JEE Important Chapter - Energetics

Let's see what happens if we pour a drop of water into an uncovered petri dish. The water evaporates after a few minutes. Let's try a different experiment in a covered petri dish. What is going to happen? After the same time interval as the first experiment, it remains on the plate.

We can ask a number of questions about these experiments. What causes these events? How quickly do they occur? Is it possible to provide a chemical explanation for these events? What is the name of this procedure? What is the name of the reversal procedure? Where have you seen it in your day-to-day life? What other effects did the initial process have? Thermochemistry, kinetics, and thermodynamics are used to study all these topics. As a result, chemical energetics plays a significant role in advanced physical chemistry.

Important Topics for Chemical Energetics

• Enthalpy Change

• Hess Law

• Born - Haber Cycle

• Bond Energies

• Reaction Pathway

• Entropy Change

• Gibbs Free Energy

• Spontaneity

• Glycolysis

Important Concepts for Chemical Energetics

Enthalpy change, ΔH 

• The thermal energy held in a chemical system is known as enthalpy. It can't be directly measured.

• Energy changes occur as a result of chemical processes (typically heat energy changes). 

• The reaction is exothermic when energy is discharged into the environment, and H is negative. The reaction is endothermic and H is positive if energy is taken in from the environment. 

• During an exothermic reaction, the temperature of the environment rises, while during an endothermic reaction, it falls.

• The energy required to break one mole of a particular gaseous bond and produce atoms is referred to as bond energy. 

• As the actual enthalpy of a bond changes depending on which molecule it is in, bond enthalpies are given as a mean average in the data book. 

• Breaking bonds requires energy, hence ΔH is positive (endothermic). When bonds are formed, energy is released, hence ΔH is negative (exothermic).

• The equation below can be used to calculate the enthalpy change of a reaction:

H = -mcΔT 

where, H stands for enthalpy change (J)

m is the mass of the environment (g)

c stands for specific heat capacity (J g-1 K-1)

ΔT - temperature change (in kelvin or degrees Celsius)

• Standard terms used in Enthalpy

Lattice Energy

• The enthalpy change that occurs when one mole of a substance is produced from its gaseous ions is known as lattice formation enthalpy. 

• The enthalpy shift that occurs when one mole of an ionic substance is broken down to form its gaseous ions is known as lattice dissociation enthalpy. ΔH is negative when gaseous ions are converted into a solid lattice.

• Lattice enthalpy is influenced by two factors:
  Ionic charge: Increased ionic charge increases the attraction between positive and negative ions, resulting in a greater and more negative lattice formation enthalpy.

Ionic radius: A smaller ionic radius means the ions are closer together in the lattice, resulting in a greater attraction between them, resulting in a bigger, more negative lattice formation enthalpy.

Hess’ Law

• According to Hess' Law, the enthalpy shift that occurs as a result of a chemical transition is unaffected by the route it follows. This is due to the fact that the reactants and products have the same enthalpy.

• Any energy released during the intermediate formation process will be used to break bonds, allowing the product to form. The enthalpy change of route 1 is the same as route 2 in the diagram below:
  
  

• Hess' Law can be combined with the enthalpy change of production and combustion to determine the enthalpy change of a reaction. 

• It's crucial to pay attention to the directions of the enthalpy change arrows in the following cycles:

• To find the enthalpy change of a reaction, using the enthalpy change of combustion:

• To find the enthalpy change of reaction, using the enthalpy change of formation:
  
  

Average Bond Energies

• When all reactants and products are gaseous, bond energies can be utilised to calculate the enthalpy change of a process.

• Method 1: ΔH = total energy required to break bonds minus total energy generated during bond formation. The average bond enthalpies for each bond must be multiplied by the number of that bond included in the equation when totaling the energy released or created.

• The second method is to use an enthalpy cycle.
  Example:  CO(g) + H2O(g) → CO2(g) + H2(g)
  
  

Using the diagram and following:
ΔH + 436 + 2(805) = 1077 + 2 (464)

ΔH = 1077 + 2(464) - 436 - 2H = 1077 + 2(464) - 436 - 2H = 1077 + 2(464) - (805)

ΔH = -41 kJ mol-1.

Born-Haber Cycles

• Born-Haber cycles are used to calculate lattice enthalpy. The following enthalpy changes are used in Born-Haber cycles:

• Enthalpy of formation on the lattice or enthalpy of dissociation on the lattice

• Atomisation's enthalpy shift

• Formation of enthalpy change

• The energy necessary to remove one electron from one mole of gaseous atoms in order to generate one mole of gaseous 1+ ions is known as the first ionisation energy. This is only used for metals in Born-Haber cycles.

• The energy produced when each atom in one mole of gaseous atoms receives an electron, generating one mole of gaseous 1-ions. This only applies to non-metals in Born-Haber cycles.

• A Born-Haber cycle for lithium fluoride and the computation of the lattice enthalpy (ΔHӨLE) are shown below:
  
  

Reaction Pathway Diagrams

• The following are diagrams of exothermic and endothermic reaction pathways:

Entropy change, ΔS

• The degree of disorder in a system is measured by entropy. When there is more chaos and energy is spread out, a system becomes more stable.

• From solid to liquid to gas, entropy increases, and watery substances have higher entropy than solids. This is because the energy spreads out as the particles get more disordered as their states change.

• As gases have significantly higher entropy than liquids or solids due to their disordered mobility, the entropy change will be positive (entropy increases) if the number of gaseous moles increases throughout a process.

• Increasing the temperature may create a state shift, increasing the entropy.

• Entropy increases when the temperature rises without a change in the state because the particles have more kinetic energy and are more disordered as they move faster.

• The following equation can be used to calculate the change in entropy of a reaction:

ΔS = ΣSϴ(products) - ΣSϴ(reactants) 

Where Σ means ‘the sum of’ and the standard entropies of the reactants and products have been given.

A negative result indicates a drop in entropy, whereas a positive result suggests an increase in entropy.

Gibbs Free Energy Change, ΔG

• This equation can be used to compute Gibbs free energy change:
  ΔG = ΔH - TΔS; where
  ΔG - Gibbs free energy change (kJ mol-1)

ΔH - enthalpy change (kJ mol-1)

T - temperature (K)

ΔS - entropy change (kJ K-1 mol-1) 

• When ΔG is less than or equal to 0, a response or process is spontaneous/possible. 

• If ΔG is greater than zero, the reaction is not spontaneous at the calculated temperature. 

• When you have H and S, rewrite the equation as ΔH - TΔS 0 and rearrange it to find the minimal temperature at which a reaction is spontaneous.

Spontaneity

• When ΔH is negative and ΔS is positive: TΔS is negative since H is negative and TΔS is positive in the equation. As ΔG is always negative regardless of temperature, the reaction is spontaneous at all temperatures.

• When ΔH is positive and ΔS is negative: the equation says that -TΔS is positive because ΔH is positive and TΔS is negative. ΔG is always positive and the reaction is never spontaneous because both terms are positive regardless of temperature.

• When ΔH and ΔS are both positive: Because ΔH is positive and TΔS is positive in the equation, -TΔS is negative. As the temperature rises, -TΔS becomes increasingly negative. -TΔS will overwhelm H at high temperatures, and ΔG will be less than 0, resulting in a spontaneous reaction. At low temperatures, the reaction will not be spontaneous.

• When ΔH and ΔS are both negative: TΔS is positive since ΔH is negative and TΔS is negative. At high temperatures, -TΔS becomes increasingly positive and outweighs ΔH, indicating that the reaction is not spontaneous. At low temperatures, the reaction will occur spontaneously.

Energetics of Glycolysis

• Glycolysis is the process of breaking down glucose to produce energy. It generates two pyruvate molecules, ATP, NADH, and water. 

• The process occurs in a cell's cytoplasm and does not require oxygen. 

• It can be found in both aerobic and anaerobic bacteria.

• The energetics of glycolysis include the formation of two molecules of glyceraldehyde 3-phosphate from one glucose molecule in the second stage of glycolysis, and the formation of two molecules of pyruvate as end products of glycolysis. 

• As a result, two molecules of glyceraldehyde 3-phosphate are used to calculate glycolysis energy.

Solved Examples from The Chapter

Question 1: Calculate the enthalpy change of combustion for a reaction in which 0.65 g of propan-1-ol was totally combusted and 150 g of water was heated from 20.1 to 45.5oC.

Solution: 

• Calculate the amount of energy that was utilised to heat the water.
  Q = m x cp x ΔT,

∴ Q = 150 x 4.18 x 25.4,

∴ Q = 15925.8 J

• Calculate how many moles of alcohol were combusted.
  moles of propan-1-ol = mass/ Mr (molecular weight) = 0.65 / 60 = 0.01083 mol

• ΔcH is the enthalpy change per mole that is calculated (the enthalpy change of combustion).
  ΔH = Q/ no of moles

∴ ΔH = 15925.8/0.01083

∴ ΔH = 1470073 J mol-1 = 1470 kJ mol-1 to 3 sf

• Finally, add the sign to represent the energy change: if the temperature rises, the reaction is exothermic and is denoted by a negative sign, such as –1470 kJ mol-1.

Question 2: Calculate the temperature range in which this reaction can occur.
N2(g) + O2(g) → 2NO(g).

Solution: 

• ∆ H = 180 kJ mol-1 ∆S = 25 J K-1 mol-1 The reaction will be feasible when ∆ G ≤0

• Make ∆G = 0 in the following equation: ∆G = ∆H - T∆S

∴ 0 = ∆H - T∆S

• So, T = ∆H / ∆S

T = 180/ (25/1000)

= 7200K

• The T must be greater than 7200K, which is a very high temperature.

Solved Examples from Previous Year Question Papers

Question 1: When an ideal diatomic gas is heated at constant pressure the fraction of the heat energy supplied which increases the internal energy of the gas is:

(a) 2/5

(b) 3/5

(c) 3/7

(d) 5/7

Solution: 

• According to the first law of thermodynamics,
  ΔU = q + W

• When under constant pressure,
  ΔU = qp - PΔV

qp = ΔU + PΔV

• Heat Energy Fraction that Increases Internal Energy
  $= \frac{\bigtriangleup U}{q_p} \ = \frac{nC_v \bigtriangleup T}{nC_p \bigtriangleup T} \ = \frac{C_v}{C_p}$

• In the case of a diatomic gas,
  Cp = 7R/2, & Cv = 5R/2.
  Cv/Cp = 5/7.

• As a result, option (c) is the correct answer.

Question 2: Enthalpy change for a reaction depends upon

(a) physical state of reactants and products

(b) initial and final concentration

(c) number of steps in the reaction

(d) conditions of reaction

Solution: 

• ΔH of a reaction is determined by the following conditions:

(i) Reactants and products' physical states

(ii) Concentration at the start and end

(iii) The circumstances in which the reaction took place

• As a result, option (a), (b), (d) is the correct answer.

Question 3: The equilibrium constant for the reaction, 2NOCl(g) → 2NO(g) + Cl2(g)  at 400 K is . (Given ∆Ho = 77.2 kJ mol-1; ∆So = 122 J K-1 mol-1 (log1.95 = 0.292)

(a) Kc = 1.995 x 104

(b) Kc = 1.995 x 10-3

(c) Kc = 1.995 x 10-4

(d) Kc = 1.995 x 10-4

Solution: 

• ∆Go = ∆H - T∆So

• ∆Go = 77.2 x 1000 - 400 x 22

• ∆Go = 77200 - 48800

• ∆Go = 28400 Jmol

• As ∆Go = -2.303RT logKeq

• Therefore, ∆Go = 1.95 x 10-5

• As a result, option (d) is the correct answer.

Practice Questions

Question 1: What is the temperature at which methane melts?

CH4(s) → CH4(l) ∆H = 0.94 kJmol-1, ∆S = 10.3 Jmol-1 K

(a) T = 91 K

(b) T = 87 K

(a) T = 45 K

(a) T = 95 K

Answer: (a) T = 91 K

Question 2: Calculate the enthalpy of solution of NaCl if the lattice enthalpy of production of NaCl is -771 kJmol-1 and the hydration enthalpies of sodium and chloride ions are -406 kJmol-1 and -364 kJmol-1, respectively.

(a) -1 kJ mol-1

(b) +3 kJ mol-1

(c) +1 kJ mol-1

(d) -2 kJ mol-1

Answer: (c) +1 kJ mol-1

Conclusion of JEE Advanced Physical Chemistry Energetics Notes

Heat is absorbed or released in conjunction with all chemical processes. From the beginning, the intimate relationship between matter and energy has piqued people's interest and sparked discussion; it's no coincidence that fire was regarded as one of the four basic elements (together with earth, air, and water) as early as the fifth century BCE. Here we went over some of the essential ideas of energy and heat, as well as the relationship between them, in this unit.