Equilibrium - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 266
Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O 
H2CO3
H2CO3 
HCO3- + H+
While the kidney is able to compensate for many acid/base changes in our bodies, vomiting is a temporary cause of acid/base imbalance. While vomiting may allow our bodies to get rid of toxic substances, it also causes us to lose gastric acid, which influences blood pH. How would the loss of gastric acid change the pH of our blood?
Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O
H2CO3
While the kidney is able to compensate for many acid/base changes in our bodies, vomiting is a temporary cause of acid/base imbalance. While vomiting may allow our bodies to get rid of toxic substances, it also causes us to lose gastric acid, which influences blood pH. How would the loss of gastric acid change the pH of our blood?
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This is an undercover Le Chatlier shift problem. The question tells us that vomiting causes us to lose gastric acid. In the equation that we can see above, losing H+ (in HCl) would pull the reaction to the right, increasing the concentration of HCO3-. Increasing the concentration of the base HCO3- increases the pH, leading the blood to become more basic.
This is an undercover Le Chatlier shift problem. The question tells us that vomiting causes us to lose gastric acid. In the equation that we can see above, losing H+ (in HCl) would pull the reaction to the right, increasing the concentration of HCO3-. Increasing the concentration of the base HCO3- increases the pH, leading the blood to become more basic.
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Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O 
H2CO3
H2CO3 
HCO3- + H+
What happens to the pH of our blood if we hyperventilate?
Carbonic anhydrase is an important enzyme that allows CO2 and H2O to be converted into H2CO3. In addition to allowing CO2 to be dissolved into the blood and transported to the lungs for exhalation, the products of the carbonic anhydrase reaction, H2CO3 and a related compound HCO3-, also serve to control the pH of the blood to prevent acidosis or alkalosis. The carbonic anhydrase reaction and acid-base reaction are presented below.
CO2 + H2O
H2CO3
What happens to the pH of our blood if we hyperventilate?
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This is an undercover Le Chatlier shift problem. If we hyperventilate, we expel more CO2, thus pulling the reaction to the left and decreasing the concentration of H2CO3. As the concentration of the acid decreases, the blood becomes more basic, leading to respiratory alkalosis.
This is an undercover Le Chatlier shift problem. If we hyperventilate, we expel more CO2, thus pulling the reaction to the left and decreasing the concentration of H2CO3. As the concentration of the acid decreases, the blood becomes more basic, leading to respiratory alkalosis.
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If the reactants and/or products in a chemical reaction are gases, the reaction rate can be determined by measuring the change of pressure as the reaction proceeds. Consider the following reaction and pressure vs. reaction rate data below.
Trial PXY(torr) PZ(torr) Rate (torr/s) 1 100 200 0.16 2 200 200 0.32 3 200 100 0.04 4 200 150 0.14
If the volume of the container were reduced, what would happen to the rate of the reaction?
If the reactants and/or products in a chemical reaction are gases, the reaction rate can be determined by measuring the change of pressure as the reaction proceeds. Consider the following reaction and pressure vs. reaction rate data below.
| Trial | PXY(torr) | PZ(torr) | Rate (torr/s) |
|---|---|---|---|
| 1 | 100 | 200 | 0.16 |
| 2 | 200 | 200 | 0.32 |
| 3 | 200 | 100 | 0.04 |
| 4 | 200 | 150 | 0.14 |
If the volume of the container were reduced, what would happen to the rate of the reaction?
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Reducing the volume of the container increases pressure. This results in a higher frequency of gas particle collisions, thereby increasing the rate of the reaction.
Reducing the volume of the container increases pressure. This results in a higher frequency of gas particle collisions, thereby increasing the rate of the reaction.
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Consider the following equation for the production of ammonia gas from hydrogen gas and nitrogen gas.
If the volume of the vessel containing hydrogen and nitrogen is decreased, the production of ammonia .
Consider the following equation for the production of ammonia gas from hydrogen gas and nitrogen gas.
If the volume of the vessel containing hydrogen and nitrogen is decreased, the production of ammonia .
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Since decreasing the volume of the container has the effect of increasing pressure, equilibrium is shifted to the right. An increase in pressure has the result of favoring the side of the reaction with fewer moles of gas. (According to the balanced equation, there are 4 moles on the reactant side as opposed to 2 moles on the product side).
Since decreasing the volume of the container has the effect of increasing pressure, equilibrium is shifted to the right. An increase in pressure has the result of favoring the side of the reaction with fewer moles of gas. (According to the balanced equation, there are 4 moles on the reactant side as opposed to 2 moles on the product side).
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Le Chatlier's principle states that when a stressor is introduced to a system, the system will shift its equilibrium state as a way of countering the stress. What are the factors (stressors) that apply to Le Chatlier's principle?
Le Chatlier's principle states that when a stressor is introduced to a system, the system will shift its equilibrium state as a way of countering the stress. What are the factors (stressors) that apply to Le Chatlier's principle?
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Le Chatelier's principle emphasizes three main stressors: changes in temperature, pressure, and concentration of reactants and products. If any of these three stressors are added, the Keq will shift in a way that counters this added stress. Temperature and concentration changes will affect any reaction, while pressure changes will only have a marked effect on reactions involving at least one gaseous compound.
Le Chatelier's principle emphasizes three main stressors: changes in temperature, pressure, and concentration of reactants and products. If any of these three stressors are added, the Keq will shift in a way that counters this added stress. Temperature and concentration changes will affect any reaction, while pressure changes will only have a marked effect on reactions involving at least one gaseous compound.
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The Haber process for creating ammonia is written below.
This reaction takes place in a glass container, and is allowed to progress to equilibrium. Which of the following manipulations to the system will NOT shift the equilibrium to the left?
The Haber process for creating ammonia is written below.
This reaction takes place in a glass container, and is allowed to progress to equilibrium. Which of the following manipulations to the system will NOT shift the equilibrium to the left?
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Le Chatlier's principle states that when a system at equilibrium is stressed, the equilibrium will shift accordingly in order to reduce the stress. There are a variety of ways to stress the system and elicit a shift.
-
Addition or removal of a reactant or product.
-
Changing the pressure of a system containing gases.
-
Changing the temperature of the system.
By increasing the pressure of the system, the reaction will shift in the direction which results in fewer gas molecules in the container. Since there are only two gas molecules on the product side and four gas molecules on the reactant side, we would predict a pressure increase to shift the equilibrium to the right.
Increasing temperature and adding ammonia both equate to adding product, which will shift the reaction away from the products and toward the reactants. Similarly, removing a reactant, such as nitrogen, will cause more of that reactant to be produced, also shifting the reaction to the left.
Le Chatlier's principle states that when a system at equilibrium is stressed, the equilibrium will shift accordingly in order to reduce the stress. There are a variety of ways to stress the system and elicit a shift.
-
Addition or removal of a reactant or product.
-
Changing the pressure of a system containing gases.
-
Changing the temperature of the system.
By increasing the pressure of the system, the reaction will shift in the direction which results in fewer gas molecules in the container. Since there are only two gas molecules on the product side and four gas molecules on the reactant side, we would predict a pressure increase to shift the equilibrium to the right.
Increasing temperature and adding ammonia both equate to adding product, which will shift the reaction away from the products and toward the reactants. Similarly, removing a reactant, such as nitrogen, will cause more of that reactant to be produced, also shifting the reaction to the left.
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Barium fluoride dissolves in solution according to the following equation.
Enough BaF2 is added to create a saturated liter of aqueous solution.
Suppose that 1M NaF is added to the solution, such that it does not change the volume of the solution. What would you expect to change as a result of the addition of NaF?
Barium fluoride dissolves in solution according to the following equation.
Enough BaF2 is added to create a saturated liter of aqueous solution.
Suppose that 1M NaF is added to the solution, such that it does not change the volume of the solution. What would you expect to change as a result of the addition of NaF?
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By adding NaF to the equation, 1M of F- ions are added to the solution. Thinking in terms of Le Chatlier's principle, the addition of a compound on one side of the equation will cause a shift to the other side of the reaction.
It DOES NOT affect the solubility product constant. Since there is an addition to the products side of the reaction, there will be a shift to the left, and more salt (reactant) will be precipitated. As a result, the solubility of the salt has decreased, because of the addition of NaF. This is known as the common ion effect.
By adding NaF to the equation, 1M of F- ions are added to the solution. Thinking in terms of Le Chatlier's principle, the addition of a compound on one side of the equation will cause a shift to the other side of the reaction.
It DOES NOT affect the solubility product constant. Since there is an addition to the products side of the reaction, there will be a shift to the left, and more salt (reactant) will be precipitated. As a result, the solubility of the salt has decreased, because of the addition of NaF. This is known as the common ion effect.
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Barium fluoride dissolves in solution according to the following equation.
Enough BaF2 is added to create a saturated liter of aqueous solution.
Suppose you double the amount of salt in the saturated solution. What would happen as a result of adding more BaF2?
Barium fluoride dissolves in solution according to the following equation.
Enough BaF2 is added to create a saturated liter of aqueous solution.
Suppose you double the amount of salt in the saturated solution. What would happen as a result of adding more BaF2?
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Remember that the concentration of the salt is not included in the solubility product constant equation, because it is a solid. Only the ion concentrations can affect the solubility of the salt in solution. When the solution is saturated, adding more salt will do nothing to the concentration of ions in the solution.
Remember that the concentration of the salt is not included in the solubility product constant equation, because it is a solid. Only the ion concentrations can affect the solubility of the salt in solution. When the solution is saturated, adding more salt will do nothing to the concentration of ions in the solution.
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A saturated solution of scandium hydroxide, which also contains solid scandium hydroxide, is treated with 0.1N HCl. The addition of acid will the solubility of scandium hydroxide because of .
A saturated solution of scandium hydroxide, which also contains solid scandium hydroxide, is treated with 0.1N HCl. The addition of acid will the solubility of scandium hydroxide because of .
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Scandium hydroxide dissociates according to this reaction.
As 0.1N HCl is added, it will quantitatively react with hydroxide anions to produce ScCl3 and water.
Hydroxide will be removed from the above equilibrium, and the system will compensate by shifting the equilibrium to the right, according Le Chatelier's principle. As the equilibrium shifts to the right, more solid scandium hydroxide is hydrolyzed, resulting in increased solubility.
Scandium hydroxide dissociates according to this reaction.
As 0.1N HCl is added, it will quantitatively react with hydroxide anions to produce ScCl3 and water.
Hydroxide will be removed from the above equilibrium, and the system will compensate by shifting the equilibrium to the right, according Le Chatelier's principle. As the equilibrium shifts to the right, more solid scandium hydroxide is hydrolyzed, resulting in increased solubility.
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Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
In studying the acid 
, a scientist finds that heat is released when the acid dissociates in solution. If this scientist raises the temperature in the vessel after the reaction has reached equilibrium, which of the following is most likely true?
Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.
All of the bases proceed in a similar fashion.
The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.
In studying the acid
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In exothermic process heat is released during the reaction.
Heat can be considered a product in this situation, and thus increasing the amount of heat in the vessel after the system has reached equilibrium will drive the reaction to the left. This property is a derivative of Le Chatelier's principle.
In exothermic process heat is released during the reaction.
Heat can be considered a product in this situation, and thus increasing the amount of heat in the vessel after the system has reached equilibrium will drive the reaction to the left. This property is a derivative of Le Chatelier's principle.
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A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:
In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.
The scientist in the passage attempts to modify the conditions for the reactions by placing the reactants in a sealed, variable-volume vessel at 1atm, and allowing it to reach equilibrium. She then decreases the volume of the vessel, increasing the pressure to 5atm. Which of the following is most likely to occur?
A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:
In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.
The scientist in the passage attempts to modify the conditions for the reactions by placing the reactants in a sealed, variable-volume vessel at 1atm, and allowing it to reach equilibrium. She then decreases the volume of the vessel, increasing the pressure to 5atm. Which of the following is most likely to occur?
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By changing the volume of the container, the scientist has changed the partial pressure of nitrogen gas in the vessel. As a result, the equilibrium shifts to the left, increasing the concentration of the reactants.
According to Le Chatelier's principle, decreasing pressure will shift equilibrium away from any gaseous compounds. The only gas in the reaction is the nitrogen product. Decreasing volume will, thus, shift equilibrium toward the reactants.
By changing the volume of the container, the scientist has changed the partial pressure of nitrogen gas in the vessel. As a result, the equilibrium shifts to the left, increasing the concentration of the reactants.
According to Le Chatelier's principle, decreasing pressure will shift equilibrium away from any gaseous compounds. The only gas in the reaction is the nitrogen product. Decreasing volume will, thus, shift equilibrium toward the reactants.
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Consider the following saturated solution. Assume it is at equilibrium.
Adding sodium chromate to the above solution would the solubility of lead chromate due to .
Consider the following saturated solution. Assume it is at equilibrium.
Adding sodium chromate to the above solution would the solubility of lead chromate due to .
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Adding sodium chromate increases the concentration of chromate ion in the solution, which shifts the reaction to the left due to the common ion effect. Thus, the solubility of lead chromate would decrease, as there would be an increased amount of solid lead chromate.
Adding sodium chromate increases the concentration of chromate ion in the solution, which shifts the reaction to the left due to the common ion effect. Thus, the solubility of lead chromate would decrease, as there would be an increased amount of solid lead chromate.
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A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.
The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:
Gibbs Free Energy Formula:
ΔG = ΔH – TΔS
Liquid-Solid Water Phase Change Reaction:
H2O(l) ⇌ H2O(s) + X
The scientist prepares two scenarios.
Scenario 1:
The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.
Scenario 2:
The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.
After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.
Scenario 3:
Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.
Scenario 4:
The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.
The same scientist in the passage measures the variables of another reaction in the lab. He knows that this reaction is spontaneous under standard conditions, with a standard free energy change of –43 kJ/mol. Using laboratory-calculated variables, he determines that the Gibbs Free Energy has a value of 0 kJ/mol. He then calculated the reaction quotient of this reaction, while knowing the equilibrium constant was 3 x 103. What is true of the reaction quotient?
A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.
The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:
Gibbs Free Energy Formula:
ΔG = ΔH – TΔS
Liquid-Solid Water Phase Change Reaction:
H2O(l) ⇌ H2O(s) + X
The scientist prepares two scenarios.
Scenario 1:
The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.
Scenario 2:
The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.
After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.
Scenario 3:
Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.
Scenario 4:
The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.
The same scientist in the passage measures the variables of another reaction in the lab. He knows that this reaction is spontaneous under standard conditions, with a standard free energy change of –43 kJ/mol. Using laboratory-calculated variables, he determines that the Gibbs Free Energy has a value of 0 kJ/mol. He then calculated the reaction quotient of this reaction, while knowing the equilibrium constant was 3 x 103. What is true of the reaction quotient?
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At equilibrium, reaction quotient and equilibrium constant are equal.
At equilibrium, reaction quotient and equilibrium constant are equal.
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Which of the following statements is false about the Keq of a reversible chemical reaction?
Which of the following statements is false about the Keq of a reversible chemical reaction?
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Keq only includes the concentrations of gases and aqueous solutions. Pure solid and liquid concentrations are left out of the equation.
Keq is given by the equation below, where the concentrations expressed are the equilibrium concentrations.
Keq is a property of a given reaction at a given temperature. It is unaffected by catalysts, which only affect rate and activation energy. As the value of Keq increases, the equilibrium concentration of products must also increase, based on the equation.
Keq only includes the concentrations of gases and aqueous solutions. Pure solid and liquid concentrations are left out of the equation.
Keq is given by the equation below, where the concentrations expressed are the equilibrium concentrations.
Keq is a property of a given reaction at a given temperature. It is unaffected by catalysts, which only affect rate and activation energy. As the value of Keq increases, the equilibrium concentration of products must also increase, based on the equation.
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For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.
Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.
The scientist in the passage is able to calculate the reaction quotient (Q) for the reaction taking place in the vessel. If the reaction is ongoing, and has not yet reached equilibrium, how will the reaction quotient compare to the reaction constant (Keq)?
(Assume the reaction is in aqueous solution and is started with 100% reactants and no products).
For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.
Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.
The scientist in the passage is able to calculate the reaction quotient (Q) for the reaction taking place in the vessel. If the reaction is ongoing, and has not yet reached equilibrium, how will the reaction quotient compare to the reaction constant (Keq)?
(Assume the reaction is in aqueous solution and is started with 100% reactants and no products).
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At equilibrium, Keq = Q. The question indicates that, starting with 100% reactants, the reaction has not yet reached equilibrium. In order to reach equilibrium, we must have a continued reduction in reactants and accumulation of products. This would necessitate an increase in Q to eventually reach the value of Keq. Essentially, Q is starting at zero and increasing to the value of Keq at equilibrium.
At equilibrium, Keq = Q. The question indicates that, starting with 100% reactants, the reaction has not yet reached equilibrium. In order to reach equilibrium, we must have a continued reduction in reactants and accumulation of products. This would necessitate an increase in Q to eventually reach the value of Keq. Essentially, Q is starting at zero and increasing to the value of Keq at equilibrium.
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For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.
Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.
Eventually, the reaction reaches equilibrium. The scientist makes a change to the reaction vessel, and again measures Q. He now finds that Q is greater than the value of the Keq he had measured when the reaction was at equilibrium. Since Q > Keq, what value is equal to the first activation energy that must be overcome as the reaction returns to equilibrium?
For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.
Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.
Eventually, the reaction reaches equilibrium. The scientist makes a change to the reaction vessel, and again measures Q. He now finds that Q is greater than the value of the Keq he had measured when the reaction was at equilibrium. Since Q > Keq, what value is equal to the first activation energy that must be overcome as the reaction returns to equilibrium?
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Because Q is now greater than Keq, we know that we need to run the reaction in reverse to come back to equilibrium, where Q = Keq. A larger Q value indicates that \[products\] must be decreased in order to equilibrate at Keq.
The first activation energy we have to overcome in the conversion of products to reactants is the difference between the energy of the products (point 5) and the first transition state (point 4) relative to the products.
Because Q is now greater than Keq, we know that we need to run the reaction in reverse to come back to equilibrium, where Q = Keq. A larger Q value indicates that \[products\] must be decreased in order to equilibrate at Keq.
The first activation energy we have to overcome in the conversion of products to reactants is the difference between the energy of the products (point 5) and the first transition state (point 4) relative to the products.
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Write the law of mass action for the given reaction.
Write the law of mass action for the given reaction.
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The law of mass action is used to compare the chemical equation to the equilibrium constant. In the equation, the product concentration are on the top, and the reactant concentrations are on the bottom. Coefficients in the balanced equation become the exponents seen in the equilibrium equation. While pure solids and liquids can be excluded from the equation, pure gases must still be included.
The law of mass action is used to compare the chemical equation to the equilibrium constant. In the equation, the product concentration are on the top, and the reactant concentrations are on the bottom. Coefficients in the balanced equation become the exponents seen in the equilibrium equation. While pure solids and liquids can be excluded from the equation, pure gases must still be included.
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The partial pressures of H2 and CH3OH are 0.5atm and 1.4atm respectively. What is the partial pressure of CO if the reaction is at equilibrium?
The partial pressures of H2 and CH3OH are 0.5atm and 1.4atm respectively. What is the partial pressure of CO if the reaction is at equilibrium?
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If the reaction is at equilibrium, we know that the law of mass action will equal the equilibrium constant given in the above information. When the reaction contains only gases, partial pressure values can be substituted for concentrations. As a result, we simply need to add the values into the equation and solve for the partial pressure of carbon monoxide (CO).
If the reaction is at equilibrium, we know that the law of mass action will equal the equilibrium constant given in the above information. When the reaction contains only gases, partial pressure values can be substituted for concentrations. As a result, we simply need to add the values into the equation and solve for the partial pressure of carbon monoxide (CO).
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The initial concentrations of this reaction are listed below.
Based on these initial concentrations, which statement is true?
The initial concentrations of this reaction are listed below.
Based on these initial concentrations, which statement is true?
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When given initial concentrations, we can determine the reaction quotient (Q) of the reaction. The reaction quotient is given by the same equation as the equilibrium constant (concentration of products divided by concentration of reactants), but its value will fluctuate as the system reacts, whereas the equilibrium constant is based on equilibrium concentrations.
By comparing the reaction quotient to the equilibrium constant, we can determine in which direction the reaction will proceed initially.
The reaction quotient with the beginning concentrations is written below.
This shows that the ratio of products to reactants is less than the equilibrium constant. Since Q is less than Keq in the beginning, we conclude that the reaction will proceed forward until Q is equal to Keq. In this manner, the denominator (reactants) will decrease and the numerator (products) will increase, causing Q to become closer to Keq.
When given initial concentrations, we can determine the reaction quotient (Q) of the reaction. The reaction quotient is given by the same equation as the equilibrium constant (concentration of products divided by concentration of reactants), but its value will fluctuate as the system reacts, whereas the equilibrium constant is based on equilibrium concentrations.
By comparing the reaction quotient to the equilibrium constant, we can determine in which direction the reaction will proceed initially.
The reaction quotient with the beginning concentrations is written below.
This shows that the ratio of products to reactants is less than the equilibrium constant. Since Q is less than Keq in the beginning, we conclude that the reaction will proceed forward until Q is equal to Keq. In this manner, the denominator (reactants) will decrease and the numerator (products) will increase, causing Q to become closer to Keq.
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A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:
In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.
Which of the following are true when the reaction reaches equilibrium?
I. The energy levels of the reactants and products will be equal
II. The reaction rate of the forward and reverse reactions will be equal
III. The concentrations of the reactants and products will be equal
A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:
In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.
Which of the following are true when the reaction reaches equilibrium?
I. The energy levels of the reactants and products will be equal
II. The reaction rate of the forward and reverse reactions will be equal
III. The concentrations of the reactants and products will be equal
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When a reaction reaches equilibrium, the forward and reverse reaction rates are equal. The arrival of a reaction at equilibrium does not speak to the concentrations. Arrival at equilibrium also does not change the inherent energy properties of the reactants and products.
When a reaction reaches equilibrium, the forward and reverse reaction rates are equal. The arrival of a reaction at equilibrium does not speak to the concentrations. Arrival at equilibrium also does not change the inherent energy properties of the reactants and products.
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