Equilibrium and Thermodynamics - AP Chemistry

Recall that the rate constant for the equilibrium reaction

is , always remembering that it is products over reactants.

Since we are at equilibrium, for every mol of we have, we have 3 mols of . Therefore,

Likewise, for every 2 mols of we have, we have 2 mols of . Therefore,

Plugging all of this into the above equation we get

Adding an inert gas will not shift the equilibrium toward the products, but neither will it shift the equilibrium toward the reactants. It has no effect because the partial pressure of each species, , , and , will still remain constant. Since the partial pressures are constant, we know from the ideal gas law that the concentration of each species remains constant. For example, for ,

To solve this problem, we must make use of the following equation:

At equilibrium, the value of is 0. Therefore, we can simplify the equation.

We must also convert temperature from degress Celsius into Kelvin.

Because formic acid is a weak acid, it will exist in equilibrium with its conjugate base, formate, while in water according to the following reaction:

If formate is added to this solution, then according to Le Chatlier's principle, the reaction will shift towards the left. Thus, the in solution will decrease and the pH will consequently increase. Furthermore, the addition of formate has no effect on the pKa.

To answer this question, it is necessary to use following equation:

We can solve for by rearranging the equation to the following:

For this question, we're asked to determine the correct equilibrium expression for a given reaction that is in equilibrium.

To begin with, we need to remember that pure solids and liquids are omitted from the equilibrium expression. Only gases and aqueous components of a reaction are included. The reason for this is because in a reaction at equilibrium, a pure solid or liquid will not have any appreciable change in concentration. Because their concentrations remain essentially unchanged, their value gets incorporated into the constant.

Also remember that an equilibrium expression is represented by the products divided by the reactants, each of which is raised to its stoichiometric coefficient.

In the case of the reaction given, we can see that it is already balanced. This means we are ready to set up the expression.

Here, the term is being squared because it has a stoichiometric coefficient of in the equilibrium reaction. Also notice that does not show up in the expression because it is a solid.

There are two concepts to consider in this problem. First, the question asks for the . Reaction 3 has being reduced so the potential for the half-reaction becomes negative. half-reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

Because formic acid is a weak acid, it will exist in equilibrium with its conjugate base, formate, while in water according to the following reaction:

If formate is added to this solution, then according to Le Chatlier's principle, the reaction will shift towards the left. Thus, the in solution will decrease and the pH will consequently increase. Furthermore, the addition of formate has no effect on the pKa.

According to Le Chatelier's principle, any changes in concentration or pressure to a system at equilibrium will cause the system to readjust to a new equilbrium. The addition of a reactant will cause the reaction to shift to the right, increasing the equilibirum concentration of the products. Thus, adding additional reactant A would increase the equlibrium concentration of product C.

Removing species A or adding species E will drive the reaction to the left, reducing the amount of species C. Since the reaction is exothermic, heat can be considered a product. Thus, increasing the temperature will also shift the reaction to the left.

Recall that the rate constant for the equilibrium reaction

is , always remembering that it is products over reactants.

Since we are at equilibrium, for every mol of we have, we have 3 mols of . Therefore,

Likewise, for every 2 mols of we have, we have 2 mols of . Therefore,

Plugging all of this into the above equation we get

Adding an inert gas will not shift the equilibrium toward the products, but neither will it shift the equilibrium toward the reactants. It has no effect because the partial pressure of each species, , , and , will still remain constant. Since the partial pressures are constant, we know from the ideal gas law that the concentration of each species remains constant. For example, for ,

To solve this problem, we must make use of the following equation:

At equilibrium, the value of is 0. Therefore, we can simplify the equation.

We must also convert temperature from degress Celsius into Kelvin.

To answer this question, it is necessary to use following equation:

We can solve for by rearranging the equation to the following:

For this question, we're asked to determine the correct equilibrium expression for a given reaction that is in equilibrium.

To begin with, we need to remember that pure solids and liquids are omitted from the equilibrium expression. Only gases and aqueous components of a reaction are included. The reason for this is because in a reaction at equilibrium, a pure solid or liquid will not have any appreciable change in concentration. Because their concentrations remain essentially unchanged, their value gets incorporated into the constant.

Also remember that an equilibrium expression is represented by the products divided by the reactants, each of which is raised to its stoichiometric coefficient.

In the case of the reaction given, we can see that it is already balanced. This means we are ready to set up the expression.

Here, the term is being squared because it has a stoichiometric coefficient of in the equilibrium reaction. Also notice that does not show up in the expression because it is a solid.

There are two concepts to consider in this problem. First, the question asks for the . Reaction 3 has being reduced so the potential for the half-reaction becomes negative. half-reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

According to Le Chatelier's principle, any changes in concentration or pressure to a system at equilibrium will cause the system to readjust to a new equilbrium. The addition of a reactant will cause the reaction to shift to the right, increasing the equilibirum concentration of the products. Thus, adding additional reactant A would increase the equlibrium concentration of product C.

Removing species A or adding species E will drive the reaction to the left, reducing the amount of species C. Since the reaction is exothermic, heat can be considered a product. Thus, increasing the temperature will also shift the reaction to the left.

Recall that the rate constant for the equilibrium reaction

is , always remembering that it is products over reactants.

Since we are at equilibrium, for every mol of we have, we have 3 mols of . Therefore,

Likewise, for every 2 mols of we have, we have 2 mols of . Therefore,

Plugging all of this into the above equation we get

Adding an inert gas will not shift the equilibrium toward the products, but neither will it shift the equilibrium toward the reactants. It has no effect because the partial pressure of each species, , , and , will still remain constant. Since the partial pressures are constant, we know from the ideal gas law that the concentration of each species remains constant. For example, for ,

To solve this problem, we must make use of the following equation:

At equilibrium, the value of is 0. Therefore, we can simplify the equation.

We must also convert temperature from degress Celsius into Kelvin.