Question

2H_{2(g) +}O_{2 (g)} <=> 2H₂O (g)

the above equilibrium experiences a sudden decrease in pressure. which one of the following will then occur?

1) dec in mole of h2

2) incr in mole of H20

3) incr in mole of 02

4) Total mole would decrease

The answer is 3). Can you explain why

Answer 1

To answer this question we need to use the Le Chateliers principle.

Le Châtelier's principle, or "The Equilibrium Law", can be used to predict the effect of a change in conditions on a chemical equilibrium.

When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to partially counteract the effect of the applied change and a new equilibrium is established.

Changes in pressure to an equilibrium reaction only applies to reactions involving gases, although not necessarily all species in the reaction need to be in the gas phase.

According to Le Châtelier, if the pressure is decreased, the position of the equilibrium will move in such a way that the pressure increases again. It can do that by producing more gaseous molecules.

A general gaseous reaction is given below:

2A(g)+B(g)⇌C(g)+D(g)

In this type of a reaction when pressure is decreased, the position of equilibrium will move towards the left-hand side of the reaction, because then more number of moles of gas is produced. Pressure is caused by gas molecules hitting the sides of their container. The more molecules in the container, the higher the pressure will be.

In your question the right answer will be increase in moles of O₂, Because this action will increase the pressure of the system.

because in the other three cases a decrease in pressure if followed with these actions will decrease the pressure further which is cannot keep the reaction in equilibrium

Dec in mol of H2 will decrease pressure, increase in mol of H2O will also decrease pressure, If total moles decrease that will also decrease pressure.