Chemical Equilibrium

Take the products of the rate constants of the forward reactions and divide them by the product of the rate constants of the reverse reactions. Ex: For a three-step reaction: K(eq) = k(1)*k(2)*k(3)/k(-1)*k(-2)*k(-3) K(eq) = equilibrium constant k(1) = forward reaction of reaction 1 k(2) = forward reaction of reaction 2 k(3) = forward reaction of reaction 3 k(-1) = reverse reaction of reaction 1 k(-2) = reverse reaction of reaction 2 k(-3) = reverse reaction of reaction 3

Describe how the equilibrium constant can be determined through rate constants of a multi-step reversible reaction.

K(eq) = k(f)/k(r) K(eq) = equilibrium constant k(f) = rate constant of the forward reaction k(r) = rate constant of the reverse reaction

Describe how the equilibrium constant can be determined through rate constants of a single-step reversible reaction.

A. [Reactants] decreases [Products] increases B. [Reactants] increases [Products] decreases C. K(eq) increases [Reactants] decreases [Products] increases D. K(eq) decreases [Reactants] increases [Products] decreases E. The concentration of the side with the lesser amount of moles increases. The concentration of the side with the greater amount of moles decreases. F. The concentration of the side with the lesser amount of moles decreases. The concentration of the side with the greater amount of moles increases.

Describe the changes to a endothermic chemical reaction in equilibrium if the following stresses were added: A. Add more reactants B. Add more products C. Increase temperature D. Decrease temperature E. Increase pressure and decrease volume F. Decrease pressure and increase volume

A. [Reactants] decreases [Products] increases B. [Reactants] increases [Products] decreases C. K(eq) decreases [Reactants] increases [Product] decreases D. K(eq) increases [Reactants] decreases [Products] increases E. The concentration of the side with the lesser amount of moles increases. The concentration of the side with the greater amount of moles decreases. F. The concentration of the side with the lesser amount of moles decreases. The concentration of the side with the greater amount of moles increases.

Describe the changes to a exothermic chemical reaction in equilibrium if the following stresses were added: A. Add more reactants B. Add more products C. Increase temperature D. Decrease temperature E. Increase pressure and decrease volume F. Decrease pressure and increase volume

Answer: C. An exothermic reaction produces heat. Decreasing the temperature favors product formation, resulting in an increase in the forward reaction rate with a decrease in the reverse rate.

FeI (aq) + I2 (g) <---> FeI3 (aq) If this reaction were exothermic, what effect would decreasing the temperature have on equilibrium? A. The forward reaction rate and the reverse reaction rate both increase B. The forward reaction rate decreases while the reverse reaction rate increases C. The forward reaction rate increases while the reverse reaction rate decreases D. The forward reaction rate and the reverse reaction rate both decrease

Answer: A. Both increasing the pressure of the container and decreasing the volume would favor the side with fewer moles of gas, which is the product side. Meaning (B) & (C) are wrong. (D) would not disturb equilibrium - the significance of decreasing the volume of the container in most equilibria is that there is an increase in pressure; in this case, however, the pressure remains constant despite the change in volume.

FeI (aq) + I2 (g) <---> FeI3 (aq) Which of the following would increase the formation of product? A. Decreasing the volume of the container B. Decreasing the pressure of the container C. Increasing the volume of the container D. Decreasing the volume of the container while maintaining constant pressure

Answer: B. ΔG° = -RT*ln(K(eq)) R = 8.314 J/mol*K T = 298 K ΔG° = -4.955 kJ/mol -4.955 kJ/mol = -(8.314 J/mol*K)(298 K) * ln(K(eq)) -4.955 kJ/mol = -2477.57 J/mol * ln(K(eq)) -4.955 kJ/mol = -2.478 kJ/mol * ln(K(eq)) 2 = ln(K(eq)) e^2 = e^ln(K(eq)) K(eq) = 7.4

For a certain chemical process ΔG° = -4.955 kJ/mol. What is the equilibrium constant K(eq) for this reaction? (Note: R = 8.314 J/mol*K) A. K(eq) = 1.0 B. K(eq) = 7.4 C. K(eq) = 8.9 D. K(eq) = 10

Entropy Gibbs Free Energy

For a reversible reaction at a given temperature, the reaction will reach equilibrium when a system's _________________________ is at a maximum and the ______________________ of a system is at a minimum.

[A] increases [B] increases [C] decreases [D] decreases

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to add more of product C into the system?

[A] increases [B] increases [C] decreases [D] decreases.

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to add more of product D into the system?

[A] decreases [B] decreases [C] increases [D] increases

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to add more of reactant A into the system?

[A] decreases [B] decreases [C] increases [D] increases.

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to add more of reactant B into the system?

[A] increases [B] increases [C] decreases [D] decreases,

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to compress the system (increase pressure and decrease volume)?

[A] decreases [B] decreases [C] increases [D] increases,

For the following generic chemical reaction: A + 2B <---> 3C + D What changes would occur if we were to decompress the system (decrease pressure and increase volume)?

We are looking for the reaction quotient. Q = [product]/[reactant] Q = 0.075 M/1.5 M Q = 0.05 = 5.0 x 10^-2 1. K(eq) = 5.0 x 10^-2 K(eq) = Q Direction: The reaction is in equilibrium Free Energy Change: ΔG = 0 2. K(eq) = 5.0 x 10^-3 K(eq) < Q Direction: The reaction is moving in the direction of the reverse reaction. Free Energy Change: ΔG is positive (+ΔG) 3. K(eq) = 5.0 x 10^-1 K(eq) > Q Direction: The reaction is moving in the direction of the forward reaction. Free Energy Change: ΔG is negative (-ΔG)

Given that [product] = 0.075 M and [reactant] = 1.5 M, determine the direction of reaction and the sign of the free energy change for reactions with the following K(eq) values: (Note: Assume that the reaction has only one product and one reactant, and that the stoichiometric coefficient for each is 1) 1. K(eq) = 5.0 x 10^-2 2. K(eq) = 5.0 x 10^-3 3. K(eq) = 5.0 x 10^-1

A. K(eq) = [C]/([A]^2 * [B]) [A] = 2 M [B] = 6 M [C] = 8 M K(eq) = 8 M/((2 M)^2 * 6 M) K(eq) = 8 M/(4 M * 6 M) K(eq) = 8 M/24 M K(eq) = 1/3 B. K(eq) = [C]/([A]^2 * [B]) [A] = 3 M [B] = ? [C] = 12 M K(eq) = 1/3 1/3 = 12 M/((3 M)^2 * [B]) 1/3 = 12 M/(9 M * [B]) 9 M * 1/3 = 12 M/[B] 3 M = 12 M/[B] [B] * 3 M = 12 M [B] = 12 M/3 M [B] = 4 M

Given the following generic reaction and concentrations: 2A (g) + B (g) <---> C (g) [A] = 2 M [B] = 6 M [C] = 8 M a) What is the equilibrium constant of this reaction? b) After adding more reactant A, the new equilibrium concentration of A is 3 M, and the new equilibrium concentration of C is 12 M. What is the new equilibrium concentration of B?

Reactant (A + B + heat <-----> C + D) Product (A + B <-----> C + D + heat)

Heat acts as a ___________________ in endothermic reactions, while it acts as a ______________________ in exothermic reactions.

Answer: A. The larger the value of K(eq), the larger the ratio of products to reactants. Therefore, if K(eq) >> 1, there are significantly larger concentrations of products than reactants at equilibrium. Even with a large K(eq), the reaction will ultimately reach equilibrium far toward the products side and is therefore reversible, eliminating (D).

If K(eq) >> 1: A. the equilibrium mixture will favor products over reactants B. the equilibrium mixture will favor reactants over products C. the equilibrium concentrations of reactants and products are equal D. the reaction is essentially irreversible

If K(eq) < 1, then the concentration of the reactants is greater than the concentration of our products at equilibrium. The reaction favors the reverse reaction, since the concentration of reactants is greater than the concentration of the products.

If a reaction in dynamic equilibrium has a K(eq) < 1, what would that tell us about the concentration of our reactants and products at equilibrium? Would the reaction favor the forward reaction or the reverse reaction.

If K(eq) = 1, then the concentration of the reactants equals the concentration of our products at equilibrium. The reaction favors neither the forward or reverse reactions, since the concentration of the reactants and products are the same.

If a reaction in dynamic equilibrium has a K(eq) = 1, what would that tell us about the concentration of our reactants and products at equilibrium? Would the reaction favor the forward reaction or the reverse reaction.

If K(eq) > 1, then the concentration of the products is greater than the concentration of our reactants at equilibrium. The reaction favors the forward reaction, since the concentration of products is greater than the concentration of the reactants.

If a reaction in dynamic equilibrium has a K(eq) > 1, what would that tell us about the concentration of our reactants and products at equilibrium? Would the reaction favor the forward reaction or the reverse reaction.

Answer: A. [H2] = 3 moles/1 L = 3 M [N2] = 1 mole/1 L = 1 M [NH3] = 0.05 moles/1 L = 0.05 M K(eq) = [NH3]^2/[N2][H2]^3 K(eq) = 0.05 M^2/1 M * 3 M^3 K(eq) = 0.0025/(1 * 27) K(eq) = 9.26 x 10^-5 ---> 1 x 10^-4 = 0.0001

In a sealed 1 L container, 1 mole of N2 (g) reacts with 3 moles of H2 (g) to form 0.05 moles of NH3 at equilibrium. Which of the following is closest to the K(eq) of the reaction? N2 (g) + 3 H2 (g) <---> 2 NH3 (g) A. 0.0001 B. 0.001 C. 0.01 D. 0.1

Shifted in the direction of the forward reaction. An decrease in pressure in the vessel will cause the reaction to favor the side that has more moles of gas. There are more moles of CO than there are of O2, which means the reaction shifted in favor of the forward reaction.

In the reaction 2 C (s) + O2 (g) <---> 2 CO (g), the pressure of the reaction vessel decreased. Which direction did the equilibrium shift?

Shifted in the direction of the reverse reaction. In an exothermic reaction, heat is treated like a product in the reaction. If heat is added into the exothermic reaction, the additional heat will result in an increase of reactants, which means the reaction shifted in favor of the reverse reaction.

In the reaction CH4 (g) + 2 O2 (g) <---> CO2 (g) + 2 H2O (l) + heat, the reaction vessel is warmed. Which direction did the equilibrium shift?

Shifted in the direction of the forward reaction. An increase in pH indicates an increase in basic properties. Since HSO4- is the conjugate base of H2SO4, this means the reaction shifted in favor of the forward reaction.

In the reaction H2SO4 (aq) <---> H+ (aq) + HSO4- (aq), the pH has been increased. Which direction did the equilibrium shift?

Shifted in the direction of the reverse reaction. If a reactant is removed at equilibrium, then the concentration of products will decrease while the concentration of reactants will increase in order to restore equilibrium. Since water is a reactant, removing it will cause the reaction to shift in the direction of the reverse reaction.

In the reaction H3PO4 (aq) + H2O (l) <---> H3O+ (aq) + H2PO4- (aq), water is removed (without changing temperature). Which direction did the equilibrium shift?

Thermodynamic pathway = Blue curve -Requires more energy to reach transition state -Greater change in free energy -Lower energy state of final products (more stable products) Kinetic pathway = Green curve -Requires less energy to reach transition state -Lower change in free energy -Higher energy state of final products (less stable products)

In the reaction diagram, identify which curve represents the thermodynamic pathway and which curve represents the kinetic pathway. Explain your reasoning.

Answer: B. At extremely high temperatures, reactants or products may decompose, which will affect the equilibrium and potentially destroy the desired products. (A) implies that reactions have limits, which is true; however, this does not make increasing temperature unfavorable. (C) is false because increasing temperature would also increase pressure, assuming constant volume. (D) is incorrect because it refers to properties of irreversible reactions, which would not be involved in an equilibrium between products and reactants.

Increasing temperature can alter the K(eq) of a reaction. Why might increasing temperature indefinitely be unfavorable for changing reaction conditions? A. The equilibrium constant has a definite limit that cannot be surpassed B. The products or reactants can decompose at high temperatures C. Increasing temperature would decrease pressure, which may or may not alter reaction conditions D. If a reaction is irreversible, its K(eq) will resist changes in temperature.

Smaller activation energy needed to reach transition state. Products are less stable, thus the products are higher in free energy.

On a reaction coordinate diagram, how would the kinetic pathway appear as compared to the thermodynamic pathway?

Answer: D. Since the reaction is spontaneous, ΔG° is negative. Whenever ΔG° is negative, then K(eq) > 1.

Pure sodium metal spontaneously combusts upon contact with room temperature water. What is true about the equilibrium constant of this combustion reaction @ 25°C? A. K(eq) < 0 B. 0 < K(eq) < 1 C. K(eq) = 1 D. K(eq) > 1

Answer: A. This problem asks for the free energy of a reaction at non-standard conditions, which can be determined with the equation ΔG = ΔG° + RT * lnQ.

Suppose ΔG° = -2000 kJ/mol for a chemical reaction. At 300 K, what is the change in Gibbs Free Energy in kJ/mol? A. ΔG = -2000 kJ/mol + (300K)(8.314 J/mol*K)(lnQ) B. ΔG = -2000 kJ/mol - (300 K)(8.314 J/mol*K)(lnQ) C. ΔG = -2000 kJ/mol - (300 K)(8.314 J/mol*K)(logQ) D. ΔG = -2000 kJ/mol - (300 K)(8.314 J/mol*K)(logQ)

K(eq) = [C]/[B]*[A] 2.1 x 10^-7 = [C]/[0.2 M - x]*[0.1 M - x] 2.1 x 10^-7 = [C]/[0.2 M]*[0.1 M] 0.02 M * 2.1 x 10^-7 = [C] [C] = 4.2 x 10^-9 M [A](i) = 0.1 M or 1 x 10^-1 [B](i) = 0.2 M or 2 x 10^-1 K(eq) = 2.1 x 10^-7 [A](i) and K(eq) are within 6 orders of each other. [B](i) and K(eq) are within 6 orders of each other. Neither [A](i) nor [B](i) are within at most 2 orders of K(eq). We can assume that x is negligible.

The following reaction has a K(eq) of 2.1 x 10^-7. Given an initial concentration for A equal to 0.1 M and an initial concentration of B equal to 0.2 M, what is the equilibrium concentration of C? Is the approximation that x is negligible valid for this calculation? A (aq) + B (aq) <---> C (g) + D (s)

1. The concentration of pure solids and pure liquids do not affect the equilibrium constant expression. 2. K(eq) is temperature dependent for a particular reaction 3. The larger the value of K(eq), the farther right the equilibrium position will be. The smaller the value of K(eq), the farther left the equilibrium position will be. 4. If the equilibrium constant for a forward reaction is equal to K(eq), then the reverse reaction will have an equilibrium constant of 1/K(eq).

What are 4 things to keep in mind when determining the equilibrium constant of a reaction?

-Changing the concentration of the reactants -Changing the concentration of the products -Changing the pressure and volume of a system -Adding or removing heat

What are the four methods that the equilibrium of a system can be shifted?

Thermodynamic product Kinetic product

What are the two types of products that can be formed in a reaction?

Kinetic: Low temperatures with low heat transfer. Thermodynamic: High temperatures with high heat transfer.

What conditions favor formation of a kinetic product? A thermodynamic product?

States that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

What does Le Chatelier's Principle.

K(eq) > 1 - [Products] > [Reactants] @ equilibrium - ΔG° (standard change in free energy) is negative K(eq) = 1 - [Products] = [Reactants] @ equilibrium - ΔG° (standard change in free energy) = 1 K(eq) < 1 - [Products] < [Reactants] @ equilibrium - ΔG° (standard change in free energy) is positive

What does it say for a reaction when: K(eq) > 1 K(eq) = 1 K(eq) < 1

- Q < K(eq) -[Reactants] > [Reactants] @ equilibrium -[Products] < [Products] @ equilibrium -Forward reaction is increased to restore equilibrium -ΔG (change in free energy) is negative, reaction is spontaneous for the forward reaction (nonspontaneous for the reverse reaction) -Q = K(eq) -[Reactants] and [Products] are present in equilibrium proportions -Forward and reverse reactions equal one another -ΔG (change in free energy) = 0, reaction is in dynamic equilibrium -Q > K(eq) -[Reactants] < [Reactants] @ equilibrium -[Products] > [Products] @ equilibrium -Reverse reaction is increased to restore equilibrium -ΔG (change in free energy) is positive, reaction is spontaneous for the reverse reaction (nonspontaneous for the forward reaction)

What does the reaction quotient and equilibrium constant tell about a reaction when: - Q < K(eq) -Q = K(eq) -Q > K(eq)

An irreversible reaction is one that only proceeds in the forward direction, while a reversible reaction can proceed in the forward and reverse direction. An irreversible reaction proceeds until completion of the reaction, while a reversible reaction proceeds until it reaches equilibrium.

What is the difference between a reversible reaction and an irreversible reaction?

Dynamic equilibrium: Equilibrium where the forward and reverse reactions are occurring at the same rate. Static equilibrium: Equilibrium where the forward and reverse reaction have stopped.

What is the difference between dynamic equilibrium and static equilibrium?

The equilibrium constant (K(eq)) shows the ratio of the concentrations of products to reactants when a reaction is at equilibrium. The reaction quotient (Q) shows the ratio of the concentrations of products to reactants when a reaction is not at equilibrium, and it shows where the reaction is in relation to the equilibrium constant.

What is the difference between the equilibrium constant and the reaction quotient?

Kinetic -Products form at lower temperatures -Reaction involves smaller transfers of thermal energy -Pathway is faster than thermodynamic pathway, because activation energy is lower in kinetic pathway. -Products are less stable (more reactive) than thermodynamic pathway, since the change in free energy is smaller in kinetic pathway Thermodynamic -Products form at higher temperatures -Reaction involves larger transfers of thermal energy -Pathway is slower than kinetic pathway, because activation energy is higher in thermodynamic pathway. -Products are more stable (less reactive) than kinetic pathway, since the change in free energy is larger in thermodynamic pathway.

What is the difference between the kinetic pathway and the thermodynamic pathway of a reaction?

ΔG° = -RT * ln(K(eq)) ΔG° = standard change in free energy R = Ideal Gas Constant (8.314 J/mol*K) T = Temperature (in K) K(eq) = equilibrium constant This equation is used to determine the ratio of products to reactants at equilibrium.

What is the equation that describes the relationship between the equilibrium constant and the standard change in free energy?

ΔG = ΔG° + RT * ln(Q) or... ΔG = RT * ln(Q/K(eq)) ΔG = change in free energy ΔG° = change in free energy at standard conditions R = Ideal Gas Constant (8.314 J/mol*K) T = temperature (in Kelvin) Q = reaction quotient K(eq) = equilibrium constant This equation is used to determine the spontaneity of a reaction through the reaction quotient

What is the equation used to determine the change of free energy of a reaction through the reaction quotient? (Hint: There are 2 equations for this)

Answer: D. Recall that pure solids and liquids do not appear in the equilibrium expression; thus, this K(eq) has no denominator because the only reactant is a solid, cuprous sulfate. This could also be called K(sp) because a solid is dissociating into ions in solution. The correct K(eq) should have [Cu+] squared because its stoichiometric coefficient is 2.

What is the equilibrium expression for the reaction: Cu2SO4 (s) <---> 2 Cu+ (aq) + SO4(2-) (aq) A. [Cu+]^2[SO4(2-)]/[Cu2SO4] B. 2 x [Cu+][SO4(2-)]/[Cu2SO4] C. [Cu+][SO4(2-)] D. [Cu+]^2[SO4(2-)]

Keq = [NH3]^2/([N2] * [H2]^3)

What is the formula of the equilibrium constant of the Haber reaction? N2 (g) + 3 H2 (g) <---> 2 NH3 (g)

Answer: A. Recall that equilibrium constants are either based on concentrations (K(eq)) or partial pressures (Kp). In this case, because all species are in the gas phase, we are using Kp - eliminating (B) and (D). When water is in the liquid phase, it does not appear in equilibrium expressions, as in (C). Here however, water is in the gas phase and thus should appear in the equilibrium expression.

What is the proper equilibrium expression for the reaction below? 2 NO2 (g) + 4 H2 (g) <---> N2 (g) + 4 H2O (g) A. K(p) = P(N2) * P(H2O)^4/P(NO2)^2 * P(H2)^4 B. K(c) = P(N2) * P(H2O)^4/P(NO2)^2 * P(H2)^4 C. K(p) = P(N2)/P(NO2)^2 * P(H2)^4 D. K(c) = P(N2)/P(NO2)^2 * P(H2)^4

K(eq) = ([A]^a * [B]^b)/([C]^c * [D]^d) K(eq) = equilibrium constant [A] = concentration of reactant A a = stoichiometric coefficient of reactant A [B] = concentration of reactant B b = stoichiometric coefficient of reactant B [C] = concentration of reactant C c = stoichiometric coefficient of reactant C [D] = concentration of reactant D d = stoichiometric coefficient of reactant D

What would be the formula of the equilibrium constant of the following generic reaction? aA + bB <---> cC + dD

Q = ([A]^a * [B]^b)/([C]^c * [D]^d) Q = reaction quotient [A] = concentration of reactant A a = stoichiometric coefficient of reactant A [B] = concentration of reactant B b = stoichiometric coefficient of reactant B [C] = concentration of reactant C c = stoichiometric coefficient of reactant C [D] = concentration of reactant D d = stoichiometric coefficient of reactant D

What would be the formula of the reaction quotient of the following generic reaction? aA + bB <---> cC + dD

K(eq) = [C]^3/([A] * [B]^2) K(eq) = [1 M]^3/([1 M] * [3 M]^2) K(eq) = 1/(1 * 9) K(eq) = 1/9 K(eq) = 0.11

What would be the value of the equilibrium constant of the following generic reaction, given the following concentrations at equilibrium? A + 2B <---> 3C [A] = 1 M [B] = 3 M [C] = 1 M

Nonspontaneous

When a reaction is at equilibrium, any further movement of either reactions (forward or reverse) will be _____________________.

When the rate of the forward reaction is equal to the rate of the reverse reaction.

When does a reversible reaction reach dynamic equilibrium?

- If K(eq) is not significantly larger than 1 - If the initial concentration of reactants and the K(eq) are not within 2 orders (10^2) of each other.

When doing equilibrium calculations, when is it appropriate to consider the amount of reactants that have reacted to be negligible?

Answer: B. Equilibrium of a reaction can be changed by adding/removing heat, changing concentrations of the reactants or products, changing the pressure of the system, and changing the volume of the system. Adding or removing a catalyst only affects the kinetics of a reaction, thus would be the correct answer.

Which of the following actions does not affect the equilibrium position? A. Adding or removing heat B. Adding or removing a catalyst C. Increasing or decreasing concentrations of reactants D. Increasing or decreasing volumes of reactants

Answer: B. Statement I is false because the addition of a catalyst could increase the rate constants of both the forward and reverse reactions. Statement II is true because - for products to come into existence - reactants must be used up. Statement III is also true: all K values are temperature dependent.

Which of the following is true of equilibrium reactions? I. An increase in k(1) results in a decrease in k(-1) II. As the concentration of products increases, the concentration of reactants decreases III. The equilibrium constant is altered by changes in temperature A. I only B. II and III only C. I and III only D. I, II, and III

Answer: A. A negative ΔH value indicates an exothermic reaction, meaning that the forward reaction produces heat. Removing heat by decreasing the temperature is similar to removing any other product of the reaction. To compensate for this loss, the reaction will shift to the right, causing an increase in the concentrations of C and D, as well as a decrease in the concentrations of A and B.

Which of the following statements best describes the effect of lowering the temperature of the following reaction? A + B <---> C + D ΔH = -1.12 kJ/mol A. [C] and [D] would increase B. [A] and [B] would increase C. ΔH would increase D. ΔH would decrease

Answer: C. For a process to progress forward spontaneously, Q must be less than K(eq) and will therefore have a tendency to move in the direction toward equilibrium. A spontaneous reaction's free energy is negative by convention.

Which of the following statements is true of a process that is spontaneous in the forward direction? A. ΔG > 0 and K(eq) > Q B. ΔG > 0 and K(eq) < Q C. ΔG < 0 and K(eq) > Q D. ΔG < 0 and K(eq) < Q

Because the activation energy in kinetic pathways are smaller than in thermodynamic pathways, since they take place at lower temperatures.

Why are kinetic pathways of chemical reactions faster than thermodynamic pathways?

Because the higher amounts of free energy to reach the transition state in the thermodynamic pathway allows for the reaction to overcome any steric hindrances, which leads to the formation of a more stable product.

Why are the products of thermodynamic pathways of chemical reactions more stable than in kinetic pathways?

K(eq) = [CH3OH]/([CO]*[H2]^2) K(eq) = [H2PO4-]*[H3O+]/[H3PO4]

Write the equilibrium constant expression for the following reactions: CO (g) + 2 H2 (g) <---> CH3OH (g) H3PO4 (aq) + H2O (l) <---> H2PO4- (aq) + H3O+ (aq)

K(eq) = [NO2]^2/[N2O4] [N2O4] = 3 moles/0.5 L = 6 M K(eq) = 6 x 10^-6 6 x 10^-6 = [NO2]^2/6 M 3.6 x 10^-5 M = [NO2]^2 sqrt(3.6 x 10^-5 M) = sqrt([NO2]^2) [NO2] = 0.006 M

3 moles of N2O4 is placed in a 0.5 L container and allowed to reach equilibrium according to the following reaction: N2O4 (g) <---> 2 NO2 (g) What is the equilibrium concentration of NO2, given K(eq) for the reaction is 6 x 10^-6?

Answer: C. This scenario likely describes a situation in which a reaction has reached equilibrium very far to the right (with high product concentration and low reactant concentration). This reaction must be reversible because the reaction did not proceed all the way to the right. Any reaction in equilibrium has equal forward and reverse rates of reaction.

A reaction is found to stop just before all reactants are converted to products. Which of the following could be true about this reaction? A. The reaction is irreversible, and the forward rate is greater than the reverse rate. B. The reaction is irreversible, and the reverse rate is too large for products to form C. The reaction is reversible, and the forward rate is equal to the reverse rate D. The reaction is reversible, and the reverse rate is greater than the forward rate

Answer: B. Adding sodium acetate increases the number of acetate ions present. According to Le Chatelier's principle, this change will push this reaction to the left, resulting in a decrease in the number of free H+ ions. Because pH is determined by the hydrogen ion concentration, a decrease in the number of free protons will increase the pH. An acid's Ka (equilibrium constant for acid dissociation)will remain constant under a given temperature and pressure, eliminating (C) and (D).

Acetic acid dissociates in solution according to the following equation: CH3COOH <-----> CH3COO- + H+ If sodium acetate is added to a solution of acetic acid in excess water, which of the following effects would be observed in the solution? A. Decreased pH B. Increased pH C. Decreased pK(eq) (pKa) D. Increased pK(eq) (pKa)

*Remember, for gases, it is easier to measure pressures than it is to measure concentrations. ΔG = ΔG° + RT * ln(Q) R = 8.314 J/mol*K T = 298 K ΔG° = -33 kJ/mol Q = [NH3]^2/([N2] * [H2]^3) Q(p) = P(NH3)^2/(P(N2) * P(H2)^3) Q(p) = 1/(1 * 1) Q(p) = 1 ΔG = -33 kJ/mol + (8.314 J/mol*K * 298 K) * ln(1) ΔG = -33 kJ/mol + (8.314 J/mol*K * 298 K) * 0 ΔG = -33 kJ/mol + 0 ΔG° = -33 kJ/mol Basically, ΔG = ΔG° Because ΔG is negative, this means the reaction is spontaneous in the forward direction. This means the reaction will move in the forward direction until it reaches equilibrium.

Calculate the change in free energy for the following reaction at 25°C given the following partial pressures: N2 (g) + 3 H2 (g) <-------> 2 NH3 (g) ΔG° = -33 kJ/mol N2 = 1.0 atm H2 = 1.0 atm NH3 = 1.0 atm Is the reaction moving forward, backward, or is in equilibrium?

*Remember, for gases, it is easier to measure pressures than it is to measure concentrations. ΔG = ΔG° + RT * ln(Q) ΔG° = -33 kJ/mol R = 8.314 J/mol*K T = 298 K Q = [NH3]^2/([N2] * [H2]^3) Q(p) = P(NH3)^2/(P(N2) * P(H2)^3) Q(p) = 4^2/(4 * 4^3) Q(p) = 16/(4 * 64) Q(p) = 0.0625 ΔG = -33 kJ/mol + (8.314 J/mol*K * 298 K) * ln(0.0625) ΔG = -33 kJ/mol + 2477.572 J/mol * (-2.773) ΔG = -33 kJ/mol + 2.478 kJ/mol * (-2.773) ΔG = -33 kJ/mol -6.87 kJ/mol ΔG = -39.87 kJ/mol Since ΔG is negative, the reaction is proceeding through the forward reaction.

Calculate the change in free energy for the following reaction at 25°C given the following partial pressures: N2 (g) + 3 H2 (g) <-------> 2 NH3 (g) ΔG° = -33 kJ/mol N2 = 4.0 atm H2 = 4.0 atm NH3 = 4.0 atm Is the reaction moving forward, backward, or is in equilibrium?

ΔG° = -RT * ln(K(eq)) R = 8.314 J/mol*K T = 1000 K ΔG° = 106.5 kJ/mol 106.5 kJ/mol = -(8.314 J/mol*K * 1000 K) * ln(K(eq)) 106.5 kJ/mol = -8314 J/mol * ln(K(eq)) 106.5 kJ/mol = -8.314 kJ/mol * ln(K(eq)) -12.81 kJ/mol = ln(K(eq)) e^-12.81 kJ/mol = e^ln(K(eq)) K(eq) = 2.7 x 10^-6 K(p) = 2.7 x 10^-6

Calculate the equilibrium constant for the following reaction at 1000 K: N2 (g) + 3 H2 (g) <-----> 2 NH3 (g) ΔG° = +106.5 kJ/mol

ΔG° = -RT * ln(K(eq)) R = 8.314 J/mol*K T = 298 K ΔG° = -33 kJ/mol -33 kJ/mol = -(8.314 J/mol*K)*(298 K) * ln(K(eq)) -33 kJ/mol = -2477.57 J/mol * ln(K(eq)) -33 kJ/mol = -2.478 kJ/mol * ln(K(eq)) 13.32 kJ/mol = ln(K(eq)) e^13.32 = e^ln(K(eq)) K(eq) = 6.1 x 10^5 K(p) = 6.1 x 10^5

Calculate the equilibrium constant for the following reaction at 298 K: N2 (g) + 3 H2 (g) <-----> 2 NH3 (g) ΔG° = -33 kJ/mol

ΔG° = -RT * ln(K(eq)) R = 8.314 J/mol*K T = 464 K ΔG° = 0 kJ/mol 0 kJ/mol = -(8.314 J/mol*K * 464 K) * ln(K(eq)) 0 kJ/mol = -3857.696 J/mol * ln(K(eq)) 0 kJ/mol = -3.858 kJ/mol * ln(K(eq)) 0 kJ/mol = ln(K(eq)) e^0 = e^ln(K(eq)) K(eq) = 1 K(p) = 1

Calculate the equilibrium constant for the following reaction at 464 K: N2 (g) + 3 H2 (g) <-----> 2 NH3 (g) ΔG° = 0 kJ/mol

Answer: A. Carbon dioxide gas evolves and leaves the bottle, which decreases the total pressure of the reactants. Le Chatelier's principle explains that a decrease in pressure shifts the equilibrium to increase the number of moles of gas present. This particular reaction will shift to the left, which in turn will decrease the amount of carbonic acid and increases the amount of carbon dioxide and water. Oxygen and nitrogen are not highly reactive and are unlikely to combine spontaneously with carbon dioxide or carbonic acid, as in (C) and (D).

Carbonated beverages are produced by dissolving carbon dioxide in water to produce carbonic acid: CO2 (g) + H2O (l) <---> H2CO3 (aq) When a bottle containing carbonated water is opened, the taste of the beverage gradually changes as the carbonation is lost. Which of the following statements best explains this phenomenon? A. The change in pressure and volume causes the reaction to shift to the left, thereby decreasing the amount of aqueous carbonic acid B. The change in pressure and volume causes the reaction to shift to the right, thereby decreasing the amount of gaseous carbon dioxide C. Carbonic acid reacts with environmental oxygen and nitrogen D. Carbon dioxide reacts with environmental oxygen and nitrogen

Q: -Concentration of reactants or products -Pressure -Volume K(eq): -Temperature

Changing which properties of a system will alter the reaction quotient (Q)? What about equilibrium constant (K(eq))?

Answer: C. Ka = [products]/[reactants], with each species raised to its stoichiometric coefficient. A compound with a Ka greater than 10^-7 contains more H+ cations than HA- anions at equilibrium, which makes it an acid. This means that the compound in question is likely to react with a compound that is basic. Of the four answer choices, NH3 is the only base.

Compound A has a K(a) (equilibrium constant of acid dissociation) of approximately 10^-4. Which of the following compounds is most likely to react with a solution of compound A? A. HNO3 B. NO2 C. NH3 D. N2O5

Answer: C. Reaction 2 is the reverse of reaction 1. This means that K(eq) for reaction 2 is the inverse of K(eq) of reaction 1, so the answer is 1/0.1 = 10.

Consider the following reactions: 3 A + 2 B <---> 3 C + 4 D (Reaction 1) 4 D + 3 C <---> 3 A + 2 B (Reaction 2) If K(eq) for reaction 1 is 0.1, what is K(eq) for reaction 2? A. 0.1 B. 1 C. 10 D. 100

1. K(eq) = 1.0 x 10^-12 [A](i) = 1 M [A](i) and K(eq) are within 12 or 13 orders from each other. They are not within at most 2 orders of each other. Yes, x will be negligible. 2. K(eq) = 1.0 x 10^-2 [A](i) = 0.1 M or 1.0 x 10^-1 M [A](i) and K(eq) are within 1 order from each other. They are within at most 2 orders of each other. No, x will not be negligible. 3. K(eq) = 1.0 x 10^-3 [A](i) = 0.001 M or 1.0 x 10^-3 M [A](i) and K(eq) are within 0 orders from each other. They are within at most 2 orders of each other. No, x will not be negligible. 4. K(eq) = 1.0 x 10^-15 [A](i) = 0.001 or 1.0 x 10^-3 [A](i) and K(eq) are within 12 orders from each other. They are not within at most 2 orders of each other. Yes, x will be negligible. *The concentration of a reactant that converts to product can be considered negligible if K(eq) is at 10^-2 less than the initial concentration of the reactant.

Consider the hypothetical reaction: A <---> B + C. For each of the following, determine if the amount of reactant A that has converted to product at equilibrium will be negligible compared to the starting concentration of A. 1. K(eq) = 1.0 x 10^-12 [A](i) = 1 M 2. K(eq) = 1.0 x 10^-2 [A](i) = 0.1 M 3. K(eq) = 1.0 x 10^-3 [A](i) = 0.001 M 4. K(eq) = 1.0 x 10^-15