J.R. S. answered 12/07/23
Ph.D. University Professor with 10+ years Tutoring Experience
The reaction we are concerned with is ...
H2CO3(aq) ==> H2O(l) + CO2(g)
According to LeChatelier, to push the equilibrium to the left (more H2CO3), you would...
Increase the pressure (fewer moles of gas on the left)
Decrease the volume (same effect as increasing pressure)
Increase the amount of CO2 (increase in product shifts reaction toward reactant)
Decrease the temperature (lower temp means greater solubility of CO2 which pushes rxn to left)