What happens to the following reaction at equilibrium if the pressure is decreased?
2H"subtext"2(g) + O"subtext"2(g) 2H"subtext"2O(g)
You could save a lot of time by using this:
2H2 + O2 ==> 2H2O. Everyone knows that the number two following H and O are subscripts. We CAN write them as subscripts but no one wants to take the time to do it.
2H2 + O2 ===> 2H2O
Here is the rule to follow:
An increase in pressure for a gaseous system in equilibrium causes the equilibrium to shift in the direction of FEWER moles. Decrease in P does the opposite.
so it shifts to the right because Q < K correct?
I don't think so.
H2 + O2 ==> 2H2O
If P increases, it shifts to side with fewer moles of gases. There are 3 mols on the left; two moles on the right so for increase P it shifts to the right. A decrease in P must shift to the left.
To determine what happens to a reaction at equilibrium when the pressure is decreased, we need to consider Le Chatelier's principle. According to this principle, if a system at equilibrium is subjected to a change in conditions, it will shift in a way that minimizes the effect of that change.
In the given reaction: 2H2(g) + O2(g) ⇌ 2H2O(g), the pressure reduction will be the change in condition.
When the pressure is decreased, the system will try to balance by shifting in the direction that produces more moles of gas. In this case, the reaction would shift to the side with more moles of gas.
To determine which side has more moles, we count the number of moles of gas molecules on each side of the equation. On the left side, we have 2 moles of H2 and 1 mole of O2, totaling 3 moles of gas. On the right side, we have 2 moles of H2O. Therefore, the left side has more moles of gas.
To minimize the decrease in pressure, the reaction will shift to the left side by reducing the formation of H2O. This means that more H2(g) and O2(g) will react to form more H2O(g). As a result, the concentration of H2O(g) will decrease while the concentrations of H2(g) and O2(g) will increase.
Overall, at equilibrium, when the pressure is decreased, the reaction compensates by shifting to the side with more moles of gas (the left side in this case) to counteract the decrease in pressure.