Le Chatelier's Principle - AP Chemistry

According to Le Chatelier's principle, when you compress a system, its volume decreases, so partial pressure of the all the gases in the system increases. The system will act to try to decrease the pressure by decreasing the moles of gas.

In an endothermic reaction, heat can be treated as a reactant. Thus, if you add more reactant (heat), the system will shift to get rid of the extra reactant and shift to the right to form more products.

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.

I) Decreasing the temperature would take away heat from the system (a product), driving the reaction towards the products. II) Evaporating product would take a product away from the system, driving the reaction towards the products. III) Adding a catalyst only affects the rate of the reaction and does not effect equilibrium.

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.

Le Chatelier's principle states that the concentration of reactants/products, the addition/subtraction of heat, and changing the volume of a reaction would all be factors that affect equilibrium. A catalyst alters the reaction rate without changing equilibrium.

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.

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.

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.

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 Keq 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.

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.

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.

Recall that ion-product constant of water, , is at and .

An endothermic reaction signifies that heat is at the reactant side. By the LeChatelier's principle, increased heat to shifts the equilibrium to the right side, favoring the increase of and . This means that and both increase, decreasing pH and pOH to less than 7, each. As a result, pH + pOH is less than 14.

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.

By Le Chatelier's principle, when a concentration on the right side increases, the reaction shifts towards the left side in order to balance out the disturbance to equilibrium. In doing so, heat will be consumed along with , which will lower the temperature of the system. Since the equilibrium constant is temperature dependent its value will change, but without experimental observation it is not entirely clear how it will change.

It is true that concentration changes normally do not affect the equilibrium constant, but because in this case temperature is affected by concentration changes, that rule does not hold.

General formula for the reaction quotient:

Since the reaction starts at equilibrium the initial value of the reaction quotient is equal to the equilibrium constant.

When , the reaction is at equilibrium.

When , the reaction favors the reactant(s).

When , the reaction favors the product(s).

For this problem:

Since the reaction started at equilibrium at this problem, initially . Therefore in order to make the (and therefore favor the production of products) we must alter the above equation for to have a smaller value. There are many ways to do this. We can reduce the concentration or either or both of the products, and/or we can increase the concentration of reactant.

Incorrect answers and explanations:

Removal of decreases the concentration of reactant and would cause

Addition of increases the concentration of product and would cause

Increasing the temperature of the system, thereby providing activation energy is incorrect because this is an exothermic reaction and therefore energy (in this case in the form of heat) is considered a product and would cause . Also it is worth noting that changing the activation energy of a reaction only changes the rate of the reaction, not the direction in which it proceeds.

Addition of thereby decreasing the concentration of products would not change the value of , because adding will also decrease the concentration or the reactants by the same factor.

Correct answer and explanation:

Addition of is the only answer which will alter the value of to be less than , because addition of will increase the concentration of reactants. This increases the value of the denominator in the following equation :

Since dividing by a larger number results in a lower value for , the addition of results in (when the reaction starts at equilibrium, if the reaction didn't start at equilibrium increasing the concentration of glucose will still lower the value of , but we would need more information to know if it decreased it enough to be less than .

For this question, we're given a reversible reaction and are asked to identify conditions which would drive the reaction towards the left. In order to answer this, we'll need to understand the fundamentals of Le Chatelier's principle. The basic idea of this principle is that whenever stress is added to a system that is in equilibrium, that system will establish a new equilibrium in order to handle the added stress.

In this question, we can see that both reactant and product are gases. However, they differ in amount. For every one mole of reactant, two moles of product are made. This is an important distinction that will allow us to get our answer. Whenever a system in equilibrium is composed of gases, increased pressure will favor the gas that can exist in a smaller amount. Conversely, decreased pressure will favor the gas that can exist in a greater amount. The relevance that this has to this question is that increased pressure will actually push the reaction to the left, since it is the reactant that can exist in fewer moles as a gas. A decrease in pressure would be expected to have the opposite effect, as it would increase the amount of product and thus would drive the reaction to the right. Reducing the amount of product in the reaction vessel would also be expected to have the opposite effect, as the system would respond by producing more product in order to help offset the loss. And finally, there is no indication of whether the addition of water would have any kind of effect on the reaction equilibrium in this question, thus there is no way to tell.

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.