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So we know that $$ln \frac{k_{T_2}}{k_{T_1}} = \frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$ If we consider both were raised from $T_1$ to $T_2$, $$\frac{1}{T_1}-\frac{1}{T_2} = constant$$, So$$\frac{ ln \frac{k_{T_2}}{k_{T_1}}}{E_a} = constant$$ So the one with higher $E_a$ has a higher value for $ln \frac{k_{T_2}}{k_{T_1}}$ Thus the fractional change ...

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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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The activation energy is the kinetic energy required from the reactants to form the products.Now lets go to your second part.Since the first reaction has a bigger activation energy than the second reaction , if we add energy to the system the first reaction has better chances of happening than the second reaction because more molecules from the first ...

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This is an alternative derivation to that given by @Andrew in his answer for the same first order reaction. We suppose that the species has a probability $p$ of reacting in time interval $\Delta t$ and that this is independent of any past history, i.e. it depends only on the length of $\Delta t$ and for a sufficiently short time is proportional to this time,...

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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 $$\ce{A_{ad} -&... 1 This can be answered both conceptually and mathematically. Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions. &... 1 Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a ... 6 UPDATED: I wrote a first answer assuming that t>0 which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that t<0 in these "backward equations". Here's a corrected explanation: The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some ... 2 To see how this calculation is done, let us consider the simpler case of just the second step shown above, ie \ce{AB ->[k_{f2}] C}. Since we are interested in the situation where no C has yet formed, the reverse reaction can be ignored. I will assume that this reaction is first order in AB, so k_{f2} has units of s^{-1}. In this case, the rate ... 3 How do I even write equilibrium constant for solid solid equilibrium? First, let's write the expression for the reaction quotient:$$Q = 1$$Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity. Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you ... 2 The activation energy of a single reaction does not change with time but there might be more than one reaction happening An activation energy is always constant for a given single reaction. But that doesn't mean the reaction speed will be constant (it might depend on concentration or temperature). But what about autocatalytic reactions? There are two ... 3 Let’s ignore the air, first of all, and put it back later. When you heat the bottom of the saucepan, the first thing that happens to the water at the top is nothing. It stays cold, because water is not very conductive. The next thing that happens is that the water at the bottom becomes so hot that it is less dense than the water above it. The first thing ... 3 Single molecule reaction Some diagrams refer to a single molecule reacting. If that is the case, several quantities are not defined or ill-defined: pressure entropy of mixing temperature In those diagrams, the best label would be potential energy, and the energy difference between reactant and transition state should be labeled activation energy. ... 2 The oxidation of hydrocarbons (\ce{RH}) such as this one proceed via a chain mechanism. The initiation step is$$\ce{RH + O2 -> HO2^. + R^.}$$and the propagation$$\ce{ R^.+O2 -> RO2^. \\ RO2^. +RH -> ROOH +R^. }$$But it does not stop there as the hydroperoxide decomposes \ce{ ROOH -> RO^. + OH^.} and there are other reactions ... 0 In order to determine the rate law, you need to be able to measure either the rate at which the reactants are consumed or the products are formed. A very common error is to base the rate law on the balanced chemical equation. In fact, many high school level teachers will teach kinetics this way. Unfortunately this is a serious error. It is possible for ... 2 Concentrations are a geometric series First, let's focus on the reactions where one of the reactants is a monomer \ce{X_1}:$$\ce{X_1 + X_n <=> X_{n+1}}$$and call the equilibrium constant for this reaction K_{\mathrm{extend}}. From this, we can figure out the equilibrium constant for making a polymer from monomers:$$\ce{n X_1 <=> X_n}...

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