Textbook 5G.9

[amreen_sandhu1k]

5G.9
A sample of ozone ,amounting to 0.10 mol, is placed in a sealed container of volume 1.0 L and the reaction is allowed to reach equilibrium. Then 0.50 mol
is placed in a second container of volume 1.0 L at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same?

Can someone explain if the amount of O2 and the partial pressure of O2 will be the same or different?

[Ethan_Choi_2I]

It says without doing any calculations in the problem, but the equation to convert between partial pressure and concentration for gas is concentration = P/RT. T = temperature, R is the gas constant and P = the partial pressure.

[305607822]

The moles of O2 and the partial pressure of O2 will both be higher in the second reaction. In the second reaction, you start with a higher concentration of reactants (0.5 mol O3 compared to 0.1 mol O3 in the first reaction), which means your equilibrium concentrations of products will be higher. Since O2 is a product, you will have more O2 in the case of the second reaction.

[Likhitha Yettela 1F]

Since the concentration of O2 will be higher due to the higher starting concentration, the partial pressure must also be higher. This is because they are both using the same size container.