Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

First, let's analyze the change in moles of gas on both sides for each reaction: a. 1 mole + 3 moles \(\to\) 2 moles b. 1 mole \(\to\) 1 mole + 1 mole c. 1 mole + 1 mole \(\to\) 2 moles d. 1 mole \(\to\) 1 mole + 1 mole e. 0 moles \(\to\) 0 moles + 1 mole

Let's determine the direction of the equilibrium shift for each reaction in response to the increase in volume (decrease in pressure): a. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the reactants (left) side. b. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. c. Since the number of moles of gas is the same on either side, increasing the volume will not cause a shift in the equilibrium position. d. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. e. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. So, the shifts in the equilibrium positions for each reaction are as follows: a. Shifts to the left b. Shifts to the right c. No shift d. Shifts to the right e. Shifts to the right