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.

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.

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. 

Dynamic equilibrium means that the forward and reverse reactions are occurring at the same rate so that there is no net change in the concentrations. 

The definition of reversible reactions is that they are indeed reversible; in other words, they can proceed in the forward and reverse directions. Since the products can react as well to reform the reactants, these reactions usually do not proceed to completion and instead exist in a dynamic state. 

For the reaction:

The expression for the equilibrium constant is:

The reaction quotient is:

The reaction quotient is much larger than the equilibrium constant, so the reaction will proceed to the left.

The forward and reverse reaction rates are equal for reactions in equilibrium, however, the molar concentration of the reactants and products are usually not equal themselves. One can calculate the equilibrium constant using the equation below.

Equilibrium expressions are a ratio of the concentration of products raised to the stoichiometric coefficient, divided by reactants raised to the stoichiometric coefficient. Solids are omitted from this expression.

For the generic example , the equilibrium expression is .

For our given reaction, we can find the equilibrium expression in a similar manner. Do not forget to omit the solid component of the reaction!

For a buffer solution, you need a weak acid and its conjugate base, or a weak base and its conjugate acid. HCO₃ from the NaHCO₃ and CO_3– from K₂CO₃ are this pair.