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Kinetics of the thermal gas‐phase dimerization of hexafluoropropene
Author(s) -
Robinson Peter J.,
Skelhorne Graham G.
Publication year - 1983
Publication title -
international journal of chemical kinetics
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.341
H-Index - 68
eISSN - 1097-4601
pISSN - 0538-8066
DOI - 10.1002/kin.550150604
Subject(s) - chemistry , arrhenius equation , olefin fiber , reaction rate constant , gas phase , kinetics , thermodynamics , atmospheric temperature range , thermal , reaction rate , phase (matter) , analytical chemistry (journal) , activation energy , organic chemistry , catalysis , physics , quantum mechanics
The reversible thermal gas‐phase dimerization of hexafluoropropene to the four isomeric cyclobutanes has been studied by pressure change and by gas‐liquid chromatographic analysis in the temperature range of 645–708 K with initial pressures of olefin from 802 to 4820 mm Hg. The reaction was accurately second order at low conversions of olefin to dimers, and at higher conversions it gave a very good fit to the rate equation for opposing second‐ and first‐order reactions. The rate constants for the dimerization, calculated from initial rates of reaction, yielded the least‐mean‐squares Arrhenius equation (95% confidence limits):\documentclass{article}\pagestyle{empty}\begin{document}$$ \log _{10} (k_2 /{\rm dm}^3 {\rm mol}^{ - 1} s^{ - 1}) = (5.93 \pm 0.40) - (131.8 \pm 9.5)k{\rm J\,\,mol}^{{\rm - 1}}/RT\ln \,10 $$\end{document}where k 2 is defined by\documentclass{article}\pagestyle{empty}\begin{document}$$ \frac{{ - 1/2{\rm d}[{\rm C}_3 {\rm F}_6]}}{{dt}} = \frac{{{\rm d}[c - {\rm C}_6 {\rm F}_{12}]}}{{dt}} = k_2 [{\rm C}_3 {\rm F}_6]^2 $$\end{document}Studies carried out in a packed vessel showed no evidence of heterogeneity. The rate constants found in this work are in excellent agreement with those found at lower pressures by Atkinson and Tsiamis, and the combined results give the Arrhenius equation\documentclass{article}\pagestyle{empty}\begin{document}$$ \log _{10} (k_2 /{\rm dm}^{\rm 3} {\rm mol}^{{\rm - 1}} {\rm s}^{{\rm - 1}}) = (6.47 \pm 0.21) - (138.6 \pm 2.7){\rm kJ\,mol}^{{\rm - 1}} {\rm /}RT\ln 10 $$\end{document}

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