A sophisticated gas-phase decomposition operates exclusively as a first-order mechanism where the ra — Chemical Kinetics Chemistry Question
Question
A sophisticated gas-phase decomposition operates exclusively as a first-order mechanism where the rate constant varies wildly via the expression $k = 10^{12} e^{-2000/T}$ (with $T$ naturally in Kelvin). Find the absolute numeric temperature in Kelvin at which the half-life period of this unique reaction drops precisely to $0.693 \times 10^{-8} \text{ seconds}$. (Assuming $\ln 10 = 2.303$).
💡 Solution & Explanation
Using $t_{1/2} = 0.693/k$, plugging in the provided half-life evaluates the operating $k$: $0.693 \times 10^{-8} = 0.693 / k \Rightarrow k = 10^8 \text{ s}^{-1}$. Equating this specific $k$ to the Arrhenius function: $10^8 = 10^{12} e^{-2000/T}$. Dividing provides $e^{-2000/T} = 10^{-4}$. Taking the natural log of both sides gives $-2000/T = -4 \ln 10$. Substituting $\ln 10 = 2.303$ gives $-2000/T = -4(2.303) = -9.212$. Isolating $T = 2000 / 9.212 \approx 217.1 \text{ K}$.