How can i determine that the volume is decreasing?
Of the following equilibria, only ??? will shift to the left in response to a decrease in volume.
A) 2HI <->(g) H2 (g) + I2 (g)
B) N2 (g) + 3 H2 (g)<-> 2 NH3 (g)
C) 4 Fe (s) + 3 O2 (g)<-> 2 Fe2O3 (s)
D) 2 SO3(g)<-> 2 SO2 (g) + O2 (g)
E) H2 (g) + Cl2 (g) <-> 2 HCl (g)
The easy way to do is to say, "volume gos down, pressure goes ??. Of course the answer is up. All of these are gaseous reactions, so an increase in P will cause the equilibrium to shift to the side with the smaller number of moles of gas. So you look at each one.
A no change in number of moles so unaffected.
B. P goes up, rxn shift to the right.
C. P goes up, rxn shifts to the right.
D. P goes up, rxn shifts to the left.
E. P goes up, no change.
To determine if the volume is decreasing, you need to understand the concept of Le Chatelier's principle. According to this principle, when a system at equilibrium is subjected to a change, it will respond by shifting in a way that partially counteracts the change.
In the case of a decrease in volume, the system will try to compensate by shifting towards the side with fewer gas molecules. This is because a decrease in volume creates a higher pressure, and the system will try to reduce the pressure by favoring the direction with fewer gas molecules.
Now, let's analyze the given equilibria:
A) 2HI (g) <-> H2 (g) + I2 (g)
This reaction consists of 3 gas molecules on the left side and 1 gas molecule on the right side. A decrease in volume would increase the pressure, and according to Le Chatelier's principle, the system would shift to the side with fewer gas molecules. Therefore, this equilibrium would shift to the right, not left.
B) N2 (g) + 3 H2 (g) <-> 2 NH3 (g)
This reaction consists of 4 gas molecules on the left side and 2 gas molecules on the right side. A decrease in volume would increase the pressure, and the system would shift to the side with fewer gas molecules. Therefore, this equilibrium would shift to the right, not left.
C) 4 Fe (s) + 3 O2 (g) <-> 2 Fe2O3 (s)
This reaction involves solids, not gases. Changes in volume do not affect the concentrations of solids. Therefore, the equilibrium position would not be influenced by a change in volume.
D) 2 SO3(g) <-> 2 SO2 (g) + O2 (g)
This reaction consists of 3 gas molecules on the left side and 3 gas molecules on the right side. A decrease in volume would increase the pressure, and the system would shift to the side with fewer gas molecules. Therefore, this equilibrium would shift to the left.
E) H2 (g) + Cl2 (g) <-> 2 HCl (g)
This reaction consists of 2 gas molecules on the left side and 2 gas molecules on the right side. A decrease in volume would increase the pressure, but since the number of gas molecules remains the same on both sides, the equilibrium position would not be influenced by a change in volume.
Therefore, the only equilibrium that would shift to the left in response to a decrease in volume is D) 2 SO3(g) <-> 2 SO2 (g) + O2 (g).