} \bar{H}{vap}$ based on the value of $\Delta{}^{16} \bar{H}{vap}$. Specifically, it shows that $\Delta{}^{18} \bar{H}{vap} > \Delta{}^{16} \bar{H}{vap}$ (because $\alpha_{liq}(vap)>1$). In practical calculations, we assume that the molar enthalpy of vaporization of pure $H_2^{16}O$ is equal to the value for the natural water (i.e., $\Delta{}^{16} \bar{H}_{vap} = 40.66~kJ~mol^{-1}$), which is a good approximation given that natural water only contains about 0.2% of $H_2{}^{18}O$.
Finally, we use Eq. \ref{eq2a} with the above results to calculate the rate constants for water evaporation (${}^{16}k_f$ and ${}^{18}k_f$) and precipitation (${}^{16}k_b$ and ${}^{18}k_b$).
Tasks
Task 1 --- equilibrium isotope fractionation
Consider a gas-tight container partially filled with liquid water and air (Figure 1). Using differential equations in Eq. \ref{eq4}, implement a model that can predict the amount of $H_2{}^{16}O$ and $H_2{}^{18}O$ molecules in the water vapor and liquid water as a function of time. Specifically, demonstrate that, in an equilibrium, the ${}^{18}O/{}^{16}O$ isotope ratio in the liquid water differs from the isotope ratio in the water vapor. Run the model assuming that the air is initially dry or fully saturated with water vapor (i.e., the initial relative air humidity is 0 or 100%, respectively).
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Assumptions
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Temperature of the system and the total pressure of the gas phase are constant ($T=298~K$, $P_{tot} = 1~Atm = 101325~Pa$). You can assume that this is accomplished by the walls of the gas-tight container being flexible (to keep the pressure constant) and heat being absorbed or supplied by the surrounding, as needed (to keep the temperature constant).
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The area of the water surface in contact with the gas phase is $A_{liq} = 1~m^2$ (constant), the initial volume of the liquid water is $V_{liq} = 1~L$, and the initial ${}^{18}O$ atom fraction of the liquid water is 0.002. For the simulation with the initial air humidity of 100%, assume the initial ${}^{18}O$ atom fraction of the water vapor of 0.002.
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The initial total volume of the gas-tight container is $V_{tot}=V_{gas}+V_{liq} = 1~m^3$.
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Both air and water vapor behave as an ideal gas, i.e., their volume, partial pressure, and temperature are related through the ideal gas law ($P\cdot V = n\cdot R\cdot T$).
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The liquid and gas phases are well mixed (i.e., homogeneous). Thus, you can ignore transport processes and base your model entirely on mass balances (Eqs. \ref{eq4}) and the ideal gas law.
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Dissolution of air components such as $N_2$ or $O_2$ in water can be ignored.
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The ratio between the molar mass and density of water is the same for both $H_2{}^{16}O$ and $H_2{}^{18}O$.
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In a thermodynamic equilibrium, water vapor is by about 10\textperthousand\ depleted in ${}^{18}O$ relative to liquid water, i.e., $\delta^*_{liq}(vap) \approx -10$\textperthousand.
Task 2 --- isotope exchange between two water reservoirs
Consider an experimental setup where two open containers are enclosed in a gas-tight container filled with air (Figure 2). Both containers are partially filled with water, and the air is initially fully saturated with water vapor. Although there is no direct mixing of liquid water between water reservoirs 1 and 2, there will be an isotope exchange between them due to evaporation and precipitation (i.e., the two liquid reservoirs are "connected" through the water vapor reservoir). Your task is to model the dynamics of this exchange, i.e., predict the isotopic composition of the water vapor and the two liquid water reservoirs as a function of time. How long does it take for an isotopic equilibrium among all water reservoirs to be reached? This task was inspired by the stable isotope labeling experiment used to study the dynamics of polyphosphate in cable bacteria (Geerlings et al. 2022, https://doi.org/10.3389/fmicb.2022.883807).
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Assumptions
In addition to the assumptions listed for Task 1, assume that
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for both liquid water reservoirs, the area of the water surface in contact with the gas phase is $0.5~m^2$ (constant), the initial volume is $0.5~L$.
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the initial ${}^{18}O$ atom fraction is 0.002 for the first liquid water reservoir and 0.004 for the second liquid water reservoir.
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the initial ${}^{18}O$ atom fraction is 0.002 for the water vapor.
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the initial total volume of the gas-tight container is $V_{tot}=V_{gas}+V_{liq,1} +V_{liq,2} = 1~m^3$.
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