Question #44832

The heterogeneously catalyzed gas-phase reaction \ce{A(g) -> P(g)} follows the Langmuir-Hinshelwood mechanism. The rate of reaction $r$ is given by: $r = \frac{k K P_A}{1 + K P_A}$ where $k$ is the rate constant for the surface reaction, $K$ is the adsorption equilibrium constant for gas \ce{A}, and $P_A$ is the partial pressure of \ce{A}. An Arrhenius plot of $\ln(r)$ versus $1/T$ at a constant pressure $P_A = 1 \text{ bar}$ exhibits a non-linear curve with two distinct asymptotic linear regions. The slope of the asymptote at very low temperatures is $-12000 \text{ K}$. The slope of the asymptote at very high temperatures is $-4800 \text{ K}$. Given that the standard entropy of adsorption for gas \ce{A} is $\Delta S_{ads}^\circ = -12 R$ (where $R$ is the universal gas constant), calculate the temperature $T_{shift}$ at which the apparent activation energy of the reaction is exactly $8400 R$. (Assume $\Delta H_{ads}^\circ$ and $\Delta S_{ads}^\circ$ are independent of temperature).