Please ans all the three questions
Dear student
(1)
The given reaction is
A (s) + 2B (g) ---------------> 2C (g)
According to Le Chatelier’s principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. So there are 2 moles of gas in reactant side and 2 moles of gas on the product side hence having equal number of gas moles.
Then the position of equilibrium depends on the concentration of A i.e. According to Le Châtelier, the position of equilibrium will move in such a way as to counteract the change. Hence the position of equilibrim shift to the right by increasing the concentrtion of A.
For the remaining question please ask in a diffrent thread.
Regards
(1)
The given reaction is
A (s) + 2B (g) ---------------> 2C (g)
According to Le Chatelier’s principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. So there are 2 moles of gas in reactant side and 2 moles of gas on the product side hence having equal number of gas moles.
Then the position of equilibrium depends on the concentration of A i.e. According to Le Châtelier, the position of equilibrium will move in such a way as to counteract the change. Hence the position of equilibrim shift to the right by increasing the concentrtion of A.
For the remaining question please ask in a diffrent thread.
Regards