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Firstly a correction and general clarification: your first general assertion is not true. The metal surface acts as a heterogenous catalysts (supported by paper linked later), and the kinetics can be sufficiently modeled by the Langmuir-Hinshelwood model for a monomolecular reaction in which the rate determining step is the surface reaction
$$ \ce{A_{ad} -&...
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First passage problems
This is a first passage problem, asking when a system reaches a certain final state for the first time . If you were to use a simulation to explore this, you would erase parts of trajectories right after reaching the final state before summing up results. The Backwards Master Equation is especially suited for first passage problems (...
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$$k_\mathrm{high} = A e^{(−E_a/RT_\mathrm{high})}$$
$$k_\mathrm{low} = A e^{(−E_a/RT_\mathrm{low})}$$
$$\frac{k_\mathrm{high}}{k_\mathrm{low}} = e^{-E_a \cdot \beta}$$
If the activation energy is zero, the ratio of rate constants will be one, i.e. there is no temperature dependency of the rate. The higher the activation energy, the larger the exponent, ...
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This can be answered both conceptually and mathematically.
Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions.
&...
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Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a ...
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