New answers tagged

0 votes

How is the ln k term in the linearised Arrhenius equation acceptable?

$$ \begin{align} k &= A \exp\left(-\frac{E_\mathrm a}{RT}\right) \label{eq:1}\tag{1}\\ \ln k &= \ln A -\frac{E_\mathrm a}{RT} \tag{2} \end{align} $$ If you use the equation \eqref{eq:1} to get ...
2 votes

What happens when para-water ice is suddenly melted?

Note that this answer is not complete. Note that wherever a "*" is placed in the answer, it means it requires figures or explanations to support it. If anyone found them or anything related, ...
  • 279
-3 votes

chemical equilibrium and activity

You intuition is telling you that ions should suppress more ions from coming into solution and that is probably correct and were it the only thing happening it should be observed. In fact it is. ...
  • 1,039
1 vote

chemical equilibrium and activity

The $\ce{Ba^{2+}}$ ions will surround by anions The $\ce{SO_4^{2-}}$ will be surrounded by cations So they stabilize a lot more actually. Remember that this may take long, but thermodynamically that ...
  • 19
2 votes

Relation between activation and threshold energies

Arrhenius activation energy $E_\mathrm{A}$ is one more closely related quantity that needs to be distinguished from activation and threshhold energies. The difference between the quantities is briefly ...
  • 36k
3 votes

Relation between activation and threshold energies

The activation energy $E_a$ of a reaction is the amount of energy required to take reactants to products across some reaction coordinate. The threshold energy $E_0$ is related, but defined as the ...
  • 1,061
-1 votes

What is the difference between a single displacement reaction and a two competing reaction system?

Upon thinking about this most displacement reactions follow your scenario. If the two reactions are independent then A is not necessary to the production of BC and is not in the equation for its ...
  • 1,039
2 votes

Excited states and emission lifetimes

Thanks @Achem and @porphyrin for your comments. I do not know if an Answer is the right format for this Comment but the comments do not allow enough character to develop my argument. I think we are ...
  • 680
0 votes

Steady-state approximation for the destruction of ozone

The error in your logic is when you write: Next, I express the overall rate in terms of the disappearance of $\ce{O3}$. This means that Rate$=−\frac{d[\ce{O3}]}{dt}=k_2[\ce{O}][\ce{O3}]$. For any ...
  • 9,260
3 votes

Steady-state approximation for the destruction of ozone

Applying the steady-state and rate-determinign-step aproximations, we have: $$\ce{2 O3 -> 3 O2}$$ $$\ce{O3 + M <=> O2 + O + M}$$ $$\ce{O + O3 -> 2 O2}$$ $$R=\frac{1}{2}\frac{\mathrm{d}[\ce{...
3 votes

Excited states and emission lifetimes

Regarding porphyrin's response and PAEP's reservation about oxygen emission from an excited triplet to a ground triplet being labelled as fluorescence, the issue can be addressed by defining ...
  • 33.8k
-1 votes
Accepted

How do I interpret the dissolution rate unit?

The linear dissolution rate can be calculated from molar specific dissolution this way: $$\frac{\mathrm{d}l}{\mathrm{d}t}=\frac{\mathrm{d}V}{A \mathrm{d}t}=\frac{\mathrm{d}m}{\rho A \mathrm{d}t}=\frac{...
  • 29.9k
2 votes

Dependence of rate on the nature of reactants and other factors

The rate is given by the differentials you present and these have values depending on concentration, temperature etc. as you mention. What defines a reaction is the rate constant. In your example $$ ...
  • 27.6k
5 votes
Accepted

Excited states and emission lifetimes

Yes, singlets fluoresce and triplets phosphoresce. So the singlet lifetime is the fluorescence lifetime and is the inverse of singlet excited state decay rate constant and similarly for ...
  • 27.6k

Top 50 recent answers are included