Citric acid chelates calcium carbonate more than ethanoic acid, why does this result in a faster rate of reaction?
Acetic acid has a $\mathrm{p}K_\mathrm{a}$ value of $4.76$ according to Wikipedia.

Citric acid has a $\mathrm{p}K_\mathrm{a,1}$ value of $3.13$ and a $\mathrm{p}K_\mathrm{a,2}$ value of $4.76$, also according to Wikipedia. It is about ten times as acidic as acetic acid is.

The greater acidity translates to a lower $\mathrm{pH}$ of the resulting solution. The lower $\mathrm{pH}$ translates to bases being protonated more and quicker. Considering that the decomposition reaction of carbonate is actually:

$$\ce{HCO3- + H+ &lt;=&gt; H2CO3 &lt;=&gt;&gt; H2O + CO2 ^ }$$

The $\mathrm{pH}$ effect alone should be enough to explain a faster reactivity.

However, you can also invoke the following equilibrium which will be right-side shifted with citric acid (but obviously not so much with acetic acid):

$$\ce{CaCO3 (s) + cit^3- &lt;=&gt; Ca^2+ + CO3^2- + cit^3- &lt;=&gt;&gt; [Ca(cit)]- + CO3^2-}$$

Where $\ce{cit}$ is citrate or the anion of citric acid. Complexation of calcium will remove free calcium ions from solution, and the solubility product only considers free ions in solution. Thus by complexing the free ions, more calcium can be dissolved.

I would still perform control experiments to make sure it is actually the complexation effect of citrate and not its acidity. For example, try reaction $\ce{CaCO3}$ with acetic acid in the presence of $\ce{Na3cit}$ or $\ce{Na2Hcit}$. These should be equally good complexing agents.