# Tag Info

0

The activation energy is not the av.kinetic energy that the colliding particles need in order to successfully react. To describe the rate of a reaction there are two models: The Arrhenius model and the more sophisticate Eyring model. The Arrhenius model uses the activation energy, defined as the energy required to reach the transition state; the "energy" ...

1

I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem. Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant). As you wrote: ... 2 In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P. \begin{align} \frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}... 0 \begin{align} &\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[k_1][k_{-1}] O2 + O} \\ &\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[k_2] O2 + O2} \end{align} The rate is determined by the slowest step. $$r=k_2[\ce{O}][\ce{O3}]$$ If the concentration of\ce{ O}$which produced from the first step ... 1 You should write down the rate equations and set [O] as an intermediate so that$\ce{d[O]/dt} = 0$. This is the steady state approach and leads to$\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$which differs from your answer because when you assume a fast equilibrium you also assumed that$...

0

The Arrhenius constant, A, does not have to necessarily be the same for two reactions taking place at the same temperature, due to the fact that it depends on the order of a reaction as well as the frequency with which successful collisions will occur during a reaction. As for the second part, as k ∝ x-Ea, for 2k to be a larger value, Ea for 2k would have ...

0

The statement is basically true, but only if you take into account that any reaction has two directions, and goes towards equillibrium. Two issues still arise temperature changes the point of equillibrium in equillibrium, the reaction rate is, by the very definition of equillibrium, zero Seeing that, it's better to throw away that overgeneralistic ...

3

Well mostly it does but there can be exceptions. Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$ It has a negative temperature dependence. The rate decrease in increasing the temperature. There is a simple reason to why this happens. We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence. ...

1

It is correct to say: "If nothing else changes, an increase in temperature will increase the rate of an elementary step of a chemical reaction". All kinds of things could go wrong. If the temperature is so high that your reaction vessel melts, spilling a liquid reactant and removing it from a solid reactant, the rate will not increase. If the temperature is ...

Top 50 recent answers are included