Chemical reactions

A chemical reaction is the process through which one substance transforms into another. These reactions can be represented by chemical formulas, which define both the substances involved and the quantitative ratios in which particles react. To facilitate their study, chemical reactions are categorized based on their characteristics, allowing us to predict outcomes when two substances are mixed. This classification is essential for determining the quantities required for a reaction, assessing solution concentrations, and more.

Most chemical reactions are not "one-way streets"; they do not proceed exclusively in a single direction. Instead, an equilibrium is established within the reaction mixture, balancing the forward reaction (where reactants form products) and the reverse reaction (where products revert to reactants). This dynamic is described by the law of mass action. Due to this equilibrium, the yield of chemical reactions is often limited. To achieve complete conversion of reactants into products, it is necessary to create specific reaction conditions, a process that can be guided by Le Chatelier's principle.

Additionally, this article covers the kinetics of chemical reactions, which provides insights into the rates at which these reactions occur.

Chemical formulas

A distinction is made between the empirical and molecular formula.

• Empirical formula: represents the simplest whole-number ratio of atoms of each element in a compound (e.g., CH₂O for glucose)
• Molecular formula: gives the actual, non-simplified number of atoms of each element in one molecule of the compound (e.g., C₆H₁₂O₆ for glucose)
• Relationship: The molecular formula is always a whole-number multiple of the empirical formula.
• Calculation: The empirical formula can be determined from the percent composition of a compound.

The greater the number of atoms in a molecule, the more potential connections can exist between them. For larger molecules and lattice structures, such as metals and salts, a molecular formula alone is often inadequate for uniquely identifying the molecule. In these instances, it becomes essential to illustrate the atomic connectivity using a structural formula.

Rules for writing molecular formulas

1. All elements involved are listed with their corresponding element symbols.
2. If an element appears more than once, its number is written as a subscript after the element symbol.
3. Charges (sign and magnitude) are written as a superscript after the respective ion (e.g., Ca²⁺, SO4²⁻).
4. For large lattice structures, the smallest possible ratio unit (the formula unit) is considered.
5. Order convention:
   1. Cations are written before anions.
   2. In organic compounds, C atoms are named first, then H atoms, and finally the other elements in alphabetical order.
6. Coordination complex conventions:
   1. Square brackets enclose the coordination sphere.
   2. Inside the square brackets, parentheses are placed around each polyatomic ligand.
   3. If there is more than one of a specific ligand, the number is written as a subscript after the parentheses around the ligand (e.g., [Cr(NH₃)₃Cl₃]).

Examples

At first glance, the formulas for salts and metals look just like those for molecules. It is important to remember that these substances actually form large lattice structures!

Molecular formulas for organic molecules are often ambiguous because they don't show atomic connectivity. For example, the molecular formula C₆H₁₂O₆ (glucose) also represents many other sugars, such as galactose, fructose, mannose, and inositol!

Chemical reactions

Chemical reactions are processes in which new substances (= products) are formed from starting materials (= reactants). In essence, all substances (elements, molecules, ions, or radicals) can react with each other. These reactions are depicted using a chemical equation.

The chemical equation

• Reactants: the substances from which new products are formed; they are always positioned before the reaction arrow, typically on the left
• Products: the newly formed substances resulting from a reaction; they stand after the reaction arrow, usually on the right
• Reaction arrow
  • Connects reactants and products
  • Arrowhead indicates the direction of the reaction
  • A series of arrows may be used as shorthand notation to signify that multiple reaction steps exist between the specified reactants and products.
    • Equilibrium arrow
      • Appearance: two arrows pointing in opposite directions, written one above the other (⇄)
      • Use: for reactions that proceed in both directions
    • Resonance arrow
      • Appearance: a double-headed arrow (↔︎)
      • Use: represents resonance
    • Curved arrows
      • Appearance: curved arrows usually point in a single direction
      • Use
        • Can represent reaction sequences forming cycles (e.g., the citric acid cycle)
        • To show reactants added or byproducts formed in a specific step
        • Illustrate the movement of electron pairs during a reaction mechanism in organic chemistry
  • Notations above or below reaction arrows
    • Reaction conditions (e.g., temperature, pH value, etc.)
    • Reactants not included on the reactant side (e.g., catalysts, acids, solvents, etc.)
    • Byproducts not mentioned on the product side (e.g., escaping gases, water, precipitating solids, etc.)
  • Balance
    • Charge balance: The total charge must be the same on both sides of the reaction arrow.
    • Mass balance: The number of each atom type (indicated by their element symbols) must match on both sides.

Representation of a chemical equation

An example of a simple chemical equation is x A + y B ⇄ z C; this shorthand notation is read as follows: x particles of type A react with y particles of type B to form z particles of type C. A, B, and C here stand for any type of particle (molecules, atoms, ions...); x, y, and z should be the smallest possible whole numbers.

• Structural formula notation: the participating particles (A, B, and C) are represented using Lewis structural formulas
  • Use: especially in organic reactions and when one wants to show exactly how the electrons shift during the reaction
• Molecular formula notation: the participating particles (A, B, and C) are represented using a molecular formula
  • Use: especially suitable for inorganic reactions

Since these are reaction equations, the reactant side and the product side must be balanced.

Arrows within complex structures: Sometimes, arrows are used in structural formulas for coordination complexes. These do not indicate a reaction; they are an outdated and incorrect way of representing coordinate covalent bonds.

Reaction types

Reactions are classified according to the processes that take place during the substance conversion.

The most important reaction types are:

• Redox reactions: transfer of electrons; redox reactions are presented in detail in "Redox chemistry".
• Acid-base reactions: transfer of protons; acid-base reactions are presented in detail in "Acid-base balance".
• Precipitation reactions: reactions in which a component precipitates as an insoluble solid
• Complexation reactions: reactions involving complexes
• Organic reactions
  • Addition reactions: reactions in which atoms or groups are added across a multiple bond
  • Elimination reactions: reactions in which atoms or groups are removed from a molecule, typically forming a multiple bond
  • Substitution reactions: reactions in which a molecule exchanges one functional group for another

A reaction can sometimes fall into several of these categories.

Stoichiometric calculation

Molecular formulas and chemical equations specify the exact quantitative ratios in which substances react with each other. From this, one can deduce the "amounts of substance" that must be combined to carry out a specific reaction. Such calculations are termed "stoichiometric" (Greek: stoicheion = basic element, metron = measure). To avoid having to calculate with a large number of particles, the "mole" was defined as the basic unit for this purpose. 1 mole corresponds to the number of C atoms contained in 12g of ¹²C (carbon-12):

1 mole (unit) ≈ 6.022 × 10²³ particles = N_A (= Avogadro constant)

The mass of one mole of a substance (in grams) is numerically equal to the mass of one particle of that substance (in atomic mass units, u, or Daltons, Da). (1 u = 1 Da = 1/12 the mass of a carbon-12 atom)!

Mathematical consequences

With the help of the mole, calculations can now be made very easily. With the following equations, all stoichiometric questions can be answered, if necessary by rearranging and substituting.

• Amount of substance: n = N/N_A
  • n = amount of substance (mol), N = number of particles, N_A = number of particles in one mole (Avogadro's constant ∼ 6.022×1023 particles/mol)
  • Equivalents (Eq): The unit 'equivalent' (Eq) or 'milliequivalent' (mEq) is often used in medicine. It relates the amount of a substance (in moles) to its chemical reactivity, typically its charge. For ions, the amount in equivalents is the amount in moles multiplied by the absolute value of the ion's charge.
• Molar mass: M = m/n
  • Unit: g/mol
  • M = molar mass (g/mol), m = mass (g), n = amount of substance (mol)
• Molar concentration (molarity): c = n/V
  • Unit: mol/L (or M)
  • c = concentration (mol/l), n = amount of substance (mol), V = volume (L)
  • Mass fraction: Instead of concentration, one can also state the mass fraction, i.e., the quotient of a specific mass relative to the total mass.
    • E.g., for a mixture of 0.9 g sodium chloride with 99.1 g water, a 0.9% saline solution is obtained (0.9 g/(0.9 g + 99.1 g) = 0.009 = 0.9%)
  • Mixing solutions of different concentrations: The concentration of the mixture is calculated as the sum of the products of the individual concentrations and their respective proportions of the mixture's volume
    • E.g., for a 1:1 mixture of two solutions of different concentrations: 0.5 × 2 mol/L + 0.5 × 5 mol/L = 1 mol/L + 2.5 mol/L = 3.5 mol/L
• Mass concentration: c = m/V
  • Unit: mg/L (or g/L)
  • c = concentration (mg/L), m = mass of dissolved substance (mg or g), V = volume (L)
• Molality: m = n_solute/m_solvent
  • Unit: mol/kg (or m)
  • m = molality (mol/kg), n_solute = moles of solute (mol), m_solvent = mass of solvent (kg)
  • Unlike molarity, molality is independent of temperature and pressure changes.
• Normality: N = equivalents/V
  • Unit: Eq/L (or N)
  • N = normality (Eq/L), equivalents = number of equivalents of solute (Eq), V = volume of solution (L)
  • Related to molarity: N = molarity × n, where n is the number of equivalents per mole of the substance (e.g., n=2 for H₂SO₄).

Example calculation 1 – calculating concentration

One liter of saline solution is prepared using a ratio of 10 g NaCl per 100 mL of water. The molar mass of sodium chloride is given as 60 g/mol. This liter of solution is then diluted with 4 L of water. What is the concentration of the diluted solution?

• Find: concentration c
• Given: volume of solution V, mass of dissolved substance m, molar mass of dissolved substance M
  • Since 1 L = 1000 mL, the mass of NaCl in 1 L is (10 g/100 mL) × 1000 mL = 100 g.
  • M = m/n → n = m/M → n_NaCl = 100 g/60 g/mol ≈ 1.7 mol
  • c = n/V → in 1 L of this liquid, there are 100 g NaCl → c_NaCl ≈ 1.7 mol/L
  • When the liquid is diluted, 1.7 mol is now in a total volume of 5 L (1 L + 4 L) → c_NaCl = n/V = 1.7 mol/5 L = 0.34 mol/L

Example calculation 2 – calculating the amount of a substance

How do you determine the amount of substance in a 3 g sugar cube (= glucose)?

• Find: amount of substance n
• Given: mass m
  • The glucose molecule has the molecular formula C₆H₁₂O₆
  • 1 mole of the glucose molecule thus contains 6 mol C atoms, 12 mol H atoms, and 6 mol O atoms
  • The relative mass of glucose is 6 × 12 u (for carbon) + 12 × 1 u (for hydrogen) + 6 × 16 u (for oxygen) = 180 u
  • The mass of 1 mole of glucose is equal to the relative mass of the molecule in g → 1 mole of glucose weighs 180 g
  • Since the mass m of the portion of glucose is only 3 g, one can calculate: M = m/n → n = m/M → n = 3 g/180 g/mol ≈ 0.017 mol

Example calculation 3 – calculating reaction components

Glucose reacts with oxygen according to the following chemical equation:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

How many moles of oxygen (O₂) are needed to react 90 g of glucose?

• Find: amount of substance n, based on the stoichiometry of the reaction
• Given: stoichiometry of the reaction (1:6), mass m
  • M (the molar mass) of glucose = 180 g/mol (see example 2) → 90 g glucose = 0.5 mol = n
  • Glucose and oxygen react in a 1:6 ratio, i.e., n_O2 = 6 × n_Glucose = 6 × 0.5 mol = 3 mol

Example calculation 4 – calculating a mass component

How many g of nitrogen are in 30 g of urea?

• Find: mass m, based on the quantitative ratios in a molecular formula
• Given: mass m
  • Urea has the molecular formula CH₄N₂O, so you need 2 nitrogen atoms per urea molecule; 1 mol of urea thus contains 2 mol of N atoms
  • M_Urea= (1 × 12) + (4 × 1) + (2 × 14) + (1 × 16) = 60 g/mol
  • M = m/n → n = m/M; in total, there are n = m/M = 30 g/60 g/mol = 0.5 mol urea
  • In 0.5 mol of urea, there is 1 mol of N atoms (0.5 mol urea × 2 mol N/1 mol urea)
  • the reactant side and the product side must be

Limiting reactant and percent yield

In many reactions, reactants are not mixed in the exact stoichiometric ratios. One reactant will be completely consumed before the others, limiting the amount of product that can be formed.

• Limiting reactant: the reactant that is completely used up first in a chemical reaction
  • It determines the maximum amount of product that can be formed.
• Excess reactant: any reactant that remains after the limiting reactant has been consumed
• Theoretical yield: the maximum mass of product that can be produced from a given amount of limiting reactant, based on the reaction's stoichiometry
• Actual yield: the measured mass of product actually obtained from the reaction in a laboratory setting
  • Actual yield is almost always less than the theoretical yield
• Percent yield: a measure of the reaction's efficiency
  • Formula: percent yield = (actual yield/theoretical yield) × 100%

The chemical equation A + B → C + D is typically a simplification of the actual scenario. Once the reaction begins, the reaction vessel contains a mixture of A, B, C, and D. Consequently, the reverse reaction, C + D → A + B, can also occur, albeit less frequently. This type of reaction is known as reversible. The state where both the forward and reverse reactions occur is referred to as chemical equilibrium, represented by the equilibrium arrow (⇄): A + B ⇄ C + D.

Law of mass action

The ratio in which the forward and reverse reactions are related (i.e., whether one of them occurs much more frequently or if both happen with similar frequency) can be mathematically specified via the equilibrium constant (K) of the reaction. :

• Law of mass action
  • For the reaction aA + bB ⇄ cC + dD , the following applies: K = ([C]^c × [D]^d)/([A]^a × [B]^b).
    • K = equilibrium constant (unitless), [C] = concentration of substance C (mol/L), [D] = concentration of substance D (mol/L), [A] = concentration of substance A (mol/L), [B] = concentration of substance B (mol/L); a, b, c, d = stoichiometric coefficients (unitless integers representing the number of moles of each substance involved in the reaction)
• Interpretation of K
  • K > 1: The reaction equilibrium favors the products, indicating that the forward reaction predominates.
  • K < 1: The reaction equilibrium favors the reactants, indicating that the reverse reaction predominates.
  • The equilibrium constant K is also used to describe the kinetics of a reaction (see below).
  • In enzymatic reactions, additional insights can be drawn from the equilibrium constant and its changes, such as the effectiveness of an enzyme and potential inhibition mechanisms. For more information, see "Enzymes and biocatalysis."

Example calculation

For a reaction Fe₂O₃ + 3 CO ⇄ 2 Fe + 3 CO₂, the following concentrations are present at equilibrium: [Fe₂O₃] = 0.001 mol/L; [CO] = 0.1 mol/L; [Fe] = 0.05 mol/L, [CO₂] = 0.03 mol/L. Calculate the equilibrium constant.

• Find: equilibrium constant K
• Given: reaction equation, concentrations c
  • For this reaction, the law of mass action is K = ([CO₂]³ × [Fe]²)/([Fe₂O₃] × [CO]³).
  • → (0.03³ × 0.05²)/(0.001 × 0.1³) = 0.0675
  • The equilibrium constant K = 0.0675 indicates that, under the given conditions, the reduction of iron oxide to iron does not proceed to completion. Since K < 1, the reverse reaction Fe → Fe₂O₃ is favored in this scenario.

Solubility

The maximum amount of a substance that can dissolve is described analogously to a chemical reaction via an equilibrium. The law of mass action can also be established for this equilibrium, which in this context is referred to as the "solubility product" (K_sp). When the solubility of a substance is exceeded, a solid precipitate forms. Polar substances, which are charged or partially charged, dissolve readily in polar solvents, such as salts in water, but do not dissolve well in nonpolar solvents. Conversely, nonpolar substances dissolve poorly in polar solvents, like oil in water, but dissolve easily in nonpolar solvents.

• Reaction equation for dissolving a substance AB in a solvent: xAB_(solid) ⇄ yA⁺_(aq) + zB^-_(aq)
  • Solubility product: K_sp = [A⁺]^y× [B^-]^z
    • Unit: varies depending on the number of ions involved – for each ion, the factor mol/L is included in the calculation
    • K_sp = solubility product constant (mol/L), [ ] = concentrations (mol/L), y, z = stoichiometric factors
  • Solubility: The value of K_sp indicates the extent of dissolution.
    • K_sp large = good solubility
    • K_sp small = poor solubility
  • Energy change during dissolution
    • Exothermic: dissolution process releases energy → the mixture heats up
    • Endothermic: dissolution process requires energy → the mixture cools down

The phrase "like dissolves like" summarizes a key chemistry principle: polar substances dissolve in polar solvents, while nonpolar substances dissolve in nonpolar solvents.

Barium
Poisoning with soluble barium salts can lead to severe symptoms such as paralysis, gastrointestinal disturbances, cardiovascular problems, seizures, or even death. In an acute case, it can be treated by oral administration of Na₂SO₄ (Glauber's salt). This forms the poorly soluble BaSO₄, which is excreted in the feces. Since barium sulfate is not absorbed in the gastrointestinal tract and has no effect of its own, it is also suitable as a contrast agent for radiological examinations such as a barium swallow or a double-contrast barium enema.

Example calculation

The solubility product of barium sulfate is K_sp = 1×10^-9 mol²/L². What is the maximum concentration of barium ions when BaSO₄ dissolves in pure water?

• Find: [Ba²⁺]
• Given: solubility product K_sp(BaSO₄)
  • The dissolution equation is: BaSO_4(s)⇄ Ba²⁺_(aq) + SO₄^2-_(aq)
  • The solubility product expression is K_sp = [Ba²⁺] × [SO₄^2-]
  • When dissolving in pure water, the concentrations of both ions are equal. Let s = [Ba²⁺] = [SO₄^2-]. It follows: K_sp = s². The maximum concentration of barium ions is s = √K_sp = √(1×10^-9 mol²/L²) ≈ 0.00003 mol/L (or 3 x 10^-5 mol/L)

Determination and shifting of equilibrium positions

The position of a reaction equilibrium corresponds to the ratio of the forward and reverse reaction rates. It can be described by the ratio of products to reactants (i.e., via the law of mass action). However, one can also actively influence the position of a reaction equilibrium by changing parameters. This is described by Le Chatelier principle, or the "principle of least constraint."

Definition: If a stress is applied to a chemical equilibrium, the equilibrium shifts in such a way that the stress is relieved.

Le Chatelier's principle indicates how a system at equilibrium responds to changes in temperature, pressure, and concentration.

Redox reactions and acid-base reactions are fundamental categories of chemical reactions, and are discussed separately in "Redox chemistry" and "Acid-base disorders". The following provides a brief overview of other relevant reaction types.

Precipitation reaction

This includes all reactions in which a solid forms and, due to its low solubility, promptly precipitates. A defining characteristic of precipitation reactions is that the precipitating substance is removed from the reaction equilibrium. Depending on its solubility, this removal can be almost complete. As a result, the reaction equilibrium shifts significantly toward the product side.

Complexation reaction

This includes all reactions in which complexes are involved. Since this is a very general classification, complex reactions are further divided into two different processes.

• Complex formation
  • Definition: a reaction in which ligands form around a central atom/ion and bind via a coordinate covalent bond
  • Example: Fe²⁺ + 6 H₂O ⇄ [Fe(H₂O)₆]²⁺
  • Chelate effect: Ligands with multiple binding sites (polydentate ligands) form much more stable complexes than an equivalent number of separate (monodentate) ligands. This increased stability is the chelate effect.
• Ligand exchange reaction
  • Definition: a reaction in which a weakly binding ligand in a complex is exchanged for a more strongly binding ligand
  • Example: [Fe(H₂O)₆]²⁺ + 6 CN^- ⇄ [Fe(CN)₆]^4- + 6 H₂O

Hydrolysis

This includes all reactions in which a molecule is cleaved by a water molecule.

• General reaction equation: A-B + H₂O ⇄ A-OH + B-H
• Important examples
  • Ester hydrolysis: R₁-COO-R₂ + H₂O ⇄ R₁COOH + R₂-OH
    • Triglyceride hydrolysis: triglyceride + 3 H₂O ⇄ glycerol + 3 fatty acids
  • ATP hydrolysis: ATP⁴⁻ + H₂O → ADP³⁻ + HPO₄²⁻ + H⁺
  • Disaccharide hydrolysis: sucrose + H₂O ⇄ glucose + fructose
  • Protein hydrolysis: protein + n H₂O ⇄ amino acids
    • n = number of H₂0 molecules required to hydrolyze the protein (depends on the lenght and structure of the protein)

Basic reactions in organic chemistry

There are four overarching principles for classifying reactions in organic chemistry: additions, eliminations, substitutions, and rearrangements.

• Addition reaction
  • Definition: includes all reactions in the course of which two molecules, one of them with at least one multiple bond, combine
  • Reaction equation
    • Example: hydration, a reaction in which a water molecule is added to a double bond, thus forming an alcohol
• Elimination reaction
  • Definition: includes all reactions in the course of which individual atoms or an entire part of the molecule are removed from a molecule, typically forming a multiple bond
  • Reaction equation: R-C-C-A ⇄ R-C=C + A
    • Example: dehydration, a reaction in which a water molecule is cleaved off and a double bond is formed
• Substitution reaction
  • Definition: includes all reactions in the course of which an atom or an entire group of atoms in a molecule is exchanged for another
  • Reaction equation: A + R-B ⇄ A-R + B
• Rearrangement reaction
  • Definition: includes all reactions in the course of which the connectivity of atomic groups within a molecule is changed
  • Example: constitutional isomers such as dihydroxyacetone phosphate and glyceraldehyde-3-phosphate in phase 2 of glycolysis

Kinetics deals with the rate of chemical reactions or physical transport processes.

• Collision theory: states that the rate of a reaction depends on the collision of molecules in the reaction mixture
• Criteria for a reaction to occur
  • Concentration: The higher the concentration of reactants, the more likely the molecules will collide, and the faster the reaction.
  • Energy: Colliding particles require a minimum energy (activation energy) for the collision to lead to a reaction.
  • Spatial orientation: Particles must collide with a specific spatial orientation for the collision to lead to a reaction.

Activation energy

The Arrhenius equation describes the rate constant (k) of a reaction as a function of the activation energy (E_a) and the temperature (T).

Formula: k = A × e ^{(-Ea /R×T)}

• k = rate constant (units depend on the reaction order), A = frequency factor (units vary by reaction but commonly in s⁻¹), E_a = activation energy (J/mol), R = gas constant (8.314 J⋅K⁻¹⋅mol), T = temperature (K)
• A very high activation energy in the exponent causes the exponential factor, and thus the rate constant k, to approach zero.
• A higher temperature in the denominator of the exponent increases the exponential factor and thus also k.

Reaction rate

The reaction rate (v) is mathematically the negative derivative of the concentration of a reactant with respect to time (as reactant concentration decreases over time).

Formula: v = -dc/dt

• v = reaction rate (mol/(L·s)), dc = change in reactant concentration (mol/L), dt = change in time (s)
• Half-life (t_1/2): corresponds to the time in which exactly half of the reactant has reacted
  • In a concentration vs. time diagram, the value for t_½ can be read directly at ½c_max.

How fast a reaction proceeds (kinetics) has nothing to do with the energetics (thermodynamics) of the reaction.

Reaction order

Since the rate of a reaction is related to how frequently the particles in the reaction mixture collide, it also depends on how many particles must collide for the reaction to occur. This consideration leads to a classification of reactions via the so-called reaction order:

• Zero-order reaction: Reaction rate v does not depend on the concentration of the reactants.
  • Formula: v = k
    • v = reaction rate (mol/(L·s)), k = rate constant (mol/(L·s))
  • General reaction: A → B
  • Example: light-dependent reactions (if light is the limiting factor), enzyme-catalyzed reactions at saturation
• First-order reaction: Reaction rate v depends on the concentration of one of the participating substances.
  • Formula: v = k[A]
    • v = reaction rate (mol/(L·s)), k = rate constant (s⁻¹), [A] = concentration of substance A (mol/L)
  • General reaction: A → B + C
  • Examples: radioactive decay, many decomposition reactions
  • Half-life: independent of the initial concentration of the reactant
    • Reaction course: After t_½, 50% of the reactant is still present, after 2t_½, 25% is still present, after 3t_½, 12.5% is still present, etc.
    • Unit: seconds (s)
• Second-order reaction: Reaction rate v depends on the concentration of two reactants, or on the square of the concentration of one reactant.
  • Formula: v = k[A][B] or v = k[A]²
    • v = reaction rate (mol/(L·s)), k = rate constant (L/(mol·s)), [A] = concentration of substance A (mol/L), [B] = concentration of substance B (mol/L)
  • General reaction: A + B → C (or 2A → C)
  • Examples: most reactions involving the collision of two molecules in solution proceed this way
• Third-order reaction: Reaction rate v depends on the concentration of three participating substances (or combinations like [A]²[B]).
  • Formula: v = k[A][B][C]
    • v = reaction rate (mol/(L·s)), k = rate constant (L²/(mol²·s)), [A] = concentration of substance A (mol/L), [B] = concentration of substance B (mol/L), [C] = concentration of substance C (mol/L)
  • General reaction: A + B + C → D
  • Example: These reactions are rare and almost never occur in a single step, as it is very unlikely that three particles will collide simultaneously with the correct orientation and energy.

Coupled reactions

Reactions that involve multiple components often proceed as a series of sequential individual steps, called elementary reactions. This creates a reaction mechanism. When reactions are linked (the product of one is the reactant for the next), they are called coupled reactions. Various scenarios are possible:

• A → B → C
  • Irreversible processes that occur sequentially
  • Rate constant k₁ for reaction A to B and k₂ for reaction B to C
  • Overall equilibrium: The overall equilibrium constant (K) is the product of the individual equilibrium constants.
  • Calculation: K_overall = [C]/[A] = ([B]/[A]) × ([C]/[B]) = K₁ × K₂
• A ⇄ B
  • Reversible process, where the forward and reverse reactions are in equilibrium
  • Rate constants k₁ for the forward reaction and k_-1 for the reverse reaction
• A → B and A → C
  • Two reactions that compete (parallel reactions)
  • Rate constants k₁ for reaction A to B, k₂ for reaction A to C