Opening the book…
\[F_{T} = 1 - exp \left(- \frac{60 \ \frac{\text{kJ}}{\text{mol}} - (1)\left(45 \ \frac{\text{kJ}}{\text{mol}}\right)}{6 \left(0.008314 \ \frac{\text{kJ}}{\text{mol K}} \right)(295.15 \ \text{K})}\right) = 0.64 \nonumber\]
Thus, the rate of the reaction could be at most 64% of the maximum rate possible as a result of thermodynamic controls.
This page titled 9.8: Answers to practice problems was last modified on Sun, 12 May 2024 08:22:40 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew F Kirk via source content that was edited to the style and standards of the LibreTexts platform.
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