Explore topic-wise InterviewSolutions in .

Answer: (d) Heat

Lattice energy is defined as the amount of energy required to completely separate one mole of a solid ionic compound into gaseous constituent.

• The variables like P. V, T and ‘n’ that are used to describe the state of the system are called as state functions. 

For example, pressure, volume, temperature, internal energy, enthalpy and free energy

• A path function is a thermodynamic property of the system whose value depends on the path or manner by which the system goes from its initial to final state.

For example, Work (w) and heat (q).

Enthalpy is a thermodynamic property of a system. Enthalpy (H) is defined as sum of the internal energy (U) of a system and the product of pressure and volume of the system. 

H = U + PV 

Unit of enthalpy: KJ mol^-1.

Kelvin-Planck statement: It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.

The standard enthalpy change for the formation of one mole of a compound from its elements in their standard states is called standard molar enthalpy of formation.

The standard heat of formation of a compound is defined as “The change in enthalpy that takes place when one mole of a compound is formed from its elements, all substances being in their standard states (298 K and 1 atm pressure).

Cm = q / (T₂ - T₁)

Where C specific heat capacity, 

q = amount otheat absorbed. m = mass and T₁ ,T₂ temperatures.

Clausius statement: Heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to spend some work.

• Thermodynamics first law tells that there is an exact equivalence between various forms of energy and that heat gained is equal to heat loss.
• Practically it is not possible to convert the heat energy into an equivalent amount of work. To explain this, another law is needed which is known as second law of thermodynamics.
• The second law of thermodynamics helps us to predict whether the reaction is feasible or not and also tell the direction of the flow of heat.
• It also tells that energy cannot be completely converted into equivalent work.

Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the Universe.

∆S_universe > ∆S_system + ∆S_surroundings

Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work. 

G = H - TS, 

where, G = Gibb’s free energy 

H = enthalpy 

T = temperature 

S = entropy

The equation representing the enthalpy of formation of methane is:

C_(S) + 2H_2(g) – CH_2(g) ; 

∆H = – 78.84 kJ 

∆U = 78.84 kJ; 

∆^ng = 1 – 2 = -1 mol 

R = 8.314 x 10^-13 KJ K^-1 mol^-1 ; T = 300 K.

According to the relation,

∆H = ∆U + ∆^ng RT – 78.84 kJ 

∆U = ∆H – ∆^ng RT 

∆U = (- 78.84 kJ) – (1 mol) x (8.314 x 10^-3 KJ K^-1 mol^-1) x 300K. 

= – 78.84 – 2.49 = -8.314 KJ.

Characteristics of enthalpy:

1. Enthalpy is a thermodynamic property of a system. Enthalpy H is defined as sum of the internal energy (U) of a system and the product of pressure and volume of the system. 

That is, H = U + PV

2. Enthalpy is a state function which depends entirely on the state functions T, P and U. 

3. Enthalpy is usually expressed as the change in enthalpy (∆H) for a process between initial and final states. 

∆H = ∆U + P∆V

4. At constant pressure, the heat flow (q) for the process is equal to the change in enthalpy which is defined by the equation.

∆H = q_p

5. In an endothermic reaction heat is absorbed by the system from the surroundings that is q > 0 (positive).

Therefore, at constant Tand P, by the equation above, if q is positive then ∆H is also positive. 

6. In an exothermic reaction heat is evolved by the system to the surroundings that is, q < 0 (negative). If q is negative, then ∆H will also be negative. 

7. Unit of enthalpy is KJ mol^-1.

1. Hess’s law can be applied to calculate the enthalpy of formation of carbon monoxide. It is very difficult to control the oxidation of graphite to give pure CO.

However, enthalpy for the oxidation of graphite to CO₂ can be easily measured and enthalpy of oxidation of CO to CO₂ is also measurable.

2. The application of Hess’s law enables us to estimate the enthalpy of formation of CO.

C + O₂ → CO₂∆H° = 393.5 kJ ….. (1)

CO + 1/2 O₂ → CO₂ ∆H° = -283 kJ  ….. (2)

On inverting equation (2), we get

CO₂ → CO + 1/2 O₂ ∆H° = + 283 kJ …… (3)

On adding equations (2) and (3), we get

C + 1/2 O₂ → CO ∆H° = 393 5 + 283 = 110.5kJ

• The property that is depend on the mass or size of the system is called an extensive property.
• Example, Volume, number of moles, mass and internal energy.

• The calorific value of food is defined as the amount of heat produced in calories (or Joules) when one gram of food is completely burnt.
• Unit of calorific value (a) Cal g^-1 (b) J Kg^-1

Enthalpy of combustion of a substance is defined as “the change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. 

It is denoted as ∆H.

Molar heat capacity is defined as “the amount of heat absorbed by one mole of a substance in raising the temperature by 1 Kelvin”. 

It is denoted as C_m 

Unit of Molar heat capacity: SI unit of C_m is JK^-1 mol^-1.

The molar heat of vapourisation is defined as the change in enthalpy when one mole of liquid is converted into vapour or gaseous state at its boiling point.

q and W are not state functions because they depend upon path therefore they are path equations. ΔE = q + W, ΔE is a state function because it depends upon initial and final state of the system but not on the path.

(a) Increase in entropy

Characteristics of entropy:

1. Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.

2. In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.

3. Entropy is defined as for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).

ΔS_sys = q_rev / T

4. If heat is absorbed, then AS is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy.

5. The change in entropy of a process represented by AS and is given by the equation, ∆H_sys = S_f – S_i

6. If S_f > S_i , ∆S is positive, the reaction is spontaneous and reversible. If S_f < S_i , ∆S is negative, the reaction is non-spontaneous and irreversible.

7. Unit of entropy: SI unit of entropy is J K‘.

When one mole of a solid changes reversibly from one allotropic form to another at its transition temperature. The entropy change is given

∆S_t = ∆H_t / T_t

Where ∆H_t = molar heat of transition and T_t = transition temperature.

When one mole of liquid is boiled at its boiling point reversibly, the heat absorbed is called as molar heat of vaporization. The entropy change is given by

∆S_V = ∆H_v / T_b

where ∆H_v is molar heat of vapourisation. T_b is boiling point.

The enthalpy is defined by H = E + pV for a change in the states of system

ΔH = Δ (E + pV)

= ΔE + Δ (pV)

= ΔE + pΔV + VΔp    ....(i)

1st law of thermodynamics states that

ΔE = q + W

= q - pΔV    ....(ii)

From equations (i) and (ii)

ΔH = q - pΔV + pΔV + VΔp

= q + VΔp

When the pressure is constant, Δp = 0, and thus VΔp = 0, therefore

ΔH = q (at constant pressure)

or, ΔH = q_P

For the reaction

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l)

ΔH° = ΔH°_f (CO₂,g) + 2ΔH°_f (H₂O, 1) - [ΔH°_f(CH₄, g) + 2ΔH°_f(O₂)]

ΔH°_f (CO₂, g) + 2ΔH°_f (H₂O, 1) - ΔH°_f (CH₄, g)

Given,

S°(urea) = 173.8 J mol^-1 K^-1

S° (H₂O) = 70 J mol^-1K^-1

S° (CO₂) = 213.5 J mol^-1 K^-1 

S° (NH₃) = 192.5 J mol^-1 K^-1 

NH₂ – CO – NH₂ + H₂O → 2NH₃ + CO₂ 

∆S_r° = ∑(S°)_product – ∑(S°)_reactants 

∆S_r° = [2 S°(NH₃) + S°(CO₂)] — [S°(urea) + S°(H₂O)] 

∆S_r° = [2 x 192.5 + 213.5] – [173.8 + 70] 

∆S_r° = [598.5] – [243.8] 

∆S_r° = 354.7 J mol^-1 K^-1

Standard entropy of formation is defined as “the entropy of formation of 1 mole of a compound from the elements under standard conditions”. It is denoted as ∆S_f°. We can calculate the value of entropy of a given compound from the values of S° of elements.

∆S_f° = ∑∆_Sproducts° – ∑∆S_reactants°

In isothermal process temperature remains constant by exchanging heat with the surroundings. In adiabatic process, no exchange of energy take place with the surroundings.

Isolated system can neither exchange matter nor energy with the surroundings. Closed system can exchange energy but not matter with the surroundings.

The relation between C_P - C_v = R

(i) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

ΔH = -1380.7 kJ

(ii) C + O₂ → CO₂, ΔH = -394.5 kJ

(iii) H₂ + \(\frac{1}{2}\)O₂ → H₂O, ΔH = -286.6 kJ

We aim at 2C + 2H₂ + \(\frac{1}{2}\)O₂ → C₂H₅OH

In order to get this thermochemical equation, multiply Eq. (ii) by 2 and Eq. (iii) by 3 and subtract Eq. (i) from their sum

2C + 3H₂ + \(\frac{1}{2}\)O₂ → C₂H₅OH

ΔH = 2(-394.5) + 3(-286.6) - (-1380.7)

= -268.1 kJ

Thus the heat of formation of ethyl alcohol is

ΔH_f = -268.1 kJ mol^-1

When 1 mole of PCl₅ is formed from its elements, P(s) and Cl₂(g), 373 kJ of heat is liberated.

n = 1 - 270/300 = 1/10

Efficiency of refrigerator = 0.5n = 1/20

If Q is the heat/s transferred at higher temperture then W/Q = 1/20

or Q = 20W = 20kJ,

and heat removed from lower temperture = 19 kJ.