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The following equation can be used to determine the rate:
$Rate = \frac{-\Delta [A]}{\Delta t}$

This equation can also be used:
$Rate = k[A]$

By simple substitution, $\frac{-\Delta [A]}{\Delta t} = k[A]$ should be true. Therefore, $k$ can be represented as: $\frac{-\Delta [A]}{[A]\Delta t}$

However, calculating the rate constant for a given reaction is obviously not that simple. Why does this equation not work then, if the mathematics of it check out? I have tried using this equation for some problems, and as I expected, they do not yield the correct answer as opposed to using the integrated rate equation.

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  • $\begingroup$ Maybe it has something to do with the fact that you cannot measure $\Delta [A]$ effectively. So we integrate it so that we can put the easily measurable non $\Delta $ values. $\endgroup$
    – Papul
    Jan 13, 2015 at 6:50

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The average rate $\Delta[A]/\Delta t$ over time interval $\Delta t$ is only approximately equal to the true instantaneous rate $\rm d[A]/dt$ at time $t$. The difference between the two can be significant if the instantaneous rate changes over the chosen time interval.

Here's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.

To see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:

enter image description here

The larger the time interval, the larger the difference between $\Delta[A]/\Delta t$ and $\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.

So why not just use very small time intervals? Well, there's a limit to how small you can make $\Delta t$ if you're computing $\Delta[A]/\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals.

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  • $\begingroup$ I see, so does this mean that -Δ[A]/Δt is the average rate over a period of time, and k[A] is the rate at an exact point in time? $\endgroup$
    – null
    Jan 13, 2015 at 22:41
  • $\begingroup$ Yes. That's right. $\endgroup$
    – Fred Senese
    Jan 13, 2015 at 22:42
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That rate constant '$\kappa$' is for "instantaneous rate of reaction"(irr) which you wish to calculate from "average rate of reaction"(arr) $$\frac{d[A]}{dt}=r_{irr}=\lim_{\Delta t\to0 }r_{arr}=\lim_{\Delta t\to0 }\frac{\Delta [A]}{\Delta t}$$ So if the time duration is very small you would get a better approximation.

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