Chemical Equilibrium - AP Chemistry
Card 0 of 870
Consider the following balanced reaction.
Write the equilibrium constant expression for this reaction.
Consider the following balanced reaction.
Write the equilibrium constant expression for this reaction.
When writing an equilibrium constant expression, remember that products are on the top, and reactants are on the bottom of the expression. The coefficients for the compounds in the balanced reaction become the exponents for the compounds seen in the expression.
When writing an equilibrium constant expression, remember that products are on the top, and reactants are on the bottom of the expression. The coefficients for the compounds in the balanced reaction become the exponents for the compounds seen in the expression.
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Consider the following balanced reaction.
Write the equilibrium constant expression for this reaction.
Consider the following balanced reaction.
Write the equilibrium constant expression for this reaction.
A key point to remember is that the equilibrium constant expression only includes compounds that are in aqueous solution or in the gaseous phase. Pure solids and liquids are omitted from the expression. As a result, solid barium sulfate will not be included in the expression.
The equilibrium constant is found by the concentration of the products over the concentration of the reactants, each raised to the power of their coefficients.
In our reaction, the only reactant is a pure solid, so only the products are used in the expression.
A key point to remember is that the equilibrium constant expression only includes compounds that are in aqueous solution or in the gaseous phase. Pure solids and liquids are omitted from the expression. As a result, solid barium sulfate will not be included in the expression.
The equilibrium constant is found by the concentration of the products over the concentration of the reactants, each raised to the power of their coefficients.
In our reaction, the only reactant is a pure solid, so only the products are used in the expression.
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What is the correct equilibrium expression for this reaction?
What is the correct equilibrium expression for this reaction?
Solids are not included in any equilibrium expression. The equilibrium expression will contain only aqueous solutions and gases. In addition, the coefficients of a balanced chemical equation correspond to exponents, not coefficients, in the equilibrium expression.
For this reaction, the only relevant product is oxygen gas and the only relevant reactant is fluorine gas.
Solids are not included in any equilibrium expression. The equilibrium expression will contain only aqueous solutions and gases. In addition, the coefficients of a balanced chemical equation correspond to exponents, not coefficients, in the equilibrium expression.
For this reaction, the only relevant product is oxygen gas and the only relevant reactant is fluorine gas.
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For the following reaction, what would be the correct equilibrium expression?
For the following reaction, what would be the correct equilibrium expression?
For any equilibrium expression, solids and liquids are excluded because their concentrations are presumed to be constant during the reaction. The equilibrium expression is defined as the ratio of the concentration of products divided by the concentration of the reactants. Each reactant or product is raised to the power corresponding to its coefficient in the balanced chemical equation.
Using an arbitrary example:
Compare this to our question. Remember to exclude the solid magnesium!
For any equilibrium expression, solids and liquids are excluded because their concentrations are presumed to be constant during the reaction. The equilibrium expression is defined as the ratio of the concentration of products divided by the concentration of the reactants. Each reactant or product is raised to the power corresponding to its coefficient in the balanced chemical equation.
Using an arbitrary example:
Compare this to our question. Remember to exclude the solid magnesium!
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Determine the equation for the equilibrium constant of the following unbalanced reaction:
Determine the equation for the equilibrium constant of the following unbalanced reaction:
The equilibrium constant, 
, is the ratio of the concentration of products raised to their coefficients, over the concentration of reactants raised to their coefficients. To find its value, we first need to balance the equation and then consider only the products and reactants that actually have concentrations (i.e. aqueous and gaseous species). Liquids and solids can be omitted from the calculation.
We note that chlorine only appears once on the right, so we add a 2 coefficient to balance the equation:
Since 
is a solid (most hydroxides precipitate, unless they are paired with alkali metals, barium, or calcium) it will not be included in the equilibrium constant calculation.
The chloride ion concentration is raised to the second power because of the 2 coefficient that we added. So, putting products (species on the right side, other than the solid) on the top and reactants (species on the left) on bottom, we get:
The equilibrium constant,
We note that chlorine only appears once on the right, so we add a 2 coefficient to balance the equation:
Since
The chloride ion concentration is raised to the second power because of the 2 coefficient that we added. So, putting products (species on the right side, other than the solid) on the top and reactants (species on the left) on bottom, we get:
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Consider the following generic reaction:
If the equilibrium concentration of 
is 1M, what is the equilibrium concentration of 
?
Consider the following generic reaction:
If the equilibrium concentration of
We can set up the equilibrium expression by placing the products, raised to the power of their coefficients, in the numerator and reactants, raised to the power of their coefficients, in the denominator:
We are given the value of the equilibrium constant and the equilibrium concentration of 
. Using stoichiometric coefficients, we can determine that a variable concentration, 
, of 
will be present at equilibrium, and 
of 
will be present at equilibrium. Plugging in:
Simplify:
Now we can solve for 
(which is the equilibrium concentration of 
, as we assigned) by dividing both sides by 4 and then taking the cube root.
Rounding to 2 sig figs (to match the two numbers we were given by the problem), we get:
We can set up the equilibrium expression by placing the products, raised to the power of their coefficients, in the numerator and reactants, raised to the power of their coefficients, in the denominator:
We are given the value of the equilibrium constant and the equilibrium concentration of
Simplify:
Now we can solve for
Rounding to 2 sig figs (to match the two numbers we were given by the problem), we get:
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Which of the following reactions will be favored when the pressure in a system is increased?
I. 
II. 
III. 
Which of the following reactions will be favored when the pressure in a system is increased?
I.
II.
III.
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.
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.
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Which of the following will cause an equilibrium shift in an exothermic reaction towards the products?
I. Decreasing the temperature
II. Evaporating the product
III. Adding a catalyst
Which of the following will cause an equilibrium shift in an exothermic reaction towards the products?
I. Decreasing the temperature
II. Evaporating the product
III. Adding a catalyst
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.
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.
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Which of the following stresses to a system at equilibrium cause an increase in the production of CH3OH ? CO(g)+ 2H2(g) → CH3OH(g)
(a) H2 is added. (b) The volume is increased. (c) Argon is added. (d) Removing CO.
Which of the following stresses to a system at equilibrium cause an increase in the production of CH3OH ? CO(g)+ 2H2(g) → CH3OH(g)
(a) H2 is added. (b) The volume is increased. (c) Argon is added. (d) Removing CO.
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.
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.
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Which of the following does not affect the equilbrium of a reaction?
Which of the following does not affect the equilbrium of a reaction?
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.
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.
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What would happen to the Ksp if NH3 was added to an existing solution of Na2SO4?
What would happen to the Ksp if NH3 was added to an existing solution of Na2SO4?
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.
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.
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Which of the following would occur if NH3 was added to an existing solution of Na2SO4?
Which of the following would occur if NH3 was added to an existing solution of Na2SO4?
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.
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.
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Figure 1: Ammonia gas formation and equilibrium
Experimental data shows that the reaction shifts to the left at very cold temperatures. Using this information, what type of reaction is shown in Figure 1?
Figure 1: Ammonia gas formation and equilibrium
Experimental data shows that the reaction shifts to the left at very cold temperatures. Using this information, what type of reaction is shown in Figure 1?
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.
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.
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Figure 1: Ammonia gas formation and equilibrium
What would most likely happen if a scientist decreased the volume of the container in which the reaction occurs?
Figure 1: Ammonia gas formation and equilibrium
What would most likely happen if a scientist decreased the volume of the container in which the reaction occurs?
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.
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.
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Consider the following reaction system, which has a Keq of 1.35 * 104, taking place in a closed vessel at constant temperature.
Which of the following is NOT true about this system at equilibrium?
Consider the following reaction system, which has a Keq of 1.35 * 104, taking place in a closed vessel at constant temperature.
Which of the following is NOT true about this system at equilibrium?
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.
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.
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Which of the following stresses would lead the exothermic reaction below to shift to the right?
Which of the following stresses would lead the exothermic reaction below to shift to the right?
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.
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.
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Which of the following will not shift the equilibrium of a reaction?
Which of the following will not shift the equilibrium of a reaction?
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.
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.
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Self-ionization of water is endothermic. What is the value of the sum pH + pOH at 
?
Self-ionization of water is endothermic. What is the value of the sum pH + pOH at
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.
Recall that ion-product constant of water,
An endothermic reaction signifies that heat is at the reactant side. By the LeChatelier's principle, increased heat to
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Consider the following balanced chemical equation:
What will be the effect on the concentration of 
if the overall pressure of the system increases, but the volume remains constant? Why?
Consider the following balanced chemical equation:
What will be the effect on the concentration of
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.
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
Since this reaction is exothermic, increasing the temperature (i.e. adding heat) causes the reaction to shift to the left; therefore, the concentration of
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Consider the following reaction:
What will happen to the equilibrium constant if the concentration of 
is increased? Why?
Consider the following reaction:
What will happen to the equilibrium constant if the concentration of
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.
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
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.
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