Answer
(a) The position of equilibrium will not shift.
(b) $H_2$ being a reactant,the reaction shifts to the right side(products).
(c) An increase in temperature will shift the reaction to the left side(reactants) as it is an exothermic reaction.
(d) As in the right side has fewer 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 neon is added at constant volume but the reaction will shift towards left if neon 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 evolved in product side in an exothermic 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 left side(reactants).
(d) Lowering the volume increases the pressure. According to Le Chatelier’s principle an increase in pressure causes a shift towards the side with fewer 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 neon gas does not affect equilibrium if it is added in constant volume.
But if neon gas is added at constant pressure the the reaction shifts toward left side(reactant) as in left side there is greater number of particles/moles.
