Given is the following reaction of carbonic acid and carbon dioxide:

$$\ce{RCOOH (g) → RH (g) + CO2 (g)}$$

The reaction rate of this decarboxylation was measured in a container with constant volume at $T = \pu{277 °C}$. The following starting pressures $p(0)$ and half-time periods $t_{1/2}$ were measured whereas there was only carbon acid at the beginning.

$$ \begin{array}{lrrrrr} \hline p(0)/\pu{mbar} & 20 & 40 & 100 & 150 & 200 \\ t_{1/2}/s & 95.7 & 47.8 & 19.2 & 12.9 & 9.58 \\ \hline \end{array} $$

Problem: Why does this data imply that the concentration of the acid doesn't follow a kinematic of the first order?

Answer: A kinetic of first order would have the same half-time period for different starting concentrations. This isn't given here. So this isn't a kinetic of the first order.

Question: It's clear that a kinetic of first order doesn't depend on the starting concentration. What I don't get is: how do I see that we actually have different starting concentrations? Why should the starting concentration depend on the starting pressure?

My though: We have

$$ c = \frac{n}{V}; \qquad pV = nRT; \qquad V = \text{const}$$

so we get

$$V = \frac{nRT}{p} = \text{const}$$

and from that we get

$$c = \frac{n}{V} = \frac{np}{nRT} = \frac{p}{RT} = \text{const} \quad \to \quad c = \text{const}$$

So apparently, we have the same concentration for different pressures. So how should I see from that data, that the the starting concentration has changed?

I also assumed that we always use the same amount of moles and just change the pressure externally. Maybe they though we change the amount of moles we put in and that changes the pressure, which we then measured?

Edit: I just noticed that sometimes people give the amount of moles of a gas in bars (which is still kind of strange to me), so basically in the units of pressure. I guess that's what's going on here?

  • $\begingroup$ Regarding the mol-bar thing you asked at the end, in reality it's neither and you have to use activities instead. It just so happened that it's convenient to use bars for the partial pressures of gases as they need to be related to the standard state of pressure (1 bar). Maybe read this topic of refer to your PC textbook for more details. $\endgroup$
    – andselisk
    Commented Jan 22, 2019 at 12:50
  • $\begingroup$ I don't understand how you conclude that the concentration c is constant. You seem to head in the right direction but then you miss the key point, that c is proportional to p, and as the table indicates, p is changed! $\endgroup$
    – Buck Thorn
    Commented Jan 24, 2019 at 20:24


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