Thermodynamics

Thermodynamic state functions are functions that depend only on the initial and final state of the system. It is independent of the path taken to change from one state to another.

Thus, answer is (ii) whose value is independent of path

Adiabatic process is processes that take place without any transfer of heat between the system and surrounding. i.e, in adiabatic process q = 0

Thus answer is (iii) q = 0

[Option (i) is an isothermal process. In the isothermal process, the temperature of the system remains constant.

Option (ii) is for isobaric process; where pressure of system is kept constant.]

By definition, the enthalpy of formation of elements in their standard state is taken as zero. Therefore, The enthalpies of all elements in their standard states is zero irrespective of the element.

Thus, right answer is (ii) zero

Given,

We need to find enthalpy change for reaction

ΔH₅ = ΔH(CH₄)

ΔH₁ = ΔH(CO₂) + 2ΔH(H₂O) – ΔH(CH₄)

ΔH₂ = ΔH(CO₂)

ΔH₄ = 2ΔH(H₂O)

Thus,

ΔH(CH₄) = ΔH₂ + ΔH₄ – ΔH₁
So, ΔH= (-393.5-2×285.8+ 890.3) KJ/mol
_{= -74.8 KJ/mol}

Thus, answer is (i) –74.8 kJ mol^–1

A reaction is said to be feasible of possible when change in Gibbs free energy for that reaction at particular temperature is negative.

We know, change in Gibbs free energy,

ΔG = ΔH – TΔS

Where

ΔH is change in enthalpy

ΔS is change in entropy

T is temperature at which reaction occurs.

Given Energy is released when a forward reaction occurs (since q is on product side), therefore ΔH is negative.

It is also given that entropy change for the reaction is positive.

Thus –TΔS would be negative for all temperatures.

Therefore,

ΔG = ΔH – TΔS would always be negative.

Thus (iv) possible at any temperature is the right answer.

We know that,

ΔU = ΔH + W

Where ΔU is change in internal energy

Δ H is change in enthalpy

And W is work done on the system.

Given, 701 J of heat is taken by the system

Thus ΔH = 701 J

And 394 J of work is done by the system

Thus W = - 394

∴ ΔU = 701 – 394

= 307 J

We know that,

ΔU = ΔH - Δn_gRT

The balanced chemical equation of reaction is,

NH₂CN(s) + 3/2O₂(g) ⇌ N₂(g) + CO₂(g) + H₂O(l)

Number of moles of gaseous particles in product side is 2 (one N₂ and one CO₂)

Number of moles of gaseous particles on reactant side is 3/2 (O₂)

Thus, change in number of moles of gaseous particle,

Δn_g = 2 – 3/2 = 1/2

∴ ΔU = ΔH - Δn_gRT

Given,

ΔU = -742.7 kJ

R =8.314 J K^-1 mol^-1

= 8.414 × 10^-3 kJ K^-1 mol^-1

Δn_g =(2-2.5) =-0.5 mole

T = 298 K

Thus,

ΔH = (-742.7 kJ mol^-1) – (1/2mol)×(8.314 × 10^-3 kJ K^-1mol^-1)×298

= -742.7 - 1.239

= -743.9 kJ

We Know that,

Q = ncΔT

Where n is number of mole of substance

c is molar heat capacity of substance

ΔT is rise in temperature

Given,

c = 24 J mol^-1 K^-1

ΔT = 55 – 35 = 20°C = 20 K

m = 60 g

number of mole of Aluminium in m gram is given as

n = m/M

Where M is molar mass of aluminium.

∴ n = 60/27 = 2.22

Thus Q = ncΔT

= (2.22 mol)×(24 J mol^–1 K^–1)×(20K)

= 1065.6 J

= 1.065 kJ

Given: Δ_fusH = 6.03 kJ mol^-1 at 0^° C

C_p [H₂O(l)] = 75.3 J mol^–1 K^–1

C_p [H₂O(s)] = 36.8 J mol^–1 K^–1

To calculate the enthalpy change, we use Hess’s Law

Total ΔH = ΔH₁ + ΔH₂ + ΔH₃

Where ΔH = C_pΔT

ΔC_p = Heat capacity at constant pressure

ΔT = T₂-T₁

∴ Total ΔH = (1 mol water at 10^° C →1 mol water at 0^° C) +

(1 mol water at 0^° C → 1 mol water at 0^° C) +

(1 mol water at 0^° C → 1 mol water at -10^° C)

∴ Total ΔH = C_p[H₂O (l)] × ΔT + ΔH_freezing + C_p [H₂O (s)] × ΔT

By substituting the given values, we get

Total ΔH = 75.3 J mol^–1 K^–1 × (0-10) K +(-6.03 kJ mol^-1) +36.8 J mol^–1 K^–1× (-10) K

⇒Total ΔH = -753 Jmol^–1 -6.03 kJ mol^-1 -368 Jmol^-1

⇒Total ΔH = -0.753 kJmol^–1 - 6.03 kJ mol^-1 - 0.368 kJmol^-1 (1J=)

⇒Total ΔH = -7.151 kJmol^-1

Thus, the enthalpy change involved in the process is -7.151 kJ/mol

Note: Hess’s Law states that “ if a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e., the total enthalpy change is independent of intermediate steps involved in the change. The enthalpy depends on the initial and final stages only.

Given: Enthalpy of combustion of carbon (ΔH) = –393.5 kJ mol^–1.

Mass of CO₂ = 35.2g

Molar mass of CO₂ = 12 + 2× 16

= 44 g/mol

∴ 1 mol of carbon = 44g

Heat released when 44g CO₂ is formed = 393.5 kJ

Heat released when 35.2g CO₂ is formed = × 35.2g

⇒314.8 kJ

Thus, heat released when 35.2g CO₂ formed is 314.8 kJ

Given: For the given reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Enthalpy of formation (ΔH_{f (CO)}) = -110 kJ mol^-1

ΔH_{f (CO2)} = -393 kJ mol^-1

ΔH_{f (N2O)} = 81kJ mol^-1

ΔH_{f (N2O4)} = 9.7 kJ mol^–1

Now, to calculate Δ_rH for the reaction, we apply the formula given below:

Δ_rH = (Sum of enthalpy of formation of products) – (sum of enthalpy of reactants)

or, Δ_rH = ∑ Δ_fH (Products) - ∑ Δ_fH (Reactants)

In the given reaction,

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Products are N₂O and CO₂

Reactants are N₂O₄ and CO

∴ Δ_rH = [Δ_fH (N₂O) + 3Δ_fH (CO₂)] - [Δ_fH(N₂O₄) + 3Δ_fH (CO)]

By substituting the values given, we get

⇒Δ_rH = [81 kJ/mol + 3 × (-393 kJ/mol)] – [9.7 kJ/mol + 3 × (-110 kJ/mol)]

⇒Δ_rH = -777.7 kJ/mol

Thus, the value of Δ_rH for the reaction is -777.7 kJ/mol.

Note: Enthalpy of reaction (Δ_rH) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted.

Given: For the given reaction,

N₂(g) + 3H₂(g) → 2NH₃(g)

Enthalpy of reaction (Δ_rH^°) = –92.4 kJ mol^–1

To calculate standard enthalpy of formation of NH₃ gas, first we will divide the whole reaction by 2, so that we get 1 mol of NH₃

∴ N₂ (g) + H₂(g) → NH₃(g)

Hence, standard enthalpy of NH₃ is equal to the 1/2 Δ_rH^°

As, Δ_rH^° = –92.4 kJ mol^–1

∴ Standard enthalpy of NH₃ =

⇒Standard enthalpy of NH₃ = -46.2 kJ mol^-1

Thus, standard enthalpy of formation of NH₃ gas = -46.2 kJ mol^-1

Note: Standard enthalpy of formation is the amount of heat absorbed or evolved when 1 mole of the substance is directly obtained from its constituent elements.

Given:

CH₃OH (I) + O₂(g) → CO₂(g) + 2H₂O(l); Δ_rH^° = –726 kJ mol^–1(1)

C(graphite) + O₂(g) → CO₂(g); Δ_cH^° = –393 kJ mol^–1 (2)

H₂(g) +O₂(g) → H₂O(l); Δ_fH^°= –286 kJ mol^–1 (3)

To calculate the standard enthalpy of formation of CH₃OH(l), first we will make the formation reaction of CH₃OH by using carbon, hydrogen, oxygen.

C (s) + 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

This is the required reaction

Now, we make the required reaction from the given data.

Step 1: Add the reaction (2) and reaction (3)

C(graphite) + O₂(g) → CO₂(g); Δ_cH^° = –393 kJ mol^–1

2H₂(g) +O₂(g) → 2H₂O(l); Δ_fH^°= –572 kJ mol^–1. [2× reaction (2)]

C + 2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); Δ_fH^° = -965 kJ mol^-1

Step 2: Subtract reaction (1) from the above reaction, we get

C + 2H₂(g) +2O₂(g) → CO₂(g) + 2H₂O(l); Δ_fH^° = -965 kJ mol^-1

CO₂(g) + 2H₂O(l) → CH₃OH (l) + O₂(g); Δ_rH^°= –726 kJ mol^–1

C (s) + 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l); Δ_rH^°= –239 kJ mol^–1

(formation of required reaction)

Thus, the standard enthalpy of formation of CH₃OH(l) is –239 kJ mol^–1

Given:

Δ_vapH^° (CCl₄) = 30.5 kJ mol^–1

Δ_fH^° (CCl₄) = -135.5 kJ mol^–1

Δ_aH^° (C) = 715.0 kJ mol^–1

Δ_aH^° (Cl₂) =242 kJ mol^–1

We can write,

CCl₄(l) → CCl₄(g); ΔH = 30.5 kJ mol^–1 (1)

C(s) + 2Cl₂(g) → CCl₄(l); ΔH = -135.5 kJ mol^–1 (2)

C(s) → C(g); ΔH = 715.0 kJ mol^–1 (3)

Cl₂(g) → 2Cl(g); ΔH =242 kJ mol^–1 (4)

To get the required reaction, CCl₄(g) → C(g) + 4 Cl(g)

We follow the steps given below:

Step 1: Add t the both reaction (1) and (2), we get

C(s) + 2Cl₂(g) → CCl₄(l); ΔH = -135.5 kJ mol^–1

CCl₄(l) → CCl₄(g); ΔH = 30.5 kJ mol^–1

C(s) + 2Cl₂(g) → CCl₄(g); ΔH = -105 kJ mol^–1 (5)

Step 2: Add the both reaction (3) and (4), we get

C(s) → C(g); ΔH = 715.0 kJ mol^–1

2Cl₂ (g) → 4Cl (g); ΔH = 484 kJ mol^–1 [ 2×reaction 4]

C(s) + 2Cl₂(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol^–1 (6)

Step 3: Subtract the both reaction (5) and (6), we get

C(s) + 2Cl₂(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol^–1

CCl₄(g) → C(s) + 2Cl₂(g); ΔH = -105 kJ mol^–1

CCl₄(g) → C(g) + 4 Cl(g); ΔH = 1304 kJ mol^–1

(formation of required reaction)

Thus, the enthalpy change for the process is 1304 kJ mol^–1

⇒Calculation the bond enthalpy of C-Cl in CCl₄(g)

Bond enthalpy of C-Cl bond in CCl₄ is equal to one fourth of the energy of dissociation of CCl₄

∴ Bond enthalpy of C – Cl in CCl₄(g) =

As ΔH =1304 kJ mol^–1

= 326 kJ mol^-1

Note: Bond enthalpy is defined as the amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 am pressure and the specified temperature is called bond enthalpy.

An isolated system is the one in which neither energy nor matter can enter/leave the system.

∆S is the change in entropy. Entropy is a measure of randomness or disorder in a system. More the randomness, more is the entropy of a system.

According to the second law of thermodynamics the total entropy can never decrease over time for an isolated system.

The total entropy is constant in ideal cases where the system is in equilibrium, or undergoing a reversible process and approaches 0 at absolute zero only.

In all spontaneous processes, the total entropy always increases and the process is irreversible.

Therefore, ∆S is always greater than 0, ∆S>0.

According to Gibbs Helmholtz equation

∆G = ∆H - T∆S, where

∆G = change in Gibbs energy

∆H = change in enthalpy, given as 400 kJ mol^-1

∆S = change in entropy, given as 0.2 kJ K^-1 mol^-1

T = temperature at which reaction occurs

For a reaction to be spontaneous, ∆G < 0, means it must be negative.

If ∆G = 0, it is in equilibrium.

So, for reaction to be spontaneous, ∆G < 0; ∆H - T∆S < 0; ∆H < T∆S;

(∆H / ∆S) < T; 400 / 0.2 = 2000 K < T

This implies that the temperature must be more than 2000K for the reaction to be spontaneous.

In the reaction, chlorine molecule is being formed from chlorine atoms. Hence, bond formation takes place, leading to release of energy. As it is exothermic reaction, ∆H is negative (-ve).

In the reaction, 2 moles of atoms having higher randomness combine to form 1 mole of molecules, having lesser randomness.

Therefore, ∆S is negative (-ve ), which is represented as S_products – S_reactants.

∆U = -10.5 kJ

Also, ∆U = ∆H – ∆nRT

Where,

∆U = change in energy

∆H = change in enthalpy

∆n = moles of products – moles of reactants = 2-3 = -1

R = 8.314 × 10^-3 kJ mol^-1 K^-1 - constant

T = temperature; here room temperature is taken as 298 K

Substituting all the above values, we get

∆H = -10.5kJ + (-1 mol × (8.314× 10^-3kJ mol^-1 K^-1) × 298 K

= -12.977kJ

we know that

∆G = ∆H - T∆S, where

∆G = change in Gibbs energy

∆H = change in enthalpy, got as -12.977 kJ

∆S = change in entropy, given as -44.1 JK^-1 = -44.1 × 10^-3 kJK^-1

T = temperature at which reaction occurs = 298K

Substituting the above values

∆G = -12.977 – (298 × -44.1 × 10^-3)

= -12.977 – ( -13.14)

= 0.163kJ

As ∆G > 0, reaction doesn’t occur spontaneously. Reactions are spontaneous only when ∆G < 0.

Using,

∆G = -RTlnK

⇒ ∆G = -2.303RTlogK

∆G = change in Gibbs energy

R = 8.314 × 10^-3 kJ mol^-1 K^-1 - constant

T = temperature; given as 300K

K = equilibrium constant

Substituting the above values

∆G = -2.303 × 8.314 × 10^-3 × 300 × log 10

∆G = -5.744 kJ mol^-1

The first reaction is endothermic. It consumes or absorbs energy to form products. So, NO has more energy than its reactants. Hence, NO is unstable.

In the second reaction, it reacts with oxygen to form NO₂, and gives off energy. It’s an exothermic reaction. So, NO₂ has less energy than its reactants, and is stable. It is stabilised with minimum energy.

Hence unstable NO gets converted into stable NO₂.

It is given that an energy of 286 kJ is released per mole, when water is formed under standard conditions. This means 286 kJ of energy is absorbed by the surroundings in this process. So,

q = - ∆H = 286 kJ mol^-1

∆S = q/T

⇒ ∆S = 286 kJ/ 298 K

⇒ ∆S = 959.73 J mol^-1K^-1

T = Temperature = 298K