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Table 1: A numerical example showing the mass-dependent relations between the different reaction rates in the oxygen isotopic system. In successive columns: (1) the isotopic reaction ⁱO+^jOM $\to {}^{i}$ O^jOM; (2) the reduced mass of the reactants for M=5; (3) the reduced mass ratio $\Delta \mu _{\rm Y-MX}^{a }$ normalized to the reaction ¹⁶O+¹⁶OM for a=1; (4) the product of the relative abundance with $^{16}\rm O=0.9976$ , $^{17}\rm O=4\times 10^{-4}$ and $^{18}\rm O=2 \times 10^{-3}$ ; (5) the product of $(3)\times (4)$ . The ¹⁷O/¹⁶O ratio ( x'₁₇/x'₁₆) of the OOM molecule (noted [¹⁷O/¹⁶O] $_{\rm OOM })$ is calculated from the values of the column (5); that is: $x'_{16} = \{1/2 [4.128 \times 10^{-4} + 2.129 \times 10^{-3}$ $+ 4.070 \times 10^{-4 }+ 2.073 \times 10^{-3}] + 9.952 \times 10^{-1}\}$ . Similarly for x'₁₇ and x'₁₈. This gives: [¹⁷O/¹⁶O] $_{\rm OOM} = (x'_{17}/x'_{16}) = 4.119 \times 10^{-4}$ and [¹⁸O/¹⁶O] $_{\rm OOM} = (x'_{18}/x'_{16}) = 2.111 \times 10^{-3}$ corresponding to $\delta ^{17}$ O = ([ $4.119 \times10^{-4 }/ (4 \times10^{-4}/0.9976)]-1)\times1000 = +27.3\mbox{\textperthousand}$ and similarly $\delta ^{18}\rm O = +53.2\mbox{\textperthousand}$ . The slope in the diagram $\delta ^{17}$ O versus $\delta ^{18}$ O is equal to 0.513.
Table 2: A numerical example showing the non-mass-dependent relations between the different reaction rates in the oxygen isotopic system. In successive columns: (1) the isotopic reaction ⁱO+^jOM $\to {}^{i}$ O^jOM; (2) the reduced mass ratio $\Delta \mu _{\rm Y-MX}^{a }= [\mu _{\rm X-MY }$ / $\mu _{\rm X-MX} ]^{a}$ for a=0.1 and M=5; (3) $\Delta \mu _{\rm Y-MX}^{b }$ for b= 0.2; (4) $\Delta \mu _{\rm Y-MX}^{c }$ for c= 0.1; (5) the reaction rate ratio R. Two situations are distinguished: (1) for ⁱO+ⁱOM ( $i\equiv j$ ), $R = \Delta \mu _{\rm Y-MY}^{a} \ \Delta \mu _{\rm Y-MY}^{b }$ and for (2) ^jO+ⁱOM ( $i\ne j$ ), $R=\{\Delta \mu _{\rm Y-MX}^{a}$ 1/2 [ $\Delta \mu _{\rm Y-MX}^{b }+\kappa \ \Delta \mu _{\rm Y-MX}^{c}$ ]; calculations are performed for $\kappa = 0.8$ (6) The product of the relative abundance with $^{16}\rm O=0.9976$ , $^{17}\rm O=4\times 10^{-4}$ and $^{18}\rm O=2 \times 10^{-3}$ (7) the product of $(5)\times (6)$ . The ¹⁷O/¹⁶O ratio of the OOM molecule (noted [¹⁷O/¹⁶O] $_{\rm OOM })$ is calculated from the values of the column [7]; that is for the abundance of ¹⁶O: {1/2 [ $3.585 \times 10^{-4} + 1.789 \times 10^{-3} + 3.587\times 10^{-4 }+ 1.792 \times 10^{-3}] + 9.952 \times 10^{-1}$ } and similarly for ¹⁷O and ¹⁸O. The $\delta ^{17}$ O ( $\mbox{\textperthousand}$ ) = ([¹⁷O/¹⁶O] $_{\rm OOM }$ /[0.0004/0.9976])- 1) $\times$ 1000; similarly for $\delta ^{18}$ O ( $\mbox{\textperthousand}$ ). That is: $\delta ^{17}\rm O =-101.1\mbox{\textperthousand}$ and $\delta ^{18}\rm O= -102.2\mbox{\textperthousand}$ with $\delta ^{17}$ O/ $\delta ^{18}\rm O=0.989$ .