Which part(s) of the reaction
2O3(g) <---> 3O2(g)
will be favored by an increase in the total pressure (resulting in compression)?
2. Neither is favored.
3. Unable to determine
4. reactants
What does "which part(s) of the reaction" mean?
An increase in P will cause it to shift to the left; thus, O2 will decrease and O3 will increase. I suppose that means 3 is the answer?
To determine which part of the reaction will be favored by an increase in total pressure, we can use Le Chatelier's principle. According to Le Chatelier's principle, when a system at equilibrium is subjected to a change in conditions, the system will adjust itself to minimize the effect of that change and restore equilibrium.
In this case, when the total pressure of the system is increased (resulting in compression), the system will shift in the direction that minimizes the increase in pressure.
Given the reaction: 2O3(g) ⇌ 3O2(g)
An increase in total pressure will cause the system to shift towards the side with fewer moles of gas molecules. This is because a decrease in volume (increase in pressure) favors the direction that will decrease the number of gas molecules in the system, thereby reducing the pressure.
In this reaction, the left side of the equation has 2 moles of gas (2O3), and the right side has 3 moles of gas (3O2). Therefore, an increase in total pressure (compression) will favor the side with fewer moles of gas, which is the reactants (2O3).
Therefore, the correct answer is 4. Reactants.