Consider the following equilibrium. 2 SO2 (g) + O2 (g) 2 SO3 (g) The equilibrium cannot be established when __________ is/are placed in a 1.0-L container.

the answer is 0.75mol SO2 but I don't understand where to even begin

I don't get it. It appears to me that any number one chooses can go there. Without O2 along with SO2 there is no way to establish equilibrium for SO2 alone or for O2 alone.

To determine the conditions under which the equilibrium cannot be established, we need to analyze the balanced chemical equation and evaluate the Le Chatelier's principle.

The given equilibrium equation is:

2 SO2 (g) + O2 (g) → 2 SO3 (g)

In a 1.0-L container, we can assume that the initial concentrations of SO2, O2, and SO3 are all zero.

According to Le Chatelier's principle, when the equilibrium is disturbed, the system will shift in order to minimize the disturbance and restore equilibrium.

Now let's consider the potential scenarios:

1. If there is no SO2 in the container, the forward reaction (2 SO2 + O2 → 2 SO3) cannot occur, and therefore, the equilibrium cannot be established.

2. If there is no O2 in the container, the forward reaction (2 SO2 + O2 → 2 SO3) cannot occur since there is no O2 to react with SO2. Therefore, the equilibrium cannot be established.

3. If there is no SO3 in the container, it means that the system is already at equilibrium with zero concentration of SO3. However, the equilibrium can still be established if we introduce small amounts of SO2 and O2 into the container since they will react to produce SO3 until the equilibrium concentrations are reached.

In conclusion, the equilibrium cannot be established when either SO2 or O2 (or both) are absent in the 1.0-L container.