IChO 2021 · Theory problem 2

Isotope Time Capsule

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Figure 1

Molecular entities that differ only in isotopic composition, such as \(\ce{CH4}\) and \(\ce{CH3D}\), are called isotopologues. Isotopologues are considered to have the same chemical characteristics. In nature, however, there exists a slight difference.

Assume that all of the substances shown in this Question are in a gas phase.

Let us consider the following equilibrium:

Figure 2

The entropy, \(S\), increases with increasing the number of possible microscopic states of a system, \(W\):

\[ S=k_{\mathrm B}\ln W \tag{2} \]

\(W=1\) for \(\ce{^{12}C^{16}O2}\) and \(\ce{^{12}C^{18}O2}\). In contrast, \(W=2\) for a \(\ce{^{12}C^{16}O^{18}O}\) molecule because the oxygen atoms are distinguishable in this molecule. As the right-hand side of the equilibrium shown in eq. 1 has two \(\ce{^{12}C^{16}O^{18}O}\) molecules, \(W=2^2=4\).

2.1 The enthalpy change, \(\Delta H\), of eq. 3 is positive regardless of the temperature.

Figure 3

Calculate the equilibrium constants, \(K\), for eq. 3 at very low (think of \(T\to0\)) and very high (think of \(T\to+\infty\)) temperatures. Assume that the reaction remains unchanged at these temperatures and that \(\Delta H\) converges to a constant value for high temperatures.

The \(\Delta H\) of the following process can be explained by molecular vibrations.

Figure 4

At \(T=0\ \mathrm K\), the vibrational energy of a diatomic molecule whose vibration frequency is \(\nu\ [\mathrm{s^{-1}}]\) is expressed as:

\[ E=\frac12h\nu \tag{5} \]

\[ \nu=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}} \tag{6} \]

Wherein \(k\) is the force constant and \(\mu\) the reduced mass, which is expressed in terms of the mass of the two atoms in the diatomic molecule, \(m_1\) and \(m_2\), according to:

\[ \mu=\frac{m_1m_2}{m_1+m_2} \tag{7} \]

2.2 The vibration of \(\ce{H2}\) is at \(4161.0\ \mathrm{cm^{-1}}\) when reported as a wavenumber.

Calculate the \(\Delta H\) of the following equation at \(T=0\ \mathrm K\) in units of \(\mathrm{J\,mol^{-1}}\).

Figure 5

Assume that:

  • only the vibrational energy contributes to the \(\Delta H\).
  • the \(k\) values for \(\ce{H2}\), \(\ce{HD}\), and \(\ce{D2}\) are identical.
  • the mass of \(\ce{H}\) to be \(1\ \mathrm{Da}\) and the mass of \(\ce{D}\) to be \(2\ \mathrm{Da}\).

The molar ratio of \(\ce{H2}\), \(\ce{HD}\), and \(\ce{D2}\) depends on the temperature in a system in equilibrium. Here, \(\Delta_{\ce{D2}}\) is defined as the change of the molar ratio of \(\ce{D2}\).

\[ \Delta_{\ce{D2}}=\frac{R_{\ce{D2}}}{R_{\ce{D2}}^*}-1 \tag{9} \]

Here, \(R_{\ce{D2}}\) refers to \(\dfrac{[\ce{D2}]}{[\ce{H2}]}\) in the sample and \(R_{\ce{D2}}^*\) to \(\dfrac{[\ce{D2}]}{[\ce{H2}]}\) at \(T\to+\infty\). It should be noted here that the distribution of isotopes becomes random at \(T\to+\infty\).

2.3 Calculate \(\Delta_{\ce{D2}}\) with natural \(\ce{D}\) abundance when the isotopic exchange is in equilibrium at the temperature where \(K\) in eq. 4 is \(0.300\). Assume that the natural abundance ratios of \(\ce{D}\) and \(\ce{H}\) are \(1.5576\times10^{-4}\) and \(1-1.5576\times10^{-4}\), respectively.

In general, the molar ratio of the doubly substituted isotopologue, which contains two heavy isotope atoms in one molecule, increases with decreasing temperature. Let us consider the molar ratio of \(\ce{CO2}\) molecules with molecular weights of 44 and 47, which are described as \(\ce{CO2}[44]\) and \(\ce{CO2}[47]\) below. The quantity \(\Delta_{47}\) is defined as:

\[ \Delta_{47}=\frac{R_{47}}{R_{47}^*}-1 \tag{10} \]

\(R_{47}\) refers to \(\dfrac{[\ce{CO2}[47]]}{[\ce{CO2}[44]]}\) in the sample and \(R_{47}^*\) to \(\dfrac{[\ce{CO2}[47]]}{[\ce{CO2}[44]]}\) at \(T\to+\infty\). The natural abundances of carbon and oxygen atoms are shown below; ignore isotopes that are not shown here.

\(\ce{^{12}C}\) \(\ce{^{13}O}\)
natural abundance 0.988888 0.011112
\(\ce{^{16}O}\) \(\ce{^{17}O}\) \(\ce{^{18}O}\)
natural abundance 0.997621 0.0003790 0.0020000

The temperature dependence of \(\Delta_{47}\) is determined as follows, where \(T\) is given as the absolute temperature in units of K:

\[ \Delta_{47}=\frac{36.2}{T^2}+2.920\times10^{-4} \tag{11} \]

2.4 The \(\Delta_{47}\) of fossil plankton obtained from the Antarctic seabed was \(4.50865\times10^{-5}\).

Estimate the temperature using this \(R_{47}\). This temperature is interpreted as the air temperature during the era in which the plankton lived. Consider only the most common isotopologue of \(\ce{CO2}[47]\) for the calculation.

Official English text of the problem.

Answers

  • 2.1: \(\displaystyle \lim_{T\to0}K=0,\qquad \lim_{T\to+\infty}K=2.\)
  • 2.2: \(\displaystyle \Delta H(0)=-6.208\times10^2\ \mathrm{J\,mol^{-1}}\), per mole of reaction as written.
  • 2.3: \(\displaystyle \Delta_{\ce{D2}}=0.199925\ldots\approx0.200\), corresponding to a \(20.0\%\) increase in the ratio relative to its random-distribution value.
  • 2.4: With the intended reading \(R_{47}=4.50865\times10^{-5}\), \(\displaystyle T\approx295.3\ \mathrm K\approx22.2\,^\circ\mathrm C\). If the number is read literally as \(\Delta_{47}\), there is no real temperature satisfying eq. 11.

Solution

2.1

The balanced isotope-exchange reaction is

\[ \ce{H2 + DI <=> HD + HI}, \qquad K=\frac{[\ce{HD}][\ce{HI}]}{[\ce{H2}][\ce{DI}]}. \]

The species are dihydrogen \(\ce{H2}\), [1H][1H]; deuterium iodide \(\ce{DI}\), [2H]I; hydrogen deuteride \(\ce{HD}\), [1H][2H]; and hydrogen iodide \(\ce{HI}\), [1H]I.

The thermodynamic relation is

\[ \ln K=-\frac{\Delta G^\circ}{RT} =\frac{\Delta S^\circ}{R}-\frac{\Delta H^\circ}{RT}. \]

At very low temperature, the positive reaction enthalpy makes the products energetically unfavorable. The enthalpy term dominates, so

\[ \boxed{\lim_{T\to0}K=0}. \]

At very high temperature, the stipulated finite limiting enthalpy gives \(\Delta H^\circ/(RT)\to0\), and the isotope distribution becomes random.

Let \(r\) denote the atomic \(\ce{D}/\ce{H}\) ratio. Random substitution gives

\[ \frac{[\ce{HD}]}{[\ce{H2}]}=2r, \qquad \frac{[\ce{DI}]}{[\ce{HI}]}=r. \]

The factor of two in the first ratio arises because either of the two hydrogen positions can contain deuterium. Therefore,

\[ \boxed{ \lim_{T\to+\infty}K =\frac{2r}{r}=2 }. \]

This statistical factor corresponds to an entropy contribution \(R\ln2\).

2.2

The balanced reaction is

\[ \ce{2HD -> H2 + D2}. \]

Deuterium gas, \(\ce{D2}\), has SMILES [2H][2H].

Using the prescribed atomic masses,

\[ \mu_{\ce{H2}}=\frac{1\times1}{1+1}=\frac12\ \mathrm{Da}, \]

\[ \mu_{\ce{HD}}=\frac{1\times2}{1+2}=\frac23\ \mathrm{Da}, \qquad \mu_{\ce{D2}}=\frac{2\times2}{2+2}=1\ \mathrm{Da}. \]

Since the force constants are identical, both frequency and wavenumber are proportional to \(\mu^{-1/2}\). Consequently,

\[ \widetilde\nu_{\ce{HD}} =4161.0\sqrt{\frac{1/2}{2/3}} =4161.0\frac{\sqrt3}{2} =3603.53\ \mathrm{cm^{-1}}, \]

\[ \widetilde\nu_{\ce{D2}} =4161.0\sqrt{\frac{1/2}{1}} =\frac{4161.0}{\sqrt2} =2942.27\ \mathrm{cm^{-1}}. \]

At \(0\ \mathrm K\), each molecule contributes its zero-point vibrational energy, \(E=\tfrac12hc\widetilde\nu\). Thus,

\[ \begin{aligned} \Delta H(0) &=\frac{N_{\mathrm A}hc}{2} \left( \widetilde\nu_{\ce{H2}} +\widetilde\nu_{\ce{D2}} -2\widetilde\nu_{\ce{HD}} \right)\\ &=\frac{N_{\mathrm A}hc}{2} (4.1610\times10^5\ \mathrm{m^{-1}}) \left(1+\frac{1}{\sqrt2}-\sqrt3\right). \end{aligned} \]

Using the standard constants

\[ h=6.62607015\times10^{-34}\ \mathrm{J\,s},\qquad c=2.99792458\times10^8\ \mathrm{m\,s^{-1}}, \]

\[ N_{\mathrm A}=6.02214076\times10^{23}\ \mathrm{mol^{-1}}, \]

gives

\[ \boxed{\Delta H(0)=-620.8\ \mathrm{J\,mol^{-1}}}. \]

This is per mole of reaction as written, consuming \(2\ \mathrm{mol}\) of \(\ce{HD}\). The products have the lower total zero-point energy.

2.3

For

\[ \ce{2HD <=> H2 + D2}, \qquad K=\frac{[\ce{H2}][\ce{D2}]}{[\ce{HD}]^2}=0.300, \]

write the atomic fractions as

\[ p=1.5576\times10^{-4}\quad(\ce{D}), \qquad q=1-p\quad(\ce{H}). \]

In a random distribution, the molecular fractions of \(\ce{H2}\), \(\ce{HD}\), and \(\ce{D2}\) are \(q^2\), \(2pq\), and \(p^2\), respectively. Hence,

\[ R_{\ce{D2}}^* =\frac{p^2}{q^2} =2.42687372\times10^{-8}. \]

To account exactly for isotope conservation at \(K=0.300\), let \(x\) be the equilibrium mole fraction of \(\ce{D2}\), normalizing the total number of molecules to one. Deuterium conservation and normalization give

\[ x_{\ce{HD}}=2(p-x), \qquad x_{\ce{H2}}=1-2p+x. \]

Substitution into the equilibrium expression gives

\[ 0.300=\frac{x(1-2p+x)}{4(p-x)^2}. \]

Equivalently,

\[ (1-4K)x^2+(1-2p+8Kp)x-4Kp^2=0. \]

The physically admissible root, \(0\le x\le p\), gives

\[ x_{\ce{D2}}=2.91115995\times10^{-8}, \]

\[ x_{\ce{HD}}=3.11461777\times10^{-4}, \qquad x_{\ce{H2}}=0.9996885091. \]

Therefore,

\[ R_{\ce{D2}} =\frac{x_{\ce{D2}}}{x_{\ce{H2}}} =2.91206703\times10^{-8}, \]

and

\[ \boxed{ \Delta_{\ce{D2}} =\frac{2.91206703\times10^{-8}} {2.42687372\times10^{-8}}-1 =0.199925\ldots\approx0.200 }. \]

The dilute-deuterium approximation gives the same answer to the appropriate precision:

\[ R_{\ce{D2}}\approx K\left(\frac{2p}{q}\right)^2 =4K R_{\ce{D2}}^*, \qquad \Delta_{\ce{D2}}\approx4K-1=0.200. \]

2.4

Two printed labels require interpretation. The second carbon-table heading must be \(\ce{^{13}C}\), rather than \(\ce{^{13}O}\). Also, the measured value must be interpreted as \(R_{47}=4.50865\times10^{-5}\), consistent with the instruction “using this \(R_{47}\).” The literal alternative is checked below.

Let \(c_{12},c_{13}\) denote the carbon isotope abundances and \(o_{16},o_{17},o_{18}\) the oxygen isotope abundances.

Carbon dioxide-44 is \(\ce{^{12}C^{16}O2}\), with SMILES [16O]=[12C]=[16O]. Its random abundance is

\[ P_{44}=c_{12}o_{16}^2. \]

The possible mass-47 carbon dioxide isotopologues are:

Carbon dioxide isotopologue SMILES Random abundance
\(\ce{^{13}C^{16}O^{18}O}\), containing carbon-13, oxygen-16, and oxygen-18 [16O]=[13C]=[18O] \(2c_{13}o_{16}o_{18}=4.43423\times10^{-5}\)
\(\ce{^{12}C^{17}O^{18}O}\), containing carbon-12, oxygen-17, and oxygen-18 [17O]=[12C]=[18O] \(2c_{12}o_{17}o_{18}=1.49915\times10^{-6}\)
\(\ce{^{13}C^{17}O2}\), containing carbon-13 and two oxygen-17 atoms [17O]=[13C]=[17O] \(c_{13}o_{17}^2=1.59614\times10^{-9}\)

These molecules are linear. For different oxygen isotopes, the factor of two counts the two possible assignments to the oxygen positions.

Thus, the most common mass-47 isotopologue is \(\ce{^{13}C^{16}O^{18}O}\). Retaining only this species as instructed,

\[ \begin{aligned} R_{47}^* &=\frac{2c_{13}o_{16}o_{18}}{c_{12}o_{16}^2}\\ &=2\frac{0.011112}{0.988888} \frac{0.0020000}{0.997621}\\ &=4.505464112\times10^{-5}. \end{aligned} \]

The sample therefore has

\[ \Delta_{47} =\frac{4.50865\times10^{-5}} {4.505464112\times10^{-5}}-1 =7.07116413\times10^{-4}. \]

Rearranging eq. 11,

\[ T= \sqrt{ \frac{36.2\ \mathrm{K^2}} {\Delta_{47}-2.920\times10^{-4}} }. \]

Hence,

\[ T= \sqrt{ \frac{36.2} {7.07116413\times10^{-4}-2.920\times10^{-4}} }\ \mathrm K =295.304\ \mathrm K. \]

Thus the estimated air temperature is

\[ \boxed{T\approx295.3\ \mathrm K\approx22.2\,^\circ\mathrm C}. \]

If the printed number were instead literally \(\Delta_{47}=4.50865\times10^{-5}\), then

\[ \Delta_{47}<2.920\times10^{-4}, \]

whereas eq. 11 requires \(\Delta_{47}>2.920\times10^{-4}\) at every finite positive temperature. That reading therefore yields no real temperature.

The solution was checked against the official solution.