By increasing the concentration of one of the reactants, the reaction will compensate by shifting to the right to increase production of products.

Increasing the concentration of one of the products (such as increasing [C]), however, would have the opposite effect. Increasing the temperature of an exothermic reaction would shift the reaction to the left, while increasing the temperature of an endothermic reaction would lead to a rightward shift. Finally, decreasing the volume leads to an increase in partial pressure of each gas, which the system compensates for by shifting to the side with fewer moles of gas. In this case, the right side has three moles of gas, while the left side has two; thus decreasing volume would shift equilibrium to the left.

Le Chatelier's principle states that changes in pressure are attributable to changes in volume. If we increase the volume, the reaction will shift toward the side that has more moles of gas. If we decrease the volume, the reaction will shift toward the side that has less moles of gas. Since the product side has only two moles of gas, compared to the reactant side with four moles, the reaction would shift toward the product side, and more NH₃ would form.

With increased pressure, each reaction will favor the side with the least amount of moles of gas. In this problem we are looking for the reactions that favor the products in this scenario. I will favor reactants, II will favor products, III will favor reactants.

An increase in volume will result in a decrease in pressure at constant temperature. As a result, the equilibrium will shift toward the side with the greater total moles of gas, according to Le Chatelier's Principle. This will result in less AX₅ being produced.

The K_eq tells us that the reaction favors the products because it is greater than 1. The definition of equilibrium is that the rate of formation of products equals the rate of formation of reactants.

Ksp is dependent only on the species itself and the temperature of the solution. Adding another compound or stressing the system will not affect Ksp.

Both Na₂SO₄ and ammonia are slightly basic compounds. Thus, adding ammonia will create a common ion effect, where less sodium sulfate will be able to dissolve and some would precipitate out of solution. 

LeChatelier’s principle states that if a stress is applied to a rxn mixture at equilibrium,
reaction occurs in the direction that relieves the stress. Therefore, adding H2 will produce
more methanol. By increasing the volume and decreasing the pressure, there will be a net
reaction in the direction that increases the number of moles of gas. So since there are 3
moles of gas on the products side and only 1 mole of gas on the reactants side, if the volume  is increased less methanol will be produced. Since Ar is an inert, it will have no effect on the amount of methanol produced. Removing CO will cause a shift in the reaction from right to left causing less methanol to be produced.

This is an application of Le Chatlier's Principle. When you take away heat from the reaction, the reaction shifts toward the left in order to compensate from the heat loss. The reaction may then be rewritten to include energy as a reactant.

Since the energy is on the reactant side, the reaction is endothermic.

A catalyst does not change the equilibrium of a reaction. Catalysts will affect reaction rate by lowering activation energy, but will ultimately have no effect on the amount of reactants and products present when equilibrium is reached.

Adding or removing reactants or products will result in a shift in equilibrium according to Le Chatelier's principle. Similarly, changing the temperature of a reaction will affect equilibrium in different ways depending on the enthalpy of reaction; increasing the temperature of an exothermic reaction will increase the reactant concentration, while increasing the temperature of an endothermic reaction will increase the products.

To answer this question we need to combine our knowledge of a few different subjects. Under normal cicrumstances, when determining the effects of a system pressure change, we compare the number of moles of gas on either side of the equilibrium. In this case, there are 2 moles of gas on either side, which means that neither side is favored in terms of pressure changes.

However, we consider that in this case, the pressure is increasing while the volume remains constant. Since the total moles of gas cannot be changed in a closed system, we have to conclude that increased pressure results in increased temperature. If  is greater than , then  must also be greater than .

Since this reaction is exothermic, increasing the temperature (i.e. adding heat) causes the reaction to shift to the left; therefore, the concentration of  increases.