What happens to the concentration of NO2(g)when the total pressure on the equilibrium reaction
2 Pb(NO3)2(s) <--> 2 PbO(s) + 4NO2(g) + O2(g) is increased (by compression)?
1. increases
2. Unable to determine
3. remains the same
4. decreases
There are 5 mols on the right and only 2 on the left so the NO2 would decrease, correct?
Yes but your reasoning is faulty. True that there are 4 mols gas on the right but there are no mols gas on the left.
To determine what happens to the concentration of NO2(g) when the total pressure on the equilibrium reaction is increased by compression, we need to consider Le Chatelier's principle.
Le Chatelier's principle states that when a stress is applied to a system at equilibrium, the equilibrium will shift in a way that minimizes the effect of the stress. In this case, the stress is an increase in total pressure.
When the total pressure on the equilibrium reaction is increased by compression, the system will try to reduce the pressure by shifting the equilibrium in the direction that produces fewer gas molecules.
Looking at the reaction, we can see that there are 4 moles of NO2(g) on the right side of the equation. By decreasing the concentration of NO2(g), the total number of moles of gas will decrease, thereby reducing the pressure.
Therefore, the system will shift towards the left to produce more reactants and decrease the concentration of NO2(g). This means that the concentration of NO2(g) will decrease when the total pressure is increased by compression.
Therefore, the correct answer is 4. decreases.