I am very new to chemistry. I am using a simulator for knowing rate of chemical reaction whose unit is $\mathrm{(kmol/m^3)/s}$. I want to convert it into time taken only. I have got concentration of each chemical can I convert it into time.

  • $\begingroup$ Perhaps you are looking to express the rate of the reaction in terms of a half life. That is the time it takes for the concentration of a reactant to reach half its original value. Assuming it is a first-order reaction, t½ = 0.693 / k. If your rate constant was expressed in units of concentration per second, the half life will be in seconds. $\endgroup$
    – iad22agp
    Dec 21, 2016 at 14:35
  • $\begingroup$ The first-order rate constant is in units of per-second, not concentration per second (that is zeroth order). $\endgroup$ Dec 23, 2016 at 4:17

1 Answer 1


Well, there isn't usually a specific time that a reaction reaches completion or equilibrium. You need to choose an arbitrary limit, such as "when 90% of the reactant has been converted to product".

You can calculate your time from an integrated rate law. This will require you to know what order your reaction is (perhaps CSE can help you there).

For example, for a first-order reaction, the integrated rate law is:

$C_t = C_0e^{-kt}$

Or, in a form more convenient for you:

$t = \frac{-1}{k}ln(\frac{C_t}{C_0})$

where $k$ is the rate constant of the reaction, $C_0$ is the initial concentration, and $C_t$ is the concentration at the time $t$ you are interested in.

If you choose the arbitrary limit I mentioned earlier (convenient for demonstration, but you'd probably want more product formed that 90%), at your "completion" time there would be 10% reactant left. The math would look like:

$\frac{C_t}{C_0} = \frac{10%}{100%} = \frac{1}{10}$

$t = \frac{-1}{k}ln(\frac{1}{10}) = \frac{1}{k}ln(10) \approx \frac{2.303}{k}$

where you have now found the time it takes for the reaction to proceed to your chosen point. Note that if you choose 100% conversion, the remaining reactant concentration would be 0. This gives you $ln(0)$, which is not defined for real numbers. To get a more meaningful understanding, you can take the limit as the reactant concentration approaches zero (the reaction approaching completion), and you will find:

$t_{completion} = \lim_{C_t \to 0}\frac{-1}{k}ln(\frac{C_t}{C_0})$ $= \frac{-1}{k}\lim_{u \to 0}ln(u) = \frac{-1}{k}(-\infty) = \infty$

And thus by our most useful models, it takes literally forever for every single molecule to react. This is true for any reaction order. This is why we cannot easily say when a reaction is "done", and we must pick an arbitrary point.

Note that all the math examples I presented only apply to first-order reactions.

To find the time your reaction "takes" you will have to: 1) choose an arbitrary point to call completion, 2) know the order of the reaction, and 3) know the rate constant of the reaction.

P.S. If you wanted, you could set $\frac{C_t}{C_0} = \frac{1}{2}$ in the equation solved for $t$, and you will derive the half-life formula $t_{1/2} = \frac{ln2}{k}$ mentioned elsewhere.


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