Answer
(a) The position of equilibrium will not shift.
(b) $H_{2}O$ being a reactant,the reaction shifts to the right side(products).
(c) An increase in temperature will shift the reaction to the right side(products) as it is an endothermic reaction.
(d) As in the right side has greater number particles, the reaction shifts right(products).
(e) A catalyst will have no effect on equilibrium,so there is no effect of catalyst on the position of equilibrium.
(f) The equilibrium remains unchanged if an inert is added at constant volume but the reaction will shift towards right side if an inert gas is added at constant pressure.
Work Step by Step
(a) $C(s)$ being a solid, it has no effect on the equilibrium.
(b) According to Le Chatelier's Principle, when a reactant is added to a reaction, the equilibrium shifts to the right side(products).
(c) Heat is absorbed in product side in an endothermic reaction.On increasing temperature according to Le Chatielier's principle the reaction shift in that direction so that the effect of excess heat is minimized and the reaction shift in right side(products).
(d) An increase in the volume decreases the pressure. According to Le Chatelier’s principle an decrease in pressure causes a shift towards the side with greater number particles/moles,and the reaction shifts toward right side(products).
(e) A catalyst does not have an effect on the location of equilibrium.In presence of catalyst Equilibrium will only be reached early.
(f) The addition of an inert gas does not affect equilibrium if it is added in constant volume.
But if inert gas is added at constant pressure the the reaction shifts toward right side(product) as in right side there is greater number of particles/moles.
