Thermodynamics

Chemical kinetics tells us about the rate of the reaction and its also give an idea about various factor which affects the rate of chemical reaction.

Thermodynamics is the branch that deals with the transformation of energy from one form to another. It also gives an idea about whether a reaction will be going to take place or not and it also gives an idea about the extent of the reaction.

Closed System: - A system in which there is only the exchange of energy between system and surroundings.

Open System: - A system in which there is an exchange of both matter and energy is termed as an Open System.

Reason: -

(i) As the system is covered, there will be no exchange of matter and the system cannot be considered as an open system.

(ii) As mentioned above there will be only the exchange of energy between system and surroundings in a closed system.

(iii) In a closed vessel there will be no exchanged of matter but there will be an exchange of energy which will be taken place through the wall of the copper vessel. So, it can be considered a closed system.

(iv) In closed vessels, there will be exchange of energy, but insulation will restrict the exchange of energy between the system and surroundings.

Reason: - For defining the state of the system you do not have to define all the variable of the system, you have to define only independent variable, which is known as State Function. State function are pressure, volume, temperature and amount.

Relation between above mention state function is mentioned below

PV = nRT

Where P = Pressure

V = Volume

n = Number of Moles

R = Gas Constant

T =Temperature

Reason: - Specific Heat is define as amount of heat require to raise the temperature of the substance by 1 degree. It is an intensive property, so change in volume or any other property will not have any impact on its value.

Reason: -

A. As it is provided that there is combustion of 1 mole of butane.

B. The heat released during the combustion of butane is 2658 kJ.

C. In this reaction, there is a combustion of 1 mole of butane and the energy released is 2658 kJ.

D. In this reaction, there is also combustion of 1 mole of butane but the sign of Δ_cH is positive, which is indicating that we have given that much amount of energy.

Reason: C (s) + 2H₂ (g) → CH₄ (g)

Δ_f = Δ_f + Δn_gRT

Δn_g= 1-2= -1

Substituting the value of Δn_g, above equation will become

Δ_f = Δ_f- RT

As temperature and gas constant both have a positive value, so the value of Δ_f will become more negative.

Reason: As it is an adiabatic process so q=0 and as it is free expansion of gas is there so w=0.

From the first law of Thermodynamics

ΔU = q + w

This means that ΔU will be 0 which indicates that temperature will remain constant.

From the above plot, it is easily understood that the area under the curve is more for irreversible compression.

Freezing is an exothermic release so there will be release in energy, which is absorbed by surrounding.

ΔS _(system)=- and ΔS _{(surroundings)}=

Therefore, the entropy of system will be decrease and that of surrounding will be decrease.

Reason:

C (graphite) + O₂ (g) → CO₂ (g); Δ_rH = x kJ mol^–1..…(i)

C (graphite) + O₂ (g) → CO (g);Δ_rH = y kJ mol ^–1……(ii)

CO (g) + O₂ (g) → CO₂ (g);Δ_rH = z kJ mol^–1 ……(iii)

Adding (ii) and (iii) will be result in equation (I)

Which means that y + z = x

Reason: Same bonds are formed in both the reaction but in the second reaction there is breaking of bond take place in which energy is absorbed.

Example:

Now, from the above example it is clear that enthalpy of a formation of compound can be positive or negative.

Reason: Sublimation is the process of conversion of solid into vapor.

Solid now can be also first converted into the liquid which require enthalpy of fusion and then it will be converted into vapor which require enthalpy of vapourization.

Reason: Gibbs free energy gives the criteria to find out the spontaneity of the reaction.

ΔG is positive for a non-spontaneous reaction.

ΔG is negative for a spontaneous reaction.

ΔG is zero indicates that reaction is at equilibrium.

Reason:

Thermodynamics is that branch of science which deals with a different form of energy and their transformation from one form to another and also deals with thermal or mechanical equilibrium but it does not account at what rate these transformations occur.

Reason: In an exothermic reaction heat is released which is indicated by the negative sign of Δ_rH.

And we know the relation that ΔH=q_p, at constant pressure.

So, if ΔH is negative which means thatq_pwill also be negative.

Reason: Spontaneous Reaction is those reactions which occur on its own without being helped by any external source.

(i) heat always flows from a higher temperature to a lower temperature.

(ii) There is always a pressure gradient require for gas to move.

(iii) This can be possible as gas moves from high pressure to low pressure.

(iv) These processes can occur on its own does not require any external assistance.

Reason:

As we all know the below relation V_f

w = -nRT ln

Here R, T, and ln are the same in both cases, sowork done will only depend on the temperature.

= = = 2

Here w₂ indicates work done at 600K

w₁indicates work done at 300k

so, from the above expression, it is clear that work done at 600K is twice the work done at 300k.

(iv)Internal energy depends on the temperature and it is given that the above process is isothermal, which indicates ΔU=0, in both the cases.

Reason:

ΔH =ΣEnthalpy of formation of product –

ΣEnthalpy of formation of reactant

= [2Δ_fH_ZnO] - [2Δ_fH_Zn+ Δ_fH_O2]

[2Δ_fH_ZnO] = ΔH + [2Δ_fH_Zn+ Δ_fH_O2]

As, above reaction is an exothermic reaction, heat will be released, so ΔH will be negative.

iii) As per the sign convention, the negative sign of ΔHindicates that the reaction is exothermic which means heat is released.

o The enthalpy or ∆H of a reaction is defined as the change in energy (more specifically heat energy) for one mole of a substance involved in a process.

It is given that, the enthalpy change for 18g of water (at 100^⁰ and 1 bar pressure) is =40.79 kJ mol^–1 in the process of vapourisation.

And for water, 18g is =1 mole (For H₂O, the calculation of molar mass: 2×1 + 16 = 18g)

Hence, enthalpy change of vapourisation for 1 mole = 40.79 kJ mol^–1 enthalpy change of vapourisation for 2 moles of water = (40.79 × 2) = 81.58kJ mol^–1

o Standard enthalpy of a process or ∆H⁰ is the enthalpy change of the process when involved substances are in their standard states.

Hence, for water ∆H_{vapourisation,} ⁰ will be equal to =40.79 kJ mol^–1

o Enthalpy of vapourisation is defined as the heat consumed by a liquid to transform into the gaseous form at a certain pressure and temperature.

o 1 mole of acetone requires less heat than 1 mole of water to vaporize. So, water consumes less heat than the same amount of acetone for vapourisation.

Hence, amongst the two liquids, water has a higher enthalpy of vapourisation (consuming higher heat energy).

Therefore, ∆H_{vapourisation} (water) > ∆H_{vapourisation} (acetone).

o The Heat of Reaction or the enthalpy of a reaction Δ_rH^Θ is the change in the enthalpy of a chemical reaction that occurs at a constant pressure.

Whereas, the standard molar enthalpy of formation Δ_fH^Θ is defined as the change in enthalpy when one mole of substances are involved in the standard state (1 atm of pressure and 298.15 K) formed from pure elements under the same standard conditions; hence it is a special case of enthalpy of reaction (in addition with the 1 mole and standard forms).

o The given reaction CaO(s) + CO₂(g) →CaCO₃(s) is clearly indicating that it is occurring in the standard form of 1 mole of each substance. And the molar enthalpy of formation Δ_fH^Θ = –178.3 kJ mol^–1 Given for CaCO₃ is also showing the standard conditions.

So,Δ_fH^Θ = –178.3 kJ mol^–1 = Δ_rH^Θ .

o Enthalpy change of a reaction is calculated as
: Σbond enthalpy of reactants- Σbond enthalpy of products

o The standard molar enthalpy of formation (for one mole) Δ_fH^Θ is given for NH₃ = – 91.8 kJ mol^–1 which is associated with the reverse exothermic reaction N₂(g) + 3H₂(g) → 2NH₃(g) i.e. heat released information of ammonia is – 91.8 kJ mol ^–1.

Hence, for the decomposition 2NH₃(g) →N₂(g) + 3H₂(g) Δ_rH^Θ will be =

- (– 91.8 kJ mol^–1 ) = + 91.8 kJ mol^–1 .

And in this case for 2 moles of NH₃ enthalpy change of the reaction will be Δ_rH = (2 X 91.8 ) = 183.6 kJ mol^–1 .

o For the reaction, A→B the formation of B goes through several intermediate reactions with different enthalpy values Δ_rH₁, ΔrH₂, Δ_rH_3....., and the overall enthalpy change is Δ_rH.

Being an extensive property means a property which depends on the quantity of the size of substance present in a system (like enthalpy, mass, volume or internal energy).
And

o According to Hess’s law: “If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.”

and that means the intermediate enthalpies will be additive in nature to give the overall enthalpy for the given reaction :

Hence, Δ_rH= Δ_rH₁ +Δ_rH₂ + Δ_rH₃ +.......... and it can also be represented as :

o Enthalpy of atomisation is defined as The enthalpy of atomization is the enthalpy change related to the total separation of all atoms in a chemical substance (either of a chemical element or of a chemical compound).

o For the given reaction CH₄(g)→C(g) + 4H (g), the enthalpy of atomisation ∆_aH^Θ =1665 kJ mol ^–1.

o And it can be seen that the decomposition of methane is occurring by the breakage of 4 C-H bonds Present in methane.

Hence, for 1 C-H bond, the bond energy will be equal to 1/4 that of the enthalpy of atomisation

= (1665/4) = 416.25 kJ mol^–1 .

o Lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its constituent ions in the gaseous state. Under standard conditions.

• Formation of an ionic compound in the gaseous state involves several processes like :

Sublimation of the metal(Δ_subH^Θ) →Ionization of the metal (Δ_iH^Θ) →Dissociation of the non-metal (Δ_dissH^Θ) →Gain of electrons by the non-metal(Δ_egH^Θ)

The enthalpy of formation is related to these enthalpies of these processes by the relation:

Δ_fH^Θ=Δ_subH^Θ+Δ_iH^Θ+ 1/2 Δ_dissH^Θ+Δ_egH^Θ+Δ_latticeH^Θ

o To calculate the lattice enthalpy of NaBr, we have to consider a set of reactions for the process:

Na(s)→Na(g) ;Δ_subH^Θ=108.4 kJ mol^–1 (i)

Na→Na⁺ + e^- ;Δ_iH^Θ= 496 kJ mol^–1 (ii)

1/2 Br₂→Br ; 1/2Δ_dissH^Θ=() = 96 kJ mol^–1 (iii)

Br + e^-→Br^- ;Δ_egH^Θ= – 325 kJ mol^–1 (iv)

And enthalpy of formation Δ_fH^Θfor NaBr (s) = – 360.1 kJ mol^–1

Enthalpies in different steps involved in the formation of NaBr (s):

Hence,

enthalpy of formation Δ_fH^Θ = processes (i) + (ii) + (ii) + (iv)

Δ_fH^Θ = Δ_subH^Θ + Δ_iH^Θ + 1/2 Δ_dissH^Θ +Δ_egH^Θ + Δ_latticeH^Θ

Therefore, Δ_latticeH^Θ = Δ_fH^Θ - Δ_subH^Θ - Δ_iH^Θ - 1/2 Δ_dissH^Θ -Δ_egH^Θ

= (–360.1 - 108.4 -96 – 496 +325) kJ mol^–1

= - 735.5 kJ mol^–1

o The spontaneity of any process mainly depends on the sign of Gibb’s free energy. A negative ∆G means the reaction will be spontaneous included contribution of the other factors like enthalpy(H) and entropy(S).

o In case of diffusion of the gases, the volume will be increased, therefore entropy will increase i.e. ∆S = positive.

Hence in the relation, ∆G= ∆H - T∆S

despite of ∆H being 0 ,

∆G will be negative (as positive value of ∆S will give a more negative T∆S)

• ∆H=0 ; ∆G= -T∆S= negative.

o As the heat has a randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system this two is related with each other through another parameter which is entropy.

o The mathematical relation which relates these three parameters is

where ΔS is the change in entropy and T stands for temperature.

o As thermal equilibrium obeys the zeroth law of thermodynamics, temperature of system and surroundings will be the same when they are in thermal equilibrium.

• Thermal equilibrium is defined as the point when two physical systems (here system and the surroundings) are brought into a connection that does not allow the transfer of matter or any kind of energy between them, such a connection is said to permit the transfer of energy as heat.

• when such a connection is made between two physical systems and the making of the connection is followed by no change of state of the either, then the two systems are said to be in relation of thermal equilibrium, and that means the system and surroundings both will have the same temperature.

o Spontaneity depends on the sign of Gibb’s free energy G of a reaction and it’s a relation with K_p, the equilibrium constant at constant pressure is: ∆_RG = - RT ln K_p

For the given reaction N₂O₄ (g) ⇋2NO₂ (g) the value of K_p = 0.98.

Hence, ∆_rG⁰ = - RT ln(0.98)

Since , ln(0.98) has a negative value , the value of ∆_rG⁰ becomes positive. Therefore, the reaction is non- spontaneous

Enthalpy or H is a state function, which means it does not depend on the path on which a process is carried out. Instead, it depends on the initial and final state of a process and changes accordingly.

In the following cyclic ( 1 → 2 →3 →1 ) process the initial and final point is the same (i.e. 1). Hence the enthalpy change or ∆H= 0, in other words, there will be no change in enthalpy.

o The standard molar entropy is the measure the disorderliness or randomness in per mole of a substance.

o The standard molar entropy of H₂O (l ) is 70 J K^–1 mol^–1. The solid form of water is ice, in which position of molecules is quite fixed; therefore the disorderliness or randomness in solid molecules is less than the liquid form where the molecular orientation is more random.

Hence, entropy of H₂O(s) <entropy of H₂O (l ) . The standard molar entropy of H₂O(s) will also be less than 70 J K^–1 mol^–1 .

o The state functions do not depend on the path on which a process is carried out, but it is determined by the state of a system:

enthalpy, entropy, temperature and free energy.

o The path functions are the ones which do not depend on the initial and a final state of a system, rather depends on the path on which the process is carried out:

heat and work.

o Molar enthalpy of vapourisation of a liquid is the amount of heat required to change the state of 1 mole of the liquid to gas.

o Acetone does not consist of any hydrogen bond-like water and that is why intermolecular attractive forces in acetone molecules will be less than that of water molecules which makes it boil (evaporation also)faster.

o On the other hand, water having strong hydrogen bonds and the high polarity also adds up in resulting it to boil at higher temperatures. Hence water has a higher molar enthalpy than acetone.

o The relation between Δ_rG and Δ_rG^Θ (standard Gibb's free energy of reaction) of a reaction is as follows:

Δ_rG = Δ_rG^Θ + RT and, where, K is the equilibrium constant.

At equilibrium , Δ_rG becomes 0

And Δ_rG^Θ = - RT lnK .

o Δ_rG^Θ can only be 0 when the value of K is =1. So, for all the other K values Δ_rG^Θ cannot be zero.

o An isolated system means no form of energy to be it heat or work can be transferred in any direction.

Hence , q=0. W =0 .

And according to 1^st law of thermodynamics: ∆U= q + w (U=internal energy)

Therefore, change in internal energy for an isolated system ∆U = 0.

Heat is independent of the path under 2 conditions:

1. When the volume of the system is kept constant-

By 1st law of thermodynamics:
q = ΔU + (-w)
and -w = pΔV

Therefore , q = ΔU + pΔV
ΔV = 0 (at constant volume)

Hence, qv = ΔU + 0 = ΔU= change in internal energy

2. When the pressure of the system is kept constant –

At constant pressure, q_p= ΔU + pΔV

But , ΔU + pΔV = ∆H

Therefore , q_p = ∆H = change in enthalpy.

o Work done in vacuum is calculated by :

-w = T_ext (V_initial - V_final )

For the given problem: P_ext = 0, because it is a free expansion.

Hence –w = 0x(5-1) = 0.

o For an isothermal expansion, no heat will be absorbed or evolved, hence q=0

By the 1^st law of thermodynamics ,

q = ∆U + (-w)

since , both w and q =0 , then the change in internal energy ∆U will also be 0.

o Heat capacity (C_p) of a substance is the amount of heat energy required in order to raise its temperature by 1 Kor 1 ^oC.

o Whereas, The specific heat is the amount of heat per mole required to raise the temperature by one degree.

1 mole of water = 18 g.

Hence, for water Heat capacity = 18 × specific heat.

i.e. C_p = 18 × C .

o C_P is the heat capacity at constant pressure and C_V is the heat capacity at constant volume.

o For an ideal gas, the difference between these two is C_P - C_V = nR, the universal gas constant and where n= no. Of moles.

o Hence for 10 moles of an ideal gas C_P - C_V = 10 R

= 10 × 8.314 J

= 83.14 J

o The heat of combustion ∆H_c of graphite (i.e. carbon) is given as = 20.7 kJ for 1g of graphite (C).

1 mole of Carbon = 12 g

Hence the molar enthalpy change (enthalpy for 1 mole )

= (20.7 × 12 )= 248.4 KJ mol^-1

Since the heat is evolved, the actual molar enthalpy change

= -248.4 KJ mol^-1

o In combustion reactions, heat is always evolved i.e. it is an exothermic reaction. Hence the sign of ∆H for the reaction will always be negative (for the process to occur).

o The enthalpy change of the given reaction can be calculated from the relation: Σbond enthalpy of reactants- Σbond enthalpy of products

For the reaction

H₂(g) + Br₂(g) →2HBr(g)

Enthalpy change

= (Bond energy of H-H bond + Br-Br bond) – (2 × bond energy of H-Br)

= (435 + 192) kJ mol^–1 – (2 × 368) kJ mol^–1

= -109 kJ mol^–1

The enthalpy of vapourisation is given for 1 mole of CCl₄ = 30.5 kJ mol^–1

Hence, for 284 g, it will be = (mole no. × 30.5) kJ

Molar mass of CCl₄ = 154 g mol^–1 that means 154 g = 1 mole.

Therefore , 284 g = (284g/154gmol^-1)= 1.84 mole.

Hence,

the heat required for the vapourisation of 284 g of CCl₄ at constant pressure = (1.84 mol ×30.5 KJ mol^-1) kJ

=56.12 KJ

The standard enthalpy of formation of H₂O (l ) can be obtained by considering the following reaction :

As Δ_fH^Θ is related with one mole of a substance at standard conditions.

For the given reaction : 2H₂(g) + O₂(g) →2H₂O(l) the standard enthalpy of reaction is ∆H_r^Θ = – 572 kJ mol^–1 , so the half of ∆H_r^Θ will be the standard molar enthalpy of formation ;

Δ_fH^Θ = 1/2 × ∆H_r^{Θ =} (-572/2) = -286 kJ mol^–1 .

o Work done in thermodynamics is obtained at constant pressure by the relation W=p∆v, i.e. the product of the constant pressure and the change in volume.

o Now, if we plot P(y-axis) vs V (x-axis ) graph for the given process, then we get the following figure :

From this graph we can obtain the be the work done on the ideal gas enclosed in the cylinder in 1 step: the area covered by P-V graph (shaded region) is the actual value of the work done is:

= length × breadth = p_ext ∆V = AV_I (or BV_II ) × (V_I - V_II )

o When an ideal gas in a compression, where the change in pressure is carried out in infinite steps i.e. through a reversible process, the work done can be calculated only through the observation of pressure vs volume plot of the process.

.

The shaded area in the above graph represents the work done in the process, by calculating the area covered we can calculate the work done.

o For the process involved in (a) Throwing a stone from the ground to roof, as the potential energy will be increasing as the stone would cover greater distance from the ground, graphically it can be represented as:

o For the reaction : (b) 1/2 H₂(g) + 1/2 Cl₂(g) ⇋ HCl(g) whose Δ_rH^Θ = –92.32 kJ mol ^–1, as it is an exothermic reaction, heat is evolved and the ∆H value (enthalpy of reaction) decreases (eventually its is negative) throughout the process. The graphical representation will be :

o Amongst these two processes, in the process or reaction (b) the potential energy/enthalpy change is contributing factor to the spontaneity. As the spontaneous reactions are the one which occurs on their own, no external work is required. And the enthalpy is decreasing through the process going from reactant tom products is a sign of spontaneity as the driving force for a chemical reaction is within itself and it is the lowering of energy which is happening in (b) and not in (a) as there is some extra work required to start the process.

It is not possible to take a decision on whether the given reaction will be spontaneous or not.

o From the given enthalpy diagram it can be said the change in enthalpy ∆H is positive for the reaction, i.e. it will be endothermic in nature.

o But when comes to the spontaneity of a reaction, enthalpy is just a factor and there are other important factors like entropy, Gibb's free energy also taken into consideration.

Hence, enthalpy alone cannot determine the spontaneity of a reaction; one must have a loo to the other contributing factors also.

o It can be said from Fig. 6.4, that the process of the expansion of the monoatomic ideal gas has been carried out in infinite steps, hence it is obviously an isothermal reversible expansion.

o Hence, the work is done in the process =

W= - 2.303nRT log (V₂/V₁)

where V₁ is the initial volume in the state (1) and V₂ is the final volume in the state (2).

For an ideal gas p₁V₁ = p₂V₂

So (given in the graph)

W= - 2.303nRT log

= - 2.303 × 1 mol × 8.314 J mol^-1 K^-1 × 298 K × log2

=- 2.303 × 8.314 × 298 × 0.3010 J

= -1717.46 J

o For the first case, this is an irreversible expansion of the ideal gas occurs, hence the amount of work done = -p_ext ∆V

= - 2 bar × (50 – 10 ) L = - 80 L bar.

Now, it is given that 1 L bar = 100 J

So , -80 L bar = (-80 × 100) = -8000 J = -8 KJ is the amount of work done in the above process.

o If the same expansion were carried out reversibly, (i.e. through infinite steps) then the internal pressure of the gas will be infinitesimally larger than the external pressure and this will be the case in every stage of expansion. Therefore, the work done will own a higher value than the earlier case.

(i) Adiabatic process – (e) No transfer of heat

Explanation: The process in which there is no transfer of heat between system and surrounding is known as an adiabatic process. The barrier or wall separating the system is called an adiabatic wall.

(ii) Isolated system – (d) No exchange of energy and matter

Explanation: The system in which there is no exchange of energy and matter between system and surrounding is known as an isolated system. Example: Tea in a thermos flask.

(iii) Isothermal change – (f) constant temperature

Explanation: When there is a slow transfer of heat from the system to surrounding that the thermal equilibrium will get maintained is an isothermal change.

(iv) Path function – (a) Heat

Explanation: Path function transit between initial to final state and work and heat are two common path functions.

(v) State function - (g) Internal energy, (k) Entropy , (l) Pressure

Explanation: state function depends only on the initial and final state and not on the path of the system. It depends on measurable properties like pressure, volume, temperature, enthalpy, entropy, mass, density.

(vi) ΔU = q - (b) at constant volume

Explanation: FREE EXPANSION

When the expansion of the gas takes place in a vacuum, it is called free expansion. No work is done of an ideal gas during free expansion whether the process is reversible or irreversible.

we know that ∆U = q + w ---(1)

Also, W= -p∆V

Substitute the value of W in equation (1)

∆U = q - p∆V

If the process is carried out at constant volume(∆V = 0)

∆U = q_v

The v in q_v denotes the heat is supplied at constant volume.

(vii) Law of conservation of energy- First law of thermodynamics

Explanation: First law of thermodynamics state that energy of an isolated system is constant. It also based on conservation law where energy can neither be created nor be destroyed.

(viii) Reversible process – (j) infinitely slow process which proceeds through series of equilibrium.

Explanation: Reversible process is the process which takes infinite time by doing a series of steps so that system and surrounding remain in equilibrium and if it brought back to initial state there is no effect on surrounding.

(ix) Free Expansion – (h) p_ext = 0

In Vaccum expansion of gas ( free expansion) p_ext = 0 . In an ideal gas, whether the process is reversible or irreversible, work done during free expansion is zero

w_irreversible= -p∆V

= 0×∆V

w_irreversible = 0

∆H = q – (i) At constant pressure

Explanation: From the first law of thermodynamics

∆U = q + W --- (1)

∆U= internal energy

q= heat

W= work done

We know W = -p∆V ---(2)

Therefore we can write eq-1 as

∆U = q - p∆V --- (3)

→ q is the heat absorb at constant pressure

Now if the system absorb q (heat), internal energy changes from U₁ to U₂ and volume increase from V₁ to V₂

∆U = U₂– U₁ –-- (4)

∆V = V₂ - V₁ --- (5)

Put equation (4) and (5) in equation (3)

U₂ - U₁= q - p (V₂ - V₁)

q_p = U₂ - U₁+ p (V₂ - V₁) --- (6)

U, P, V are state functions and U + PV is heat content or enthalpy of the system which is denoted by

H = U + PV –-- (7)

So, q = H₂ – H₁

q = ∆H

This is the heat absorbed at constant pressure.

(xi) Intensive property – (a) heat, (l) pressure, (m) specific heat

Explanation: those properties which depend on the nature of the substance and not the quantity of the system are known as an intensive property.

(xii) Extensive property – (g) internal energy, (k) entropy

Explanation: Those properties which depend on the quantity of the system are known as extensive property.

(i) A liquid vaporize – (b) ΔS = positive

Explanation: When the liquid vaporizes, the degree of the disorder increases, therefore, entropy is always positive.

(ii)The reaction is nonspontaneous at all temperatures and ΔH is positive – (c) ΔS = negative ideal gas

Explanation: When the reaction is non-spontaneous

∆S_universe˂ 0

If a reaction has positive entropy and enthalpy change, the process can be spontaneous when T∆S is more enough to overweigh ∆H.

a. Entropy change in the system can be small in which Temperature must be large.

b. Entropy change in the system can be large in which temperature must be small

Because of the above reasons, reactions are often carried out at high temperature.

(iii) Reversible expansion of an - (a) ΔS = 0

Explanation: When the system is at equilibrium or reversible, the entropy remains constant

ie why ΔS = 0

(i) + – + :(c) Reaction is Non spontaneous at all temperatures

(ii) – – + at high T :(a) Reaction is non-spontaneous at high

temperature

(iii) – + – :(b) Reaction is spontaneous at all temperature

Explanation: ∆_rG^- = ∆_rH^- - T∆_rS^-

∆G gives the criteria for spontaneity at constant temperature and pressure.

1. If ∆G is positive the process is non-spontaneous.

2. If ∆G is negative the process is spontaneous.

If a reaction has positive entropy and enthalpy change, the process can be spontaneous when T∆S is more enough to overweigh ∆H.

c. Entropy change in the system can be small in which Temperature must be large.

d. Entropy change in the system can be large in which temperature must be small

Because of the above reasons, reactions are often carried out at high temperature.

Reversible process in which the system is in perfect equilibrium with the surroundings. In chemical terms, reversible means, the reaction can proceed in any direction with a decrease in free energy, which is impossible. It is only possible when the energy of the system is minimum.

A+ B C + D

∆_rG = 0

Gibbs free reaction in which all the reactant and product are in the standard state.

∆_rG^‑= -RT ln K

∆_rG^- = -2.303RT log K

Also,

∆_rG^- = ∆_rH^- - T∆_rS^- = -RT ln K

∆_rH^- is large and positive for strong endothermic reaction and the value of K is smaller than 1.

∆_rH^- is large and negative for exothermic reaction and the value of K is larger than 1. ∆_rG^- is also negative and large.

To get a large value of K we need a strongly exothermic reaction, hence we go for completion of the reaction.

∆_rG^- depend on ∆_rS^- and if entropy changes then the value of K also get affected, which depend on whether ∆_rS^- is positive or negative.

(i)Entropy of vapourisation – (b)is always positive

Explanation: When the liquid vaporizes, the degree of the disorder increases, therefore, entropy is always positive.

(ii) K for the spontaneous process- (d) the ideal gas

Explanation: K is equilibrium constant and is always positive.∆_rH^- is large and positive for strong endothermic reaction and the value of K is smaller than 1.

∆_rH^- is large and negative for exothermic reaction and the value of K is larger than 1. ∆_rG^- is also negative and large.

To get a large value of K we need a strongly exothermic reaction, hence we go for completion of the reaction.

∆_rG^- depend on ∆_rS^- and if entropy changes then the value of K also get affected, which depend on whether ∆_rS^- is positive or negative.

(iii) Crystalline solid-state : (c) lowest entropy

Explanation: Crystalline solid state is the most ordered form so it has the lowest entropy and gaseous state is disordered state so highest entropy.

(iv) ΔU in adiabatic expansion – (a) decreases

Explanation, as there is no exchange of heat between system and surrounding, internal energy will decrease after applying some amount of pressure.

(ii) Both A and R are true but R is not the correct explanation of A.

Explanation: Combustion reaction are exothermic and are important in industries, rocketry etc. Standard enthalpy of combustion is defined as enthalpy change per mole of substance when it undergoes combustion and all reactant and product are in standard state at specific temperature.

Cooking gas in cylinders contain BUTANE (C₄H₁₀).

C₄H₁₀(g)+ O₂(g) → 4CO₂(g) + 5H₂O

Combustion of glucose

C₆H₁₂O₆(g) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

The change in enthalpy is independent of initial and final state and enthalpy change of a reaction is same whether it occurs in 1 step or series of steps.

(ii) Both A and R are true but R is not the correct explanation of A

Explanation: Spontaneous process is an irreversible process and may be reversed by some external agency like pressure, temperature etc because spontaneous process deals with the degree of increase in randomness and decrease in energy.

(i) Both A and R are true and R is the correct explanation of A

Explanation: when the irregularity occurs in a structure entropy increases, therefore when liquid crystallises into solid entropy decreases because, in crystals, molecule organised systematically.

From the first law of thermodynamics

∆U = q + W --- (1)

∆U= internal energy

q= heat absorbed

W= work done

We know W = -p∆V --- (2)

Therefore we can write equation-1 as

∆U = q - p∆V--- (3)

Now if the system absorb q_pinternal energy changes from U₁ to U₂ and volume increase from V₁ to V₂

∆U = U₂ - U₁---(4)

∆V = V₂ - V₁ --- (5)

Substitute the value of ∆U and ∆V from equation-4 and equation-5 in equation-3 ie.

∆U = q_p - p∆V

U₂ - U₁= q_p - p (V₂ - V₁)

q_p = U₂ - U₁+ p (V₂ - V₁)

U, P, V are state functions and U + PV is heat content or enthalpy of the system which is denoted by

H = U + PV

So q_p = H₂ – H₁

q_p = ∆H

Therefore,∆H = ∆U + p∆V--- (6)

At constant temperature when the heat is absorbed, we measure the changes in enthalpy

∆H = q_p (when the heat is absorbed by the system at constant pressure)

∆H is positive for an endothermic reaction.

∆H is negative for an exothermic reaction.

At constant volume (∆V = 0)

∆U = q_v

Therefore,

∆H = ∆U = q_v

Now using ideal gas law if V_A is the total volume of gas reactant and V_B is the volume of gas product and n_A and n_B are moles of gas reactant and product respectively.

pV_A = n_ART

pV_B = n_BRT

Thus, pV_B- pV_A = n_BRT - n_ART

p(V_B – V_A) = RT (n_B- n_A)

p∆V = ∆n_gRT --- (7)

Put the value of p∆V from equation-7 to equation-5

Therefore,

∆H = ∆U + ∆n_gRT

Extensive property: Those properties which depend on the quantity of the system are known as extensive property.

Mass, internal energy, heat capacity

Intensive property: those properties which depend on the nature of the substance and not the quantity of the system are known as an intensive property

Pressure, molar heat capacity, density, mole fraction, specific heat, temperature, molarity

NOTE:

Mole fraction and molarity are intensive property because:

= intensive property

That means when two extensive quantity divide by each other, they form an intensive property

Example : = = intensive

Na(s) + 1/2 Cl(g) → NaCl(s) ∆_fH

BORN-HABER CYCLE is an enthalpy diagram which is an indirect method of calculating lattice enthalpy.

One mole of NaCl(s) is prepared by the following steps:

FIRST STEP: Sublimation of Na (sodium) (∆_subH)

• Na(s) → Na(g)

SECOND STEP: Ionization of sodium (∆_iH)

• Na(g) → Na⁺ + 1e^-

THIRD STEP: Dissociation of chlorine molecule to a chlorine atom (∆_aH)

• 1/2 Cl₂(g) → Cl(g)

FOURTH STEP: Chlorine atom gains 1e^- (∆_egH)

• Cl(g) + 1e^-→ Cl^-

Na⁺(g) + Cl^- (s) → NaCl(s) ∆_latticeH

The sequence of the steps is known as BORN-HABER CYCLE. The importance of this cycle is, the sum of enthalpy changes round a cycle is zero.

Applying Hess’sLaw

Hess’s law state that if the reaction takes place in many steps, then the standard enthalpy of the reaction is the sum of the standard enthalpies of intermediate reaction, into which overall is divided at the same temperature.

∆_fH = ∆_subH + ∆_iH + ∆_aH + ∆_egH + ∆_latticeH

∆_latticeH = ∆_subH + ∆_iH + ∆_aH + ∆_egH - ∆_fH

Gibbs energy

G = H - TS

H = Enthalpy of the system

T = Temperature

S = Entropy of the system

We know

H = U + PV

Therefore G = U + PV – TS

G is a state function and an extensive property. The change in Gibbs energy for the system can be

∆G = ∆U + ∆(PV) - ∆(TS)

If the process is done at constant temperature and pressure then

∆G_sys = ∆U_sys + P∆V - T∆S_sys

∆G_sys = ∆H - T∆S_sys

∆G has the unit of energy because both ∆H and T∆S_sys are terms of energy

Unit of ∆G is joules.

∆G is the energy available to do useful work and also known Gibbs free energy

∆G gives the criteria for spontaneity at constant pressure and temperature

If ∆G is positive then non-spontaneous

If ∆G is negative then spontaneous.

Consider a cylinder having 1-mole gas with no weight and no friction having area of cross-section A. Total volume is V_i and the initial pressure is p.

Let p_ext is the external pressure and if p_ext> p, piston move down until p_ext = p, Now the final volume is V_f. The distance moved by piston be ∆l

Therefore ∆V = ∆l × A (eq-1)

∆V = V_f - V_i

Force = pressure ×area

Therefore

F = p_ext× A (eq-2)

If w is work done on the system

W = force × displacement

= p_ext × A ×∆l

From eq-1

W = p_ext×(-∆V)

W = -p_ext∆V

W = -p_ext( v_f – v_i)

If v_f> v_i work is done by the system and w is negative

If v_f< v_i work is on the system and w is positive