32. Consider the following reaction at equilibrium: \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO} 3 . What would be the effect of adding additional \mathrm{H}_{2} \mathrm{O} ?
a. It would drive the equilibrium dynamics to the right.
b. It would drive the equilibrium dynamics to the left.
c. Nothing would happen because the reactants and products are in equilibrium.
d. Reactions in both the directions will slow down.

For a chemical reaction in equilibrium, adding more of the species affect the reaction depending on whether the reaction is reversible or otherwise.

For reversible reactions, that is, reactions with the capacity to proceed on both reactants and products' sides, adding more reactants will shift the equilibrium to the right - the product's side. Similarly, adding more of the product will shift the equilibrium towards the reactants' side.

For reactions that are not reversible but in equilibrium, adding more of the reactants will shift the equilibrium to the right - towards the product. Adding more products will have no effect since the reaction cannot proceed backward.

Thus, in this case, one of the products is being added to the reaction. This means that the equilibrium will shift to the right towards the product.

More on equilibrium can be found here: https://brainly.com/question/15118952

#SPJ1