2H2O(l)=2H2(g) +O2(g)
the stress is an increase in pressure. If there is an increase in pressure, the reaction would go to the side of fewer moles. Even though the left side has no moles of gas, is the result still shifting to the left
No, in this case, the result would actually shift to the right, not to the left.
According to Le Chatelier's principle, an increase in pressure favors the side of the reaction with fewer moles of gas. In this reaction, there are no moles of gas on the left side, and there are 2 moles of gas on the right side (H2 and O2). Therefore, an increase in pressure would favor the formation of more gas molecules and push the reaction towards the right side, resulting in an increase in the production of H2 and O2 gases.
In the given reaction, 2H2O(l) = 2H2(g) + O2(g), the stress you mentioned is an increase in pressure. It is correct that an increase in pressure would typically cause the reaction to shift to the side with fewer moles of gas.
In this reaction, the left side has no moles of gas initially, while the right side has a total of 2 moles of gas (2 moles of H2). Based on the principle of Le Chatelier's principle, when pressure is increased, the equilibrium will shift to the side with fewer moles of gas to reduce the pressure.
In this case, since there are no moles of gas on the left side and 2 moles of gas on the right side, the reaction will shift to the left (toward the reactants) in order to reduce the pressure. As a result, more H2O(l) will be formed, leading to a decrease in the amount of H2(g) and O2(g).
It's important to note that this explanation relies on the assumption that the reaction is at equilibrium. If the reaction is not yet at equilibrium, the shifts and changes in concentrations would be different. To determine the exact effect, thermodynamic calculations or equilibrium constants would need to be considered.