• The study of enthalpy changes in chemical reactions is known as Thermochemistry
• Every chemical substance has a unique enthalpy, H, which is also known as its intrinsic energy or heat content
• The value of this enthalpy cannot be measured experimentally but its change during a reaction can be measured
• Every element in its free state has an enthalpy value of zero, but different compounds have different enthalpies at the same temperature and pressure
• The change in enthalpy of a reaction, ΔH, is the difference between the total enthalpy of the products and the total enthalpy of the reactants
• Mathematically, ΔH = ∑H_P – ∑H_R
• If ∑H_P > ∑H_R, ΔH is positive and the reaction is endothermic
• If ∑H_P < ∑H_R, ΔH is negative and the reaction is exothermic
• For example:
  • If the molar enthalpies of A, B, C are -20, +32, and +40 kJmol^-1 respectively, then the enthalpy change of the reaction A + 3B → 2C is given by:
  • ΔH = ∑H_P – ∑H_R
  • ΔH = 2(40) – [-20 + 3(+32)]
  • ΔH = 80 – 76 = 4 kJ
• A thermochemical equation always indicates the nature and magnitude of the enthalpy change written after the equation for a reaction. For example:
  • H_2(g) + ½O_2(g) → H₂O_(l)    ΔH = -286 kJ
  • 2H_2(g) + O_2(g) → 2H₂O_(l)    ΔH = -572 kJ
• The magnitude of ΔH varies with the number of moles of substances in the reaction as shown in the reaction above; and it is only true for the physical states of the substances in the equation. For example: H_2(g) + ½O_2(g) → H₂O_(g)    ΔH = -244 kJ

Exothermic and Endothermic Reactions

• An exothermic reaction is one in which the reactants lose heat to the surroundings in order to form the products
• ΔH is negative for an exothermic reaction because ∑H_P < ∑H_R
• A hypothetical exothermic reaction given by the thermochemical equation 2W + X → Q    ΔH = -98 kJ may be represented as:
  • 2W + X – 98 kJ → Q, OR
  • 2W + X → Q + 98 kJ
• An exothermic reaction may be represented on an energy level diagram to show the relative enthalpies of the reactants and products, the activation energy, Ea, and the enthalpy change of the reaction
  

  Photo Credit: GCSE Science

• Dissolution of NaOH in water and combustion reactions are examples of exothermic process

• An endothermic reaction is one in which the reactants absorb heat from the surroundings in order to form the products
• ΔH is positive for an endothermic because ∑H_P > ∑H_R
• An endothermic reaction given by the equation A + B → C      ΔH = +150 kJ may be represented as:
  • A + B + 150 kJ → C
  • A + B → C – 150 kJ
• An endothermic reaction may be represented on an energy level diagram
  

  Photo Credit: Twinkl

• The thermal decomposition of CaCO₃ and dissolution of NH₄Cl in water are examples of endothermic processes

Important Enthalpy Changes

1. Standard Enthalpy Change for a Reaction, ΔH^θ, is the heat absorbed or evolved when the stoichiometric quantities of reactants combine to form the product at 1 atmosphere pressure and 298 K. ΔH^θ = ∑H_P^θ – ∑H_R^θ
2. Standard Enthalpy of Formation, ΔH_f^θ, is the heat liberated or absorbed when 1 mole of a compound is formed from its elements in their standard states (1 atm and 298 K). Enthalpy of formation equations show more than 1 mole of the compound being formed. For example:
   1. ½N_2(g) + ³/₂H_2(g) → NH_3(g)    ΔH_f^θ(NH₃)
   2. 2C_(s) + 3H_2(g) + ½O_2(g) → C₂H₅OH_(l)    ΔH_f^θ(C₂H₅OH)
3. Standard Enthalpy of Combustion, ΔH_c^θ, is the heat evolved when 1 mole of a substance (element or compound) is completely burnt in oxygen gas under standard conditions.
   • The standard enthalpy of combustion of a substance is measured using a bomb calorimeter.
   • Enthalpy of combustion equations must not show more than 1 mole of the substance being burnt.
   • Enthalpy of combustion equations for carbon, methane, hydrogen, and ethanol are as follows:
     1. C + O → CO₂   ΔH_c^θ (C)
     2. CH₄ + O₂ → CO₂ + 2H₂O    ΔH_c^θ (CH₄)
     3. H₂ + ½O₂ → H₂O    ΔH_c^θ (H₂)
     4. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O    ΔH_c^θ (C₂H₅OH)
4. Standard Enthalpy of Hydrogenation of a compound is the heat change which occurs when 1 mole of an unsaturated compound is converted to the corresponding saturated compound by reaction with gaseous hydrogen under standard conditions
   
5. Standard Enthalpy of Neutralization, ΔH_n^θ, is the heat liberated when 1 mole of hydrogen ions reacts with 1 mole of hydroxide ions to form 1 mole of water under standard conditions
   • The enthalpy of neutralization for any strong acid and any strong base reaction is constant and equal to -57 kJmol^-1, irrespective of the types of strong acid and strong base
   • The value is the same and constant because the dilute solutions of strong acids and strong bases are completely ionized so that the main reaction occurring in each case is between 1 mol H⁺ and 1 mol OH^– to form 1 mole H₂O
   • Each of the following reactions has its enthalpy of neutralization equal to -57 kJmol^-1:
     1. 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O   ΔH_n^θ = -57 kJmol^-1
     2. NaOH + HCl → NaCl + H₂O      ΔH_n^θ = -57 kJmol^-1
   • NOTE: The enthalpy of reaction for (a) is -114 kJ and this corresponds to the thermochemical equation:
     • 2K⁺ 2OH^– + 2H⁺ SO₄^2- → 2K+ SO₄^2- + 2H₂O        ΔH = -114 kJ
     • Cancelling out K⁺ and SO₄^2- ions which are spectator ions and dividing all through by 2 gives:
       OH^–_(aq) + H⁺_(aq) → H₂O(l)    ΔH_n^θ = -57 kJ
   • The enthalpy of neutralization for a weak acid and a strong base (or weak base and strong acid) is always less than -57 kJmol^-1
   • The value is lower because weak acids (and weak bases) are only partially ionized and so part of the energy liberated is used in enhancing the ionization of the weak acid or base
   • The enthalpy of neutralization for a weak acid and weak base is much lower than -57 kJmol^-1
6. Enthalpy of Solution is the heat absorbed or evolved when 1 mole of a solute is dissolved in such a large amount of water that further dilution results in no detectable heat change
   • The dissolution of an ionic solid water is a two-stage process involving two enthalpy changes
   • The ionic crystal is broken down into free ions using the energy absorbed from water to overcome the lattice energy of the ionic solute
   • The energy required for this stage is known as lattice dissociation energy and it is always positive
   • The free ions are then hydrated by polar water molecules and the heat liberated during this exothermic process is known as enthalpy  of hydration
   • thus, ΔH^θ_dissolution = (positive) ΔH^θ_{lattice dissociation} + (negative) ΔH^θ_hydration
   • ΔH^θ_dissolution is positive if the ΔH^θ_{lattice dissociation} is greater than ΔH^θ_hydration
   • An increase in temperature favors this type of dissolution
   • If the enthalpy of hydration is larger than the lattice energy,the dissolution process is exothermic, and such a solute dissolves readily in water at low temperatures

Enthalpy Change and Bond Energy

• Enthalpy changes in chemical reactions arise because bonds are broken and formed in every shemical reaction
• Every bond breaking process is endothermic and every bond formation process is exothermic
• The relative magnitudes of the heat associated with the bond breaking and bond forming processes determine the endothermicity or exothermicity of the overall reaction
• Thus, the enthalpy change of a reaction is the sum of the energy associated with the bond breaking process and the energy associated with the bond formation process. That is
  • ΔH_reaction = total energy required to break bonds (positive) + total energy released during bond formation (negative)
• A large amount of energy is required to break a strong bond and a large amount of energy is released when a strong bond forms
• Bond energy is defined as the average amount of energy associated with the formation or breakage of 1 mole of a particular bond in its gaseous state.
• For diatomic molecules such as H₂ or Cl₂, bond energy is the same as bond dissociation energy (B.D.E)
• For H₂, bond energy or B.D.E. = 536 kJmol^-1. Thus, the energy to break down 1 mole of H₂(g) into its free gaseous atoms is 536 kJ
• Atomization energy is the amount of heat required to convert an element in its normal state under standard conditions into 1 mole of its gaseous atoms. For example:
  • ½H_2(g) → H_(g)     ΔH^θ_atomization = +218 kJmol^-1
  • C_(s) → C_(g)       ΔH^θ_atomization = +715 kJmol^-1

Examples

1. Calculate the enthalpy change of the reaction
   
   using the table below:
   
   • Total energy used in breaking all bonds in the reactants:
     • 2 × E(C – H) + E(C ≡ C) + 2 × E(Cl – Cl)
     • (2 × 414) + 837 + (2 × 244) = 2153 kJ
   • Total energy liberated during the formation of all bonds in the product:
     • 2 × E(C – H) + 4 × E(C – Cl) + E(C – C)
     • (2 × 414) + (4 × 338) + 347 = 2527 kJ
   • ∴ ΔH = 2153 – 2527 = -374 kJ

• Alternatively, ΔH may be defined as the difference between the total energy for breaking relevant bonds in the reactants and the total energy released when new bonds are formed in the products
• For the problem above:
  • Energy of bonds broken = E(C ≡ C) + 2E(Cl – Cl)
  • = 837 + 2(244) = 1325 kJ
  • Energy of bonds formed = 4E(C – Cl) + E(C – C)
  • = 4(338) + 347 = 1699 kJ
  • ∴ ΔH = 1325 – 1699 = -374 kJ

Hess’ Law

• Hess’ law is a special case of the first law of thermodynamics which states that energy cannot be created or destroyed but may be converted from one form to another
• Hess’ law states that the enthalpy change of a reaction is constant and independent of the pathway taken or the number of stages that the reaction goes through
• From Hess’ law, if the ΔH for X → Y is Q kJ, then for Y → X, ΔH is -Q kJ
• According to Hess’ law, the enthalpy change for the reaction A → B is the same as the sum of the enthalpy changes for reactions A → C and then C → B
• Using the following enthalpy diagram
  
• By Hess’ law, ΔH₁ = ΔH₂ + ΔH₃
• Hess’ law is particularly important because it can be used to calculate enthalpies of formation that are difficult to determine experimentally

Examples

1. The enthalpy of combustion of ethyne is -1300 kJmol^-1. If the enthalpies of combustion of carbon and hydrogen are -394 and -286 kJmol^-1 respectively, calculate the enthalpy of formation of ethyne
   • Construct an enthalpy diagram for the problem to evolve two different routes
     
   • In route 1, carbon and hydrogen combine to form C₂H₂ which then undergoes combustion to give CO₂ and H₂O. The total energy involved = ΔH_f + (-1300) kJ
   • In route 2, carbon and hydrogen undergo combustion to form CO₂ and H₂O directly. The total energy involved = 2(-394) + (-286) = -1074 kJ
   • By Hess’ law ΔH_f – 1300 = – 1074
   • ∴ ΔH_f = + 226 kJmol^-1

• The equation ΔH = ∑H_P – ∑H_R is also an application of Hess’ law which may be used to solve the problem above as follows:
• Note that:
  • Standard enthalpy of combustion of carbon is the same as standard enthalpy of formation of CO2
  • Standard enthalpy of combustion of hydrogen is the same as standard enthalpy of formation of water
  • Standard enthalpy of formation of a substance is the same as the molar enthalpy of formation, which in turn is the same as the heat content associated with the substance
• Inserting the appropriate heat contents for the known substances in the combustion equation for ethyne, we have
  • C₂H_2(g) + ⁵/₂O_2(g) → 2CO_2(g) + H₂O_(g)       ΔH = -1300 kJ
    ΔH_f             0           2(-394)   (-286)
  • Using ΔH = ∑H_P – ∑H_R
  • – 1300 = 2(-394) + (-286) – (ΔH_f – 0)
  • ∴ ΔH_f = +226 kJmol^-1

Entropy

• Entropy is defined as the degree of disorder or randomness in a system
• It is denoted by the symbol, S, and a change in entropy is denoted by ΔS
• Every substance has a certain amount of entropy but a crystalline substance at -273 °C or 0 Kelvin has an entropy value of zero because its particles are essentially motionless at this temperature
• Of the three states of matter, the gaseous state is the most disordered or the state with the highest entropy
• The solid state has the least entropy because its particles are tightly packed together

S_solid < S_liquid < S_gas

• Like enthalpy, entropy is a state function and so entropy change, ΔS, is independent of the reaction pathway.
• Thus, entropy change, ΔS, is the difference between the total entropy of products and the total entropy of the reactants

ΔS = ∑S_P – ∑S_R

• For the reaction 2SO_2(g) + O_2(g) → 2SO_3(g), if the standard entropies for SO₂, O₂ and SO₃ are 248, 205, and 256 JK^-1mol^-1 respectively at 25 °C, then the standard entropy change of the reaction, ΔS^θ, is given by
  • ΔS^θ = 2(256) – [2(248) + 205] = -189 JK^-1
• For a reversible change, such as melting of ice, which occurs at a constant temperature, the entropy change, ΔS, associated with the process is equal to ΔH
  • ΔH is the heat absorbed by the ice and T is the Kelvin temperature at which heat is absorbed
  • Qualitatively, it if often possible to tell whether a reaction proceeds with a decrease or an increase in entropy by inspecting the reaction equation
  • Any change that brings about a decrease in the degree of molecular complexity tends to produce an increase in entropy, i.e., ΔS is positive for the process because ∑S_P > ∑S_R
  • For example:
    • H₂O_(s) → H₂O_(l)
    • H_2(g) → 2H_(g)
  • If the number of moles of the substance undergoing change is constant but the final state allows the substance a greater degree of freedom, the entropy change, ΔS is positive
  • For example:
    • H_2(g) at 1 atm → H_2(g) at 0.5 atm
• Changes which produce gases from liquids or solids are accompanied by increases in entropy because the entropy of a gas is much larger than that of a liquid or solid
  • For example:
    • CaCO_3(s) → CaO_(s) + CO_2(g)
  • If a change results in a decrease in the number of moles of gas, the change will bring about a decrease in the entropy of the system as the molecular complexity of the final state increases
  • For example:
    • N_2(g) + 3H_2(g) → 2NH_3(g)
      4 moles            2 moles
• Dissolution of solids such as KCl in water results in an increase in entropy because the regular arrangement of the crystal structure is replaced by a random distribution of the ions in solution
  
  • If a process is nor accompanied by any change in the number of moles of gas, the entropy change will be small and practically equal to zero
    • H_2(g) + I_2(g) → 2HI_(g)

Free Energy Change and Spontaneous Reactions

• A spontaneous change is one which occurs naturally on its own without any external aid
• Although most exothermic reactions are spontaneous, there are many endothermic reactions that are also spontaneous
• Thus, enthalpy change is not a unique criterion for ascertaining whether whether a reaction will be spontaneous or not
• The fact that endothermic reactions such as melting of ice could be spontaneous suggests that entropy is an important factor for determining the spontaneity of a reaction
• Entropy change of a system cannot on its own be used as a singular criterion for predicting spontaneity of a chemical reaction
• Apart from ΔH and ΔS, temperature also affects spontaneity of a reaction
• Gibb’s free energy, G, relates to the maximum amount of work obtainable from a system
• It is defined as a function of enthalpy, entropy, and temperature according to the equation G = H – TS
• Like ΔH and ΔS, ΔG = ∑G_P – ∑G_R because it is a function of state which depends only on the initial and final state of the system undergoing change
• At constant temperature and pressure
  • G_P – G_R = H_P – H_R – T(S_P – S_R) OR ΔG = ΔH – TΔS
• The free energy change, ΔG = ΔH – TΔS, gives the combined effects of enthalpy and entropy changes on the spontaneity of a chemical or physical change at constant temperature, T.
• For any system, the value of ΔG is equal to the maximum amount of energy that can be obtained in the form of useful work when the process or change takes place reversibly
• ΔG is the singular criterion for determining the spontaneity or the feasibility of a reaction
  • If ΔG is negative (i.e. TΔS > ΔH), the reaction is spontaneous
  • If ΔG is positive (i.e. TΔS < ΔH), the reaction is not spontaneous
  • If ΔG is zero, the reaction is at equilibrium
• Thus, a change can only be spontaneous if it is accompanied by a decrease in the free energy change or if its ΔG is negative
• The equation ΔG = ΔH – TΔS may be used in predicting the conditions that must prevail for a change to be spontaneous
  • If ΔH is negative and ΔS is positive, the reaction is necessarily spontaneous
    • ΔG = ΔH – TΔS becomes (-ΔH) – [T(+ΔS)], which must turn out negative
  • If ΔH is positive, the reaction is necessarily non spontaneous
    • ΔG = ΔH – TΔS becomes +ΔH – [T(-ΔS)], which must out positive
  • If ΔH and ΔS are both positive, the change will ce spontaneous at high temperatures but not at low temperatures
    • ΔG = ΔH – TΔS becomes (+ΔH) – [T(+ΔS)]
• For ΔG to be negative, TΔS must be larger than ΔH.
• This is the case with the melting of ice, H₂O_(s) → H₂O_(l), which will not occur at low temperatures
• When both ΔH and ΔS are negative, the reaction will be spontaneous at low temperatures but not at high temperatures
  • H₂O_(l) → H₂O_(s) which occurs only at low temperatures below 0 °C