LeChatlier’s Principle?

C(s) + CO₂(g) ⇌ 2CO(g) …. ΔH = +119 kJ

a) The equilibrium position shifts to the RIGHT.
When CO is removed, according to Le Chatelier’s Principle, the equilibrium position shifts to the right in order to produce more CO.

b) The equilibrium position shifts to the RIGHT.
According to Le Chatelier’s Principle, adding heat would favor the endothermic reaction. As the forward reaction is endothermic, the equilibrium shifts to the right.

c) The equilibrium position shifts to the RIGHT.
When CO₂ is added, according to Le Chatelier’s Principle, the equilibrium position shifts to the right in order to consume more CO₂.

d) The equilibrium position shifts to the RIGHT.
In the equation, there is 1 mole of gaseous molecules on the left but 2 moles on the right. The increase in volume of the container leads to the decrease in total pressure inside the container. According to the Le Chatelier’s principle, the equilibrium position shifts to the right in order to increase the number of gaseous molecules.

e) The equilibrium position is UNCHANGED.
Carbon is solid. The effective concentration of carbon is constant. Addition or Removal of carbon does not affect the effective concentration of carbon. Therefore, removal of carbon causes no change to the equilibrium position.

a) The reaction will shift to the right to replenish to some extent the CO that was removed.

b) This reaction is endothermic. Heat appears on the reactant side. Thus, raising the temperature will cause the reaction to produce more CO which will absorb the
additional heat.

c) Adding CO2 puts a stress on the reactant side, so this will cause the reaction to shift right and produce more CO.

d) Increasing the volume will decrease the pressure. The system will move in a direction to increase the pressure. Since there are 2 molecules of CO gas obtained
for every one molecule of CO2 gas, the reaction will shift right which increases the pressure since more molecules are obtained. The solid C will not affect the pressure.

e) C being a solid does not appear in the equilibrium constant expression. Thus, no effect on the equilibrium system.