I want to ask a question about the sum of a rate constant to yield the total decay rate.

From previous studies, the rate of a reaction can be expressed evidently as

$$\ce{v_{rate} = k[A]}$$

where $\ce{A}$ is a molecule in a reaction.

I studied today about unimolecular photochemistry and that a molecule in its excited state $\ce{S_1}$ can do a number of processes:

  • Fluoresence (emit a photon and return to its ground state), $\ce{F}$
  • Intersystem crossing (convert to a triplet state and then relax), $\ce{ISC}$
  • Internal conversion (release heat to the environment and relax back to the ground state), IC
  • Photochemistry (be involved in a reaction), PC

and according to the lecture material, the total decay rate can be expressed as:

$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$

i.e. the sum of the rate constants.

But I thought that from my studies, the total decay rate is expressed as:

$$\ce{v_{total \ decay \ rate} = (k_F + k_{ISC} + k_{IC} + K_{PC})[A]}$$

so why do I not include the concentration term as part of the total decay rate?

My thought was that $\frac{v_{\ total \ decay \ rate}}{[A]}$ is expressed as the total decay rate and hence it can be expressed as the sum of the rate constants, but I want to check that my logic is correct.

Why is the total decay rate expressed as a sum of all the rate constants of the expressions above?

  • $\begingroup$ Because it's not exactly a first rate reaction. More like pseudo-zeroth order, depending on illumination. $\endgroup$
    – Mithoron
    Nov 28, 2019 at 22:14
  • $\begingroup$ @Mithoron can you explain this a bit further - I'm a bit confused here apologies! $\endgroup$
    – vik1245
    Nov 28, 2019 at 23:43
  • 1
    $\begingroup$ It looks to me like an example of the common misuse of "rate" when they mean "rate constant". $\endgroup$
    – Andrew
    Nov 29, 2019 at 1:34
  • 2
    $\begingroup$ In your usage v is rate (conc /time) and k is rate constant in your case 1/time and as @Andrew writes you are confusing them. And yes rate constants do add as all process start from the same excited state. As you write $k=k_F+k_{isc}$ etc $\endgroup$
    – porphyrin
    Nov 29, 2019 at 17:18

1 Answer 1


I am simply confusing two different variables here, rate and rate constants.

In this situation, the sum of the rate constants is the sum of the individual rate constants of the processes occuring in the reaction i.e.

$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$

This is consistent with my current knowledge.

It should be referred to as "total decay rate constant"

Of course, the rate equation can still be written here as:

$$\ce{v_{rate} = k[A]}$$


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