Respuesta :
This is a question about LeChatelier’s rules. When it comes to volume changes, systems will shift equilibrium in ways that try to keep pressure stable. This usually means changing the total number of gas particles.
For the first reaction, there are two molecules of gas in both the reactants and the products. There is no benefit to favouring one side over the other because the total amount of gas will always be the same. Therefore decreasing the volume will have little or no effect on the equilibrium.
For the second reaction, there are two molecules of gas on the left and one on the right. If volume is increased, the equilibrium will shift in a way that produces more gas molecules in order to keep pressure equal. Therefore the equilibrium will shift left.
For the third reaction, there are no molecules of gas on the left and one on the right. By the same logic as the second reaction, the equilibrium will shift right.
For the first reaction, there are two molecules of gas in both the reactants and the products. There is no benefit to favouring one side over the other because the total amount of gas will always be the same. Therefore decreasing the volume will have little or no effect on the equilibrium.
For the second reaction, there are two molecules of gas on the left and one on the right. If volume is increased, the equilibrium will shift in a way that produces more gas molecules in order to keep pressure equal. Therefore the equilibrium will shift left.
For the third reaction, there are no molecules of gas on the left and one on the right. By the same logic as the second reaction, the equilibrium will shift right.
No net shift in the reactants and products is an equilibrium constant. In the first reaction, there is no effect, in second shifts to the left and third shifts to right.
What is the equilibrium shifts?
Equilibrium shift can be explained by the rules of Le and the change in the volume affects the system as the number of the particles gets varied.
For the first reaction [tex]\rm CO(g) + H_{2}O(g) \Leftrightarrow CO_{2}(g) + H_{2}(g)[/tex] when the volume is decreased then the pressure will increase then the equilibrium will shift towards the fewer and then more number of the gas moles and hence, will have no net effect.
In the second reaction [tex]\rm PCl_{3}(g) + Cl_{2}(g) \Leftrightarrow PCl_{5}(g)[/tex] when the volume is increased then the pressure will decrease and the reaction will shift towards more moles of the gas present that is the left side.
In the third reaction, [tex]\rm CaCO_{3}(s) \Leftrightarrow CaO(s) + CO_{2}(g)[/tex] when the volume is increased then the pressure will decrease and the equilibrium will shift towards the side where more moles are present. Hence the increase in the volume will shift the equilibrium towards the right.
Therefore, a decrease in volume will have no effect, while the increase in volume will shift the reaction towards the left and right.
Learn more about the equilibrium constant here:
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