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Entropy: It is a measure of degree of randomness or disorderness in an isolated system. It is represented by the symbol 'S',

\(\Delta\)S_system = \(\frac{+q_{sys.rev.}}{T}\)

Its unit is J K⁻¹ mol⁻¹

The entropy of system increase with increase in temperature.

The energy of a system is defined as its capacity to perform the work. A system with higher energy can perform more work.

∆_fH^Θ of O₂(g) = 0 by convention

∆_rH^Θ = \(\sum\)∆_fH^Θ_(products) - \(\sum\)∆_fH^Θ _(reactants)

Substituting the given values:

−1323 = [2 × (−393.5) + 2 × (−249)] − [∆_fH^ΘC₂H₄]

−1323 = [−787 − 498] − [∆_fH^ΘC₂H₄]

−1323 = −1285 − ∆_fH^ΘC₂H₄

 \(\therefore\) ∆_fH^ΘC₂H₄ = 1323 − 1285

= 38 kJ mol^-1

∆E = 12000 − 9500 = 2500 kJ

100 g of sucrose has energy = 1632 kJ

\(\therefore\) 1000 g of sucrose has energy

= 1632 × 10 = 16320 kJ

Now 2500 kJ of energy of lost in one day

\(\therefore\) 16320 kJ of energy is lost in

= \(\frac{1}{2500}\)x 16320

= 6.5 days.

The energy of a system has many different forms as follows :

• Kinetic energy which arises due to motion, like rotational, vibrational and translational.
• Potential energy which arises due to position and state of a matter. If depends upon the temperature of the system.
• Heat energy (or thermal energy) which is transferred from the hotter body to the colder body.
• Radiant energy which is associated with electro-magnetic or light radiation.
• Electrical energy produced in the galvanic cells.
• Chemical energy stored in chemical substances.

All these various forms of energy can be converted from one form to another without any loss.

There are various forms of energy like kinetic energy, potential energy, heat or thermal energy, radiant energy, electrical energy and chemical energy.

All these forms of energy are interconvertible. For example, a body at very high level possesses higher potential energy. When it falls down, potential energy is converted into kinetic energy.

Falling of water from high level is used to drive turbines converting potential energy into kinetic energy which is further converted into electrical energy.

In galvanic cells, chemical energy is converted into electrical energy.

In electrolytic cells, chemical energy is converted into electrical energy. But during interconversion, the energy can neither be created nor destroyed, and there is a conservation of energy.

Thermodynamics : 

It is concerned with the energy changes in physical and chemical changes.

Drawbacks :

• It does not give information on the rates of physical or chemical changes.
• It does not explain mechanisms involved in physical and chemical processes.

State function : The property which depends on the state of the system and independent of the path followed by the system to attain the final state is called a state function.

For example,

pressure, volume, temperature, etc.

As energy is absorbed in the first reaction ,NO (g) is unstable. As energy is released in the second reaction, NO₂ (g) is stable. Thus, unstable NO (g) changes into the stable NO₂ (g)

Hess’s law states that enthalpy change for a reaction is same whether the reaction occurs in a single step or several steps. 

                                     OR 

If a reaction occurs in several steps then, its standard enthalpy of reaction is equal to the sum of standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. Applications: 

(i) It is used to calculate enrhalpy of the formation of the compound (ii) It is used to calculate Lattice enthalpy of an ionic solid.

Strong acids and bases dissociate completely into ions in aqueous solution and form 1 mole of water. 

Most of the processes are carried out in an open system i.e, at constant pressure.

ΔU = q + w 

ΔU = 802 + (-294) =508J

Heat absorbed by the system = +ve 

Heat evolved by the system = -ve 

Work done by the system = -ve 

Work done on the system = +ve 

(i) Order increases, entropy decreases.

(ii) Order increases, entropy decreases.

(iii) Disorder increases, entropy increases

(iv) Order increases, entropy decreases.

(v) Order increases, entropy decreases.

(vi) Order increases, entropy decreases.

(vii) Disorder increases, entropy increases.

(viii) Order increases, entropy decreases.

(b) The process of interpretation involves coupling of two reactions, getting the sum of ΔG and looking for its sign. Overall –ve value of ΔG shows that reduction of ore is possible.

Given : 

ΔH = - 42.0 kJ mol^-1,

T = 298 K, 

ΔU = ?

\(\frac{1}{2}\)N_2(g) + \(\frac{3}{2}\)H_2(g) ⟶ NH_3(g)

Δn = 1 – (\(\frac{1}{2}\)+\(\frac{3}{2}\)) = -1 mol

ΔH = ΔU + ΔnRT

∴ ΔU = ΔH – ΔnRT

= - 42 – (- 1) × 8.314 × 298 × 10^-3

= - 42 + 2.477

= - 39.523 kJ

∴ ΔU = -39.523 kJ

It states that the enthalpy change of the reaction remains the same whether the reaction is carried out in one step or several steps, i.e., ∆H = ∆H₁ + ∆H₂ + ∆H₃ +…………

No, energy of the universe remains constant (law of conservation of energy). Entropy of the universe is continuously increasing.

A spontaneous process should continue taking place by itself after initiation . But this is not so in the given case because water will go up so long as the pump is working

 No because after breaking of C-H bonds one by one, the electronic environments change. The reported value is the average of of the bond dissociation energies of the four C-H bonds.

(i) H₂O(l) → H₂O(g)

Δn_(g) = 1 − 0 = 1

\(\because\) ΔH = ΔU + Δn_gRT

or ∆U = ∆H − ∆n_gRT

Given, ∆H = 41 kJ mol⁻¹,

T = 100 + 273 = 373 K

\(\therefore\) ∆U = 41 kJ mol⁻¹ − 1 × 8.3 J mol⁻¹K⁻¹ × 373 K

= 37.904 kJ mol⁻¹

(ii) H₂O(l) → H₂O(s)

There is negligible change in volume.

\(\therefore\) p∆V = ∆n_gRT = 0

\(\therefore\) ∆H = ∆U

Or ∆U = 41 kJ mol⁻¹

Given,

N = 10 moles

P_i = 5 atm

P_f = 1 atm

T = 300K

R = 8.134 J K^-1 mol^-1

H = 1m

\(\because\) W_exp = −2.303 nRT log\(\frac{P_i}{P_f}\)

= −2.303 × 10 × 8.314 × 300 log\(\frac{5}{1}\)

= −40.15 × 10³ J

If mass (M) can be lifted by this work through a height of 1m, then

Work done = Mgh

or M = \(\frac{Work\,done}{g\,h}\)

= \(\frac{40.15\times10^3\,kgm^2s^{-2}}{9.81\,ms^{-2}\times1m}\) [\(\because\) J = kg m²s^-2]

= 4092.76 kg

Given : Enthalpy of solution of CuSO₄ in 10 mol 

H₂O

= Δ_solnH = ΔH₁ = -54.5 kJ mol^-1

Enthalpy of solution of CuSO₄ in 100 mol

H₂O = ΔH₂

= - 68.4 kJmol^-1

Mass of water = 1620 g

For dilution,

Δ_dilH = ?

Now,

80 g H₂O = \(\frac{180}{18}\) =  10 mol H₂O

And,

1620 g H₂O = \(\frac{1620}{18}\) = 90 mol H₂O

Hence for heat of dilution,

CUSO₄(10H₂O) + 90H₂O_(I) → CUSO₄(100H₂O)

Δ_dilH = ?

∴ Δ_dilH = ΔH₂ -ΔH₁

= - 68.4 – (54.5)

= -13.9 kJmol^-1

∴ Heat of dilution = Δ_dilH = -13.9 kJ mol^-1

Given : Heat of solution of AgF = Δ_solnH

= -ΔH₁ = -20.5 kJmol^-1

Heat of hydration of AgF = Δ_hydH = ΔH₂

= -930 kJ mol^-1

Lattice energy of AgF = Δ_LH = ΔH₃ = ?

For heat of solution,

AgF_(s) + aq → AgF_(aq) ΔH₁

For heat of hydration,

Ag⁺_(g) + F^-_(g) + aq → Ag⁺_(aq) + F^-_(aq)ΔH₂

For Lattice energy,

Ag⁺_(g) + F^-_(g) → AgF_(s)

Δ_LH = ?

From above equations,

∴ ΔH₃ = ΔH₂ – ΔH₁

= -930 – (-20.5)

= -909.5 kJmol^-1

∴ Lattice energy of AgF_(s) = -909.5 kJ mol^-1

Given : Enthalpy of hydration of Li⁺_(g)

= Δ_hydH₁

= -500 kJmol^-1

Enthalpy of hydration of Br^-_(g) = Δ_hydH₂

= -350 kJ mol^-1

Lattice energy of LiBr_(s) = Δ_LH₃ = 807 kJ mol^-1

Enthalpy of solution of LiBr_(s) = Δ_solnΔH = ?

The thermochemical equation for the dissolution of LiBr_(s) forming a solution is,

LiBr_(s) + aq → Li⁺_(aq) + Br^-_(aq)(I) Δ_solH = ?

This takes place in two steps as follows :

(i). LiBr_(s) → Li⁺_(g) + Br^-_(g) Δ_LH₃

(ii).(a). Li⁺_(g) + aq → Li⁺_(aq) Δ_hydH₁

(b) Br^-_(g) + aq → Br^-_aq Δ_hydH₂

Hence by adding equations (i) and (ii) (a) and (b) we get equation .

∴ Δ_solH = Δ_LH₃ + Δ_hydH₁ + Δ_hydH₂

= 807 + (-500) + (-350) 

= -43 kJ mol^-1

∴ Heat of solution of LiBr_(s) = Δ_solH = -43 kJ mol^-1

\(\because\) 22.4 dm³ of gas at STP = 1mol

\(\therefore\) 7.5 dm³ of a gas STP = \(\frac{1}{22.4}\)x 7.5

= 0.33 mol

\(\because\) For 15°C rice, 0.33 moles of the gas at constant volume requires heat = 110 J

\(\therefore\) For 1°C rise, 1 mole of the gas at constant volume will require heat

= \(\frac{110}{15\times0.33}\)

= \(\frac{110}{4.95}\) = 22.222 J

\(\therefore\) C_v = 22.222 JK⁻¹ mol⁻¹ = 22.22 J K⁻¹mol⁻¹

\(\because\) C_p − C_v = R

or C_p = C_v + R

= 22.222 J K⁻¹ + 8.314 J K⁻¹ mol⁻¹

= 30.536 = 30.53 J K⁻¹ mol⁻¹

y = \(\frac{C_p}{C_v}\) = \(\frac{30.53}{22.22}\) = 1.37

\(\therefore\) The gas is triatomic.

According to this law, "the entropy of a perfectly crystalline substance approaches zero as the absolute zero of temperature is approached."

According to this law, for a spontaneous process in an isolated system, the change in entropy is positive.

Thermochemistry : Thermodynamic study of heat changes or enthalpy changes during the chemical reactions is called thermochemistry.

Consider a reaction, 

Reactants → Products The heat changes ΔH for the reaction may be represented as,

ΔH_reaction  = ΣH_products  – ΣH_reactants

Where H represents enthalpy. The energy released or absorbed during a chemical change appears in the form of heat energy.

Dissolving salt in water: When we pour table salt in water, the salt dissolves on its own, the reaction consumes some energy from the water, lowering its temperature slightly. Even through enthalpy increases, entropy increases even more. So, this reaction is both endothermic and spontaneous at standard temperature and pressure.

Endothermic reaction :

• In endothermic reaction heat is absorbed from suroundings. 
• Sum of enthalpies of products is greater than sum of enthalpies of reactants i.e. ΣPH > ΣRH 
• Heat of reaction, ΔH is positive. 
• Products are less stable than reactants. C_(s) + O_2(g)  → CO_2(g) ΔH = -394 kJ 
• This reaction requires supply of thermal energy.

 Exothermic reaction :

• In exothermic reaction heat is given out to surroundings.
• Sum of enthalpies of products is less than sum of enthalpies of reactants. i.e. Σ_PH < Σ_RH
• Heat of reaction, ΔH is negative.
• Products are more stable than reactants.
• N_2(g) + O_2(g) → 2NO ΔH = + 180 kJ
• This reaction does not require supply of thermal energy.

Enthalpy or heat of reaction : The enthalpy of a chemical reaction is the difference between the sum of the enthalpies of products and that of the reactants with every substance in a definite physical state and in the amounts (moles) represented by the coefficients in the balanced equation.

Explanation : Consider the following general reaction,

aA + bB → cC + dD

The heat of the reaction ΔH is the difference in sum of enthalpies of products and sum of enthalpies of reactants.

∴ ΔH = ΣH_produts – ΣH_reactants 

= [cH_c + dH_D] – [aH_A + bH_B] 

= Σ_PH – Σ_RZ

Where H represents enthalpy of the substance. For endothermic reaction, 

ΔH is positive, (ΔH > 0).

For exothermic reaction, 

ΔH is negative, 

(ΔH < 0).

(i) Exothermic process : A process taking place with the evolution of heat is called exothermic process.

For this process, 

Q is -ve, 

ΔH is -ve.

(ii) Endothermic process : A process taking place with the absorption of heat (from the surroundings) is called endothermic process. 

For this process, 

Q is +ve, 

ΔH is +ve.

(a)and (d) : Reversible ; 

(b) and (c ): Irreversible. 

•  The amount of heat added to a system at higher temperature causes less disorder than when the heat is added at lower temperature.
• Since disorder depends on the temperature at which heat is added, ΔS relates reciprocally to temperature.
• This can also be explained from equation, ΔS = \(\frac{Q_{rev}}{T}\)

(i) Open and closed system

(ii) Adiabatic and isothermal process

(iii) State function and path function

Open system :

1. An open system can exchange both matter and energy with the surroundings. 

2. In this, the total amount of matter does not remain constant. 

3. Example : Hot water kept in an open beaker.

Closed system :

1. A closed system can exchange only energy, but not matter with the surroundings.

2. In this, the total amount of matter remains constant.

3. Example : Hot water kept in a closed glass flask.

We begin by writing the equation that represents this reaction. Recall that "complete combustion," or burning, is a reaction with oxygen from the atmosphere, forming carbon dioxide and water:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) 

ΔG° = (1 mol)[ΔGf° for CO₂(g)] + (2 mol)[ΔGf° for H₂O(l)] – (1 mol)[ΔGf° for CH₄(g)] – (2 mol)[ΔGf° for O₂(g)] = (1 mol) (–394.4 kJ/mol) + (2 mol) (–237.0 kJ/mol) – (1 mol) (–50.8 kJ/mol) – (2 mol) (0) = –817.6 kJ 

The negative value of ΔG° indicates that the reaction is spontaneous. This matches ourexperiences in everyday life, where we have seen that natural gas burns spontaneously.

(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.

Explanation :

1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc. 

2. The intensive properties are not additive.

(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.