Reaction of barium peroxide forming barium carbonate I stumbled across this reaction:

$$\ce{BaO2 + H2CO3 -&gt; BaCO3 + H2O2}$$

This reaction seems quite similar to the variant with sulfuric acid

$$\ce{BaO2 + H2SO4 -&gt; BaSO4 + H2O2}$$

which is well known and works.

But as carbonic acid is a lot weaker than sulfuric acid, I started to question this reaction. Why does it even proceed in the first place? I thought that the sulfate/carbonate is really insoluble in water which is why no equilibrium forms (or does it?), however barium peroxide is itself pretty insoluble.

So, a) does the first reaction even happen and b) why?
Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged: 

$$\ce{BaO2 + 2H-O-H &lt;=&gt; H2O2 + Ba^{2+} + 2OH-}$$

$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.

Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling  $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.  

In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).



$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.