In our chemistry practical we dropped 0.07g of Magnesium into various concentrations of an acid (sulfuric or hydrochloric - only one was used, not sure which), and measured the rate of reaction by the rate of the H2 produced (used a gas syringe). We then collected data at a suitable interval of the culmulative amount of H2 produced. We drew the graph (with multiple concentrations - 0.25, 05, 0.75, 1, 2). Then we measured the steepest point (fastest rate) and took the gradient. Plotting rate against concentrations (0.25, 05, 0.75, 1, 2), we came to the conclusion that rate varies exponentially with concentration. But the rate law of a typical reaction is:

r = k * [A] * [B]

, which would mean doubling [B] would double the rate law, not have an exponential effect on the reaction. Why did we come to the conclusion that concentration has an exponential effect?

In our case the rate law, I would think, would be elementary, but we are not dealing with concentrations of both Mg and HCl, just HCl, as the Mg was solid - therefore no concentration of Mg in solution.

My question is: In a reaction between two liquids, is the rate of reaction linearly dependent on the concentration? AND: What is the rate law of our reaction? - Why did we come to the conclusion we did?

Thanks for the help!! :)

  • $\begingroup$ You need to be careful when stating what kind of a curve fits a collection of data points. For sufficiently small datasets, it is possible to interpolate any curve with good accuracy. Would you be able to provide the raw data and graphs you made so we can see for ourselves? $\endgroup$ – Nicolau Saker Neto Mar 4 '14 at 18:39
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    $\begingroup$ What is the rate-limiting step? The faster Mg evolves hydrogen, the less surface area is exposed for reaction but the faster the convection. Mg heats by reaction and cools by liquid flow. Mg surface area changes over time, residual dimensions vs. pitting. Confounded variables cannot be disentangled by varying only one at a time (Theory of Experimentation). The world is dirty and complicated. Computer speech was not solved not by modeling anatomy. $\endgroup$ – Uncle Al Mar 4 '14 at 19:04
  • $\begingroup$ @NicolauSakerNeto, OK good point. Our teacher also implied that this relationship was normal. $\endgroup$ – Swedish Architect Mar 4 '14 at 20:38

Rate equation is experimentally determined. So, you can't directly write $r=k[A][B]$ (where r is rate, k is rate constant). Probably you might have performed first order reaction. For which $r=k[A]=\frac{2.303}{t}log\frac{[R]}{[R_0]}=\frac{2.303}{t}log{e^{kt}}$.

The above equation suggests that rate is proportional to exponential function.

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  • $\begingroup$ Fair point. I was assuming too much it seems. Is there a less mathematical way of describing it? Like why is it proportional to exponential function? Thanks $\endgroup$ – Swedish Architect Mar 5 '14 at 17:29
  • $\begingroup$ Where does the 2.303 come from??? I'm used to principles of chemistry, but not the mathematics behind it.. Thanks! :) $\endgroup$ – Swedish Architect Mar 5 '14 at 20:43
  • $\begingroup$ In my statement I had assumed that we were dealing we an elementary reaction, and thus could write the rate law from the chemical equation. $\endgroup$ – Swedish Architect Mar 5 '14 at 20:58
  • $\begingroup$ @SwedishArchitect: $\frac{1}{t}In\frac{[R]_0}{[R]}=\frac{2.303}{t}log\frac{[R]_0}{[R]}$. It may be noted that concentration of pure solid remains constant and can become the cause for not being in the rate law expression as like the reactant taken in excess in pseudo first order reactions. $\endgroup$ – Immortal Player Mar 6 '14 at 3:18

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