2H2O(l)=2H2(g) +O2(g)
if water is removed, the direction of equilibrium is unaffected since liquid water is not in the equil equation. What if large amount of water is removed, is the answer still the same
As long as there is enough water(liquid) to provide the proper vapor pressure at the equilibrium temperature, then how much is removed is not in the equation. BUT there must be at least one drop there. If all of it is removed all bets are off.
In the given chemical equation, the reaction represents the decomposition of water into hydrogen gas (H2) and oxygen gas (O2). The equation shows that two moles of water (H2O) yield two moles of hydrogen gas (H2) and one mole of oxygen gas (O2).
When a reaction is at equilibrium, it means the forward and reverse reactions are occurring at the same rate, resulting in no net change in the concentrations of the reactants and products. In this case, if water is removed from the system, it does not directly affect the equilibrium because liquid water (H2O) is not part of the equilibrium expression.
However, if a large amount of water is removed from the system, it can disrupt the equilibrium and cause a shift in the reaction. The reason for this lies in Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change in conditions, it will adjust to counteract that change and reestablish equilibrium.
When water is removed from the system, the equilibrium will try to compensate for the decrease in water concentration by producing more water molecules. According to Le Chatelier's Principle, the reaction will shift in the direction that produces more moles of water. In this case, it means the equilibrium will shift towards the side of the reaction with more reactants, causing the production of more water molecules until a new equilibrium is established.
Therefore, if a large amount of water is removed, the equilibrium will shift towards the side of the reaction with more reactants (hydrogen gas and oxygen gas) to compensate for the decrease in water concentration. As a result, the reaction will proceed in the forward direction to produce more water until a new equilibrium is reached with the altered concentrations of the reactants and products.