Chemical Equilibrium and Kinetics for the MCAT: Everything You Need to Know

Learn key MCAT concepts about chemical equilibrium and kinetics, plus practice questions and answers

(Note: This guide is part of our MCAT General Chemistry series.)

Table of Contents

Part 1: Introduction to chemical equilibrium and kinetics

Part 2: Reaction equilibria

a) Equilibrium constant and reaction quotient

b) Le Chatelier's principle

c) Bicarbonate buffer system

Part 3: Rate Laws

a) Reaction orders

b) Determining rate laws from experimental data

c) Arrhenius equation

Part 4: Transition State Theory

a) Reaction coordinate diagrams

b) Thermodynamic and kinetic products

Part 5: High-Yield Terms

Part 6: Passage-Based Questions and Answers

Part 7: Standalone Questions and Answers

-----

Part 1: Introduction to chemical equilibrium and kinetics

How quickly do reactants in a chemical reaction turn into products? Instead of reactants becoming products, does the reaction ever proceed in the opposite direction-turning products into reactants? In this guide, we will be focusing on chemical equilibrium and kinetics. Equilibrium and kinetics govern the speed and direction of many biological processes. As a result, many of the topics presented here go hand-in-hand with other MCAT topics.

Receive Your Free MCAT Question of the Day

Get every last bit of practice in before test day with a free MCAT question delivered straight to your inbox daily.

-----

Part 2: Reaction equilibria

a) Equilibrium constant and reaction quotient 

Certain chemical reactions known as reversible reactions can progress either forwards or backward. In other words, these reactions are able to progress “forwards,” creating products from the reactants, and also progress “backwards,” creating reactants from the products. 

How does a reversible chemical reaction “decide” to progress towards the products or reactants? Imagine a chemical reaction as a see-saw with reactants on one side and the products on the other. When this see-saw is balanced, the reaction has achieved equilibrium. At this point, the forward reaction and the reverse reaction are proceeding at the same rate. Note that this does not mean that there is no progression of the forward and reverse reactions—rather, it implies that the rates at which these two reactions are equal.

The ratio of reactants and products at which the reaction has achieved equilibrium is expressed by the law of mass action. The equilibrium constant, known as K_eq, is an expression that defines where this point is. Given the reaction aA + bB ⇋ cC + dD, then:

K_eq = [C]^c[D]^d ÷ [A]^a[B]^b
Note that in the above expression, the square brackets ([ ]) denote the concentration of a certain chemical species. Lowercase letters (a, b, c, d) are stoichiometric coefficients. A reaction’s equilibrium position may be changed depending on environmental conditions, such as the temperature of the environment and the presence of any catalysts. Therefore, K_eq can change if the temperature is changed.

In chemistry, the capital letter “K” almost always denotes an equilibrium constant. This includes the acid dissociation constant, K_a; the base dissociation constant, K_b; and even the Michaelis constant, K_m. For instance, you may recall learning about the acid dissociation constant, K_a. A high K_a value indicates that the acid is strong. Let’s take a look at the Ka expression for the dissociation of acid, represented as HA ⇋ H⁺ + A^-:

K_a = [H⁺][A^-] ÷ [HA]
Since stronger acids dissociate more readily, the dissociation reaction of a strong acid favors the products. Thus, the resulting K_a value is higher.

Another important variable is the reaction quotient, Q. The definition of Q is virtually identical to that of K_eq:

Q = [C]^c[D]^d ÷ [A]^a[B]^b

While K_eq utilizes the concentrations of species at equilibrium, Q tells us the reaction’s current position. Thus, if Q is equal to K_eq, then the reaction is at equilibrium. However, if Q is less than K_eq, it has not yet achieved equilibrium, and the reaction will proceed forward (towards the products). If Q is greater than K_eq, then the reaction has gone past equilibrium and will proceed in the reverse direction (towards the reactants).

b) Le Chatelier’s principle

Le Chatelier’s principle describes how much a reaction favors the products or reactants under different environmental conditions. In other words, the principle describes how an equilibrium position shifts. 

When exposed to stress, a system will shift to relieve that stress and re-establish equilibrium. In this context, stress is any sort of change in chemical concentration or environmental conditions, including temperature, pressure, or volume. 

• A chemical reaction will shift to the right if the concentration of reactants is increased or the concentration of products is decreased.

• When the temperature is changed, the direction in which the reaction shifts depends on whether heat is a reactant or product. In an endothermic reaction, heat acts as a reactant, whereas in an exothermic reaction, heat acts as a product. Therefore, if we have an endothermic reaction and the temperature is increased, our reaction shifts right. 

• When the pressure is changed, the directionality of the resulting shift depends on the number of moles of gas on either side of the reaction. Increasing the pressure of a system will shift the reaction to the side of fewer moles of gas, whereas decreasing the pressure will shift it to the side of greater moles of gas. If we have more moles of reactants than products, an increase in pressure will shift the reaction right. This may also be rationalized through the ideal gas law (PV = nRT). This equation tells us that as pressure increases, so does the number of moles. Therefore, the system alleviates this by shifting the reaction towards the side with fewer moles.

  c) Bicarbonate buffer system

The previously discussed concepts are foundational for one of the most essential systems in your body: the bicarbonate buffer system. The action of the bicarbonate buffer system can be expressed using the following chemical equation: 

CO_2(g) + H₂O _(l) ⇋ H₂CO_3(aq) ⇋ H⁺_(aq) + HCO₃^- _(aq)

The bicarbonate buffer system plays an incredibly important role in regulating the pH of the blood. (For more information on the physiological impacts of the bicarbonate buffer system, be sure to refer to our guide on the respiratory and cardiovascular systems.)

The bicarbonate buffer system is a highly regulated balancing act that leverages Le Chatelier’s principle. When the concentration of hydrogen ions increases, pH decreases. To counteract this, the respiration rate is increased, forcing the body to expel carbon dioxide. This shifts the reaction left, thereby reducing the hydrogen ion concentration and increasing the pH. 

Similarly, to counteract high pH, or a low hydrogen ion concentration, respiration is downregulated to trap carbon dioxide within the body. This shifts the equilibrium position of the reaction rightwards to increase the hydrogen ion concentration and decrease the pH.

SAC339 code block

----

Part 3: Rate laws

a) Reaction orders

It is often useful to quantify how quickly a reaction proceeds. The rate at which a reaction proceeds is measured by the rate of product formation or the rate at which the reactants are consumed. The reaction rate can be calculated by measuring the amount (e.g., change in concentration) of a product produced over a known period of time (e.g., in seconds).

Given the reaction aA + bB → cC + dD, the rate law, measured in molarity per second, is expressed as k[A]^x[B]^y. In this expression, k is a unique rate constant and is dependent on temperature and the presence of a catalyst.

The exponents x and y can be added together to find the reaction order. The reaction order describes the rate of product formation in proportion to the rate of reaction consumption.

Table 1   Rate law equations by reaction order

Reaction Order	Rate Law	Description
Zero-order	Rate=k	The rate of product formation is equal to the rate constant, k, and is independent of any changes to the concentrations of the reactants.
First-order	Rate=k[A]1 Rate=k[B]1	The rate of product formation is directly proportional to one reactant.
Second-order	Rate=k[A]2 Rate=k[B]2 Rate=k[A][B]	The rate of product formation is proportional to the square of the concentration of one reactant or just the concentration of both reactants.

b) Determining rate laws from experimental data

The MCAT may test your ability to determine a rate law and reaction order, given a set of experimental data. Here’s an example with data resulting from the fusion of sodium chloride (Na + Cl → NaCl).

Trial	[Na]	[Cl]	Initial Rate (M/s)
1	2.00 x 10-2	2.00 x 10-2	2.00 x 10-8
2	4.00 x 10-2	2.00 x 10-2	8.00 x 10-8
3	4.00 x 10-2	4.00 x 10-2	1.60 x 10-7

Based on this information, how could we find the rate law?
First, look for two trials during which the concentration of only one substance was changed. In trials 1 and 2, [Na] increased by a factor of 2 from 2.00 x 10^-2 to 4.00 x 10^-2. [Cl] remained unchanged. The rate increased by a factor of 4 from 2.00 x 10^-8 to 8.00 x 10^-8. With this information, we can set up the following equation to express the relationship between the change in concentration of [Na] and the change in the rate.

∆Rate = (∆[Na])^x
4 = 2^x
x = 2


Next, look for two trials during which the concentration Cl changed but Na did not. In trials 2 and 3, [Cl] increased by a factor of 2 from 2.00 x 10^-2 to 4.00 x 10^-2. The rate increased by a factor of 2 from 8.00 x 10^-8 to 1.60 x 10^-7. Let’s set up an expression similar to the one we set up for sodium.

∆Rate = (∆[Cl])^y
2 = 2^y
y = 1


Now let’s plug in the values we’ve found into our rate law formula:

rate = k[A]^x[B]^y
Rate = k[Na]²[Cl]

To find the rate constant k, simply substitute values from any of the trials into the rate law that has been derived. Then, use algebra to solve for the exact value. To keep things simple, let’s choose trial 1.

Rate = k[Na]²[Cl]
k = Rate/[Na]²[Cl]
k = (2.00 x 10^-8 M/s) / (2.00 x 10^-2M)²(2.00 x 10^-2M)
k = 2.50x 10^-3 s^-1M^-2


Finally, put it all together by plugging in our k value into our rate law formula.

Rate = (2.50 x 10^-3 s^-1M^-2)[Na]²[Cl]

c) Arrhenius equation

It is important to remember that k is an experimentally determined value and depends on environmental conditions, including temperature. Thus, if there is no experimental data provided for the temperature or environmental conditions in question, the rate constant cannot be determined.

If the activation energy of the reaction is known, then the Arrhenius equation can then be applied to determine the change in the rate constant:

k=Ae^-E_a/RT where k is the rate constant, A is a constant known as the frequency factor, E_a is the activation energy of the reaction, R is the ideal gas constant (8.314 J/k˙mol), and T is the temperature in Kelvin.

This equation can also be written in logarithmic form as: ln(k)=ln(A) -(E_a/RT)

-----

Part 4: Transition state theory

a) Reaction coordinate diagrams

Transition state theory states that molecular collisions result in the formation of an intermediary complex that has greater energy than both the reactants and products. Under this theory, molecules and atoms are modeled as particles of significant mass and energy that must make contact with one another under specific orientations. 

The transition state is made up of partially dissociated old bonds and partially formed new bonds. The difference in free energy between the transition state and the reactants is called the activation energy. Since the transition state is necessary for product formation, the activation energy is the minimum amount of energy required for a reaction to proceed. Think of activation energy as a barrier that a reaction needs to overcome. The higher it is, the slower the reaction will be. Catalysts, such as enzymes, can lower the activation energy and thereby increase the rate of the reaction. In a reaction coordinate diagram, the transition state is found at the peak of the curve. 

Figure 1    A reaction coordinate diagram. The y-axis depicts change in Gibbs free energy (ΔG).

b) Thermodynamic and kinetic products

Factors such as the presence of a catalyst and temperature can result in a reaction proceeding via a high or low energy transition state. This results in thermodynamic products and kinetic products. Thermodynamic products are favored at high temperatures, have lower energy, and are more stable. However, their formation occurs more slowly because greater activation energy is required. 

Kinetic products, on the other hand, are favored at a lower temperature, have higher energy, and are less stable. Their formation occurs more quickly because smaller activation energy is required.

Figure 2    The thermodynamic product is product A because it has a higher activation energy but lower free energy. The kinetic product is product B because it has a lower activation energy but higher free energy.

The MCAT may test your ability to determine if a certain product is a kinetic or thermodynamic product. Thus, it will be important to be able to determine activation energy and compare these quantities.

Many chemical processes and reactions take place over a series of intermediate steps. The intermediate step with the highest activation energy is known as the rate-limiting or rate-determining step, since the velocity of the overall reaction is limited by the slowest step.

-----

Part 5: High-yield terms

Bicarbonate buffer system: CO_2(g)+H₂O_(l) ⇋ H₂CO_3(aq) ⇋ H⁺_(aq)+HCO₃^-_(aq)

First-order reaction: rate of product formation is directly proportional to one reactant

Kinetic products: generally formed at lower temperatures as they require less activation energy; less stable than the thermodynamic products as they have a higher free energy

Law of mass action: K_eq = [C]^c[D]^d ÷ [A]^a[B]^b

Le Chatelier’s principle: states that when a system is exposed to stress, the equilibrium position will shift to relieve that stress

Second-order reaction: rate of product formation is proportional to the square of the concentration of one reactant or the concentration of both reactants

Thermodynamic products: generally formed at higher temperatures as they require greater activation energy; more stable than kinetic products as they have a lower free energy

Transition state theory: states that molecular collisions result in the formation of a high-energy intermediary transition state complex

Zero-order reaction: rate of product formation is equal to the rate constant, k

-----

Part 6: Passage-based questions and answers

The respiratory system is vulnerable to the inhalation of harmful compounds and substances found in everyday life. Many household cleaning products contain either bleach (NaClO) or ammonia as an active ingredient. Mixing the two compounds can form the respiratory irritant chloramine. At large amounts, chloramine inhalation may require immediate medical attention.

NH₃ + NaClO ⇋ NaOH + NH₂Cl
Reaction 1: Formation of chloramine.

Similarly, individuals with asthma may experience respiratory irritation from everyday airborne allergens, such as pollen, dust mites, and tobacco smoke. These allergens can trigger the bronchioles to constrict and trap air in the lungs. 

As respiratory mechanisms have a strong influence on the acidity of the blood, the bicarbonate buffer system (reaction 2) is used to maintain the blood’s physiological pH of around 7.4. In humans, the rate of respiration is generally regulated by chemoreceptors that monitor carbon dioxide levels.

CO_2(g) + H₂O_(l) ⇋ H₂CO_{3 (aq)} ⇋ H⁺_(aq) + HCO₃^-_(aq)
Reaction 2: The bicarbonate buffer system.

As respiratory mechanisms can affect the acidity of the blood, this buffer system is essential for maintaining the blood’s physiological pH of around 7.4. When dysfunctional, the buffer system is unable to prevent blood alkalemia and blood acidemia. 

Note: The information in this passage was created for the sole purpose of presenting an MCAT-style passage and should not be construed to be factually true.  

Question 1: The bronchioles of a patient experiencing an asthma attack are constricted. Which of the following can be expected of the bicarbonate buffer system?

A) The reaction equilibrium will shift toward reactants to decrease the acidity of the blood

B) The reaction equilibrium will shift toward reactants to prevent dehydration

C) The reaction equilibrium will shift toward products due to the increased concentration of carbon dioxide

D) The reaction equilibrium will shift toward products to decrease the acidity of the blood

Question 2: Which of the following is the most likely physiological response to a decrease in blood pH?

A) An increase in the respiration rate

B) A decrease in the respiration rate

C) An increase in the heart rate

D) A decrease in the heart rate

Question 3: The second product created in the formation of chloramine is best described as an:

A) Strong acid

B) Strong base

C) Weak acid

D) Weak base

Question 4: According to the information presented in the passage, which of the following would increase the formation of chloramine?

A) Decreasing the concentration of ammonia 

B) Increasing the concentration of ammonia

C) Decreasing the concentration of bleach

D) Increasing the concentration of sodium hydroxide

Question 5: Assume Reaction 1 is endothermic. Which of the following describes the most likely effect of increasing the temperature of the reaction environment?

A) The rate of the forward reaction would increase

B) The rate of the reverse reaction would increase

C) There would be no effect on the reaction

D) Chloramine formation would increase, but sodium hydroxide formation would decrease

Answer key for passage-based questions

1. Answer choice C is correct. According to the passage, the constriction of the bronchioles traps air in the lungs. Therefore, the concentration of carbon dioxide will rise. Under Le Chatelier’s principle, the reaction equilibrium will shift rightwards (choices A and B are incorrect). This would result in a higher concentration of products, which include hydrogen ions. Therefore, the acidity would increase and pH would decrease (choice C is correct).

2. Answer choice A is correct. A drop in pH indicates an increase in the concentration of hydrogen ions. In considering the bicarbonate buffer system, driving the reaction leftwards would result from a decrease in the concentration of carbon dioxide. Blood concentrations of carbon dioxide are most directly affected by respiratory rate (choices C and D are incorrect). Increasing the respiration rate would increase the expulsion of carbon dioxide and shift the reaction to the left (choice B is incorrect). 

3. Answer choice B is correct. The second product created in the formation of chloramine is sodium hydroxide (NaOH). Basicity can be characterized by the high concentration of hydroxide ions which the dissociation of NaOH forms (choices A and C are incorrect). Most alkali metal hydroxides are strong bases (choice D is incorrect).

4. Answer choice B is correct. This question requires consideration of Le Chatelier's principle. Decreasing the concentration of a reactant would drive the reaction in the reverse, or “leftward,” direction (choices A and C are incorrect). Increasing the concentration of a product would also shift the reaction to the left (choice D is incorrect).

5. Answer choice A is correct. In an endothermic reaction, heat can be considered a reactant. Thus, Le Chatelier’s principle may be applied. If the temperature increases, the reaction will shift to the right (choices B and C are incorrect). This shift would cause an increase in the formation of all products (choice D is incorrect). 

-----

Part 7: Standalone questions and answers

Question 1: Given the rate law below, what is the order of the reaction?

Rate = k[A]²

A) Zero-order

B) First-order

C) Second-order

D) Third-order

Question 2: Which of the following best describes a zero-order reaction?

A) The rate of product formation is independent of any changes to the concentrations of the reactants

B) The rate of product formation is inversely proportional to the concentration of the reactants

C) The rate of product formation is directly proportional to one reactant

D) The rate of product formation is proportional to the square of one reactant

Question 3: How does the free energy of a transition state complex compare to the free energy of the reactants and products?

A) The free energy of the transition state is higher than the reactants but lower than the products

B) The free energy of the transition state is lower than the reactants but higher than the products

C) The free energy of the transition state is lower than both the reactants and the products

D) The free energy of the transition state is higher than both the reactants and the products 

Question 4: Which of the following reaction conditions would encourage a reaction to proceed in the forward direction?

A) A reaction that has a reaction quotient lower than its equilibrium constant 

B) A reaction that has a reaction quotient higher than its equilibrium constant 

C) A reaction that has a reaction quotient equal to its equilibrium constant 

D) A reaction that has a reaction quotient equal to zero

Question 5: What role do catalysts play in chemical reactions?

A) They increase the concentration of products formed

B) They decrease the concentration of products formed

C) They decrease the activation energy of a reaction

D) They increase the activation energy of a reaction

Question 6: What is the expected result of increasing the pressure the given reaction occurs at?

N_(g) + O_2(g) ⇋ N₂O_{5 (g)}

A) The reaction will shift right

B) The reaction will shift left

C) There will be an increase in the concentration of elemental oxygen

D) Increasing the pressure will not affect the reaction

Question 7: Which of the following accurately describes the thermodynamic products of a reaction?I. They have a lower free energy than the kinetic products

II. They have a lower activation energy than the kinetic products

III. They are more stable than the kinetic products

IV. They tend to form at higher temperatures than the kinetic products

A) I and II

B) I, III, and IV

C) II, III, and IV

D) II and III

Question 8: The chemical reaction, Mg + O₂ → MgO₂, resulted in the data acquired below. What is the rate law for this reaction?

Trial	[Mg]	[O2]	Initial Rate (M/s)
1	1.00 x 10-3	1.00 x 10-3	2.00 x 10-6
2	2.00 x 10-3	1.00 x 10-3	4.00 x 10-6
3	2.00 x 10-3	2.00 x 10-3	8.00 x 10-6

A) Rate = 2[Mg]²[O₂]
B) Rate = 2[Mg][O₂]
C) Rate = 0.5[Mg]²[O₂]
D) Rate = 0.5[Mg][O₂]²

Answer key for standalone questions

1. Answer choice C is correct. The overall reaction order is given by summing the exponents in the rate law. The provided rate law only has one exponent, 2. Therefore, it is a second-order reaction (choices A, B, and D are incorrect). 

2. Answer choice A is correct. A zero-order reaction is expressed by the rate law:

Rate = k

Thus, the reaction rate is independent of the reactants. A first-order reaction rate of product formation is directly proportional to one reactant (choice C is incorrect). A second-order reaction rate of product formation is proportional to the square of one reactant (choice D is incorrect). 

3. Answer choice D is correct. According to the transition state theory, molecular collisions result in the formation of a high-energy intermediary complex. This complex has greater energy than both the reactants and the products (choices A, B, and C are incorrect). 

4. Answer choice A is correct. When the reaction quotient, Q, is less than the equilibrium constant K_eq, the reaction proceeds in the forward direction and favors the formation of the products (choice A is correct). When Q is greater than K_eq, the reaction shifts in the reverse direction (choice B is incorrect). When Q is equal to K_eq, the reaction is at equilibrium (choice C is incorrect). A reaction quotient must be greater than 0 (choice D is incorrect). 

5. Answer choice C is correct. Catalysts such as enzymes decrease the activation energy required for a reaction to occur (choice D is incorrect). They do not increase or decrease the concentration of products formed. Instead, this decrease in activation energy results in an increased reaction rate (choices A and B are incorrect).

6. Answer choice A is correct. When the pressure is increased, a gaseous reaction shifts to the side of fewer moles. After balancing the equation, the reaction equation becomes:

4 N_(g) + 5 O_{2 (g)} ⇋ 2 N₂O_{5 (g)}

On the right-hand side, there are 2 moles of product. On the left-hand side, there are 9 moles of reactants. Thus, the reaction will shift rightwards (choices B, C, and D are incorrect).

7. Answer choice B is correct. The thermodynamic products of a reaction are more stable than the kinetic products because their free energy is lower (choices I and III are correct). However, they form more slowly because thermodynamic products tend to have a higher activation energy than the kinetic products (choice II is incorrect). Due to this, thermodynamic products generally form at higher temperatures compared to the kinetic products, as the higher temperature provides energy to reach the high-energy transition state (choice IV is correct). 

8. Answer choice B is correct. [Mg] and the rate both increased by a factor of 2 between trials 1 and 2. Thus, the reaction must be first-order with respect to Mg (choices A and B are incorrect). [O₂] and the rate both increased by a factor of 2 between trials 2 and 3. Thus, the reaction must also be first-order with respect to O2 (choice D is incorrect). The final rate law is given by k[Mg][O₂]. To find k, substitute the values from any of the trials. To keep things simple, we’ll use trial 1.

Rate = k[Mg][O₂]
k = Rate/[Mg][O₂]
k = (2.00x10^-6)/{(1.00x10^-3)(1.00x10^-3)}
k = 2

Therefore, the complete rate law is given by:

Rate = 2[Mg][O₂]

Not ready to commit to your MCAT prep at this time?

Enter your name and email for a free MCAT question delivered straight to your inbox daily.