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(a) ∆S =+ve 

(b) ∆S=-ve

Entropy is defined as a measure of degree of randomness or disorderness in an isolated system.

It is related to heat absorbed at a constant temperature of the system in a reversible process as

S = \(\frac{q_{rev}}{T}\)

For reaction, CaCO₃ (s) → CaO(s) + CO₂ (g)

The sing of ΔS is negative so it is not spontaneous.

The equation for the formation of one mole of N₂O will be 

 NΞN (g) + ½ O=O (g) ---> N=N=O 

Calculated value of Δ_fH⁰ for this reaction will be Δ_fH⁰ = [B.E.(NΞN) + ½ B.E.(O=O)] – [B.E.(N=N) + B.E.(N=O)] 

 =[946 + ½ (498)] – [418 + 607] kj mol-1 = 170kj mol^-1 

Resonance energy = Observed Δ_fH⁰ – Calculated Δ_fH⁰ = 82 – 170 = - 88 kj mol^-1 . 

C(s) +1/2 O₂(g) → CO (g)

Option : (a) P₁V₁ = P₂V₂

Given that

(i) CH₄ (g) + 2O₂ (g) → CO₂ (g) + CH₂O(g);∆_CH^Θ = −890.3 kJ mol⁻¹

(ii) C(s) + O₂ (g) → CO₂ (g);∆_CH^Θ = −393.5 kJ mol⁻¹

(iii) H₂ (g) + \(\frac{1}{2}\)O₂ (g) → H₂O(g); ∆_CH^Θ = −285.8 kJ mol⁻¹

We have to calculate ∆_fH^Θ of the equation:

C(s) + 2H₂ (g) → CH₄ (g); ∆_fH^Θ = ?

Appling inspection method,

−eq. (i) + eq. (ii) + 2 × eq. (iii)

−CH₄(g) − 2O₂(g) + C(s) + O₂(g) + 2H₂(g) + O₂ (g) → −CO₂ (g) − 2H₂O(g) + CO₂(g)

Or C(s) + 2H₂(g)

\(\therefore\) ∆_fH^Θ = −(−890.3) + (−393.5) + 2 × (−285.8)

= 890.3 − 393.5 − 571.6

= 890.3 − 965.1

= −75.8 kJ mol⁻¹

NH₄NO₂(S) → N₂ (g) + 2H₂O (g),

∆_ng= 3 − 0 = 3

∆U = ∆H − ∆_ng RT

= −223.6 kJ mol⁻¹ − (3 mol) x (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × (373 K)

= −232.9 kJ mol⁻¹ .

No, this process is not spontaneous because a spontaneous process should occur continuously by itself after initiation.

Given :

Enthalpy change for atomisation of 10¹⁰ 

molecules = 1.94 × 10^-11 kJ

Number of NH₃ molecules dissociate = 10¹⁰

Bond enthalpy of N-H = ΔH = ?

1 mole of NH₃ contains 6.022 × 10²³ NH₃ molecules.

∵ For atomisation of 1010 molecules of NH₃

ΔH = 1.94 × 10^-11 kJ

∴ For atomisation of 6.022 × 10²³ molecules of NH₃,

ΔH = \(\frac{1.94\times 10^{-11}\times 6.022\times 10^{23}}{10^{10}}\)

= 1168 kJ mol^-1

In NH₃ three N-H bonds are broken on atomisation.

NH_3(g) → N_(g) + 3H_(g)ΔH = 1168 kJ mol^-1

∴ Average bond enthalpy of N-H bond is,

ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^-1

∴ Bond enthalpy of N-H bond = 389.3 kJ mol^-1.

When ΔH > 0 , energy factor opposes the reaction . When ΔS< 0, i.e. randomness decreases, this factor also opposes the process. As both factors oppose the process, reaction is never spontaneous.

Option : (d) 196 kJ

Option : (b) -126 kJ mol^-1

Option : (a) – 74.8 kJ mol^-1

For the reaction,

H₂ (g) + Br₂ (g) → 2HBr(g)

For reactions,

Bond enthalpy of 1 mole of H-H bond = 435 kJ mol^-1

Bond enthalpy of 1 mole of Br-Br bond = 192 KJ mol^-1

For products,

Bond enthalpy of 2 moles of H-Br bond = −2 × 368

= −736 kJ mol⁻¹

∴ Enthalpy changes for the reaction

∆_rH^Θ = 435 + 192 − 736

= −109 kJ mol⁻¹

For the reaction,

H₂ (g) + Cl₂ (g) → 2HCl(g)

For reactants

Bond energy of mole of H-H bond = 104 kcal/mol

Bond energy of mole of Cl-Cl bond = 58 kcal/mol

For products

Bond energy of moles of H-Cl bond = −2 × 103

= −206 kcal/mole

∴ Enthalpy change for the reaction:

∆_rH = 104 + 58 − 206 = −44 kcal/mole

∴ Enthalpy of formation of one mole of HCl gas

= \(\frac{-44}{2}\) = -22 kcal/mole

(a) Given : 

ΔH = -110 kJ 

ΔS = 40 JK^-1 = 0.04 kJK 

Temperature = T = 400 K 

ΔG = ?

Since ΔH is negative, 

The reaction is exothermic.

ΔG = ΔH – TΔS 

= -110 – 400 × 0.04 

= -110 – 16 

= -126 kJ

Since ΔG is negative, 

The reaction is spontaneous.

(b) Given : 

ΔH = 50 kJ, 

ΔS = -130 JK^-1 = -0.13 kJ K^-1 

Temperature = T = 250 K 

ΔG = ?

Since ΔH is positive, 

The reaction is endothermic. 

ΔG = ΔH – TΔS 

= 50 – 250 × (-0.13) 

= 50 + 32.5 

= 82.5 kJ

Since ΔG > 0, 

The reaction is non-spontaneous.

∴ (a) ΔG = -126 kJ; 

The reaction is exothermic and spontaneous.

(b) ΔG = 82.5 kJ; 

The reaction is endothermic and non-spontaneous.

Br₂(l), Cl₂, ∆_fH^Θ = 0

Br₂ exists as liquid at 298 K and liquid bromine is the elementary state.

Since ∆H is +ve and ∆S is -ve, ∆G will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.

(i) (a) Enthalpy change accompanying the formation of 1 mole of a compound from its constituent elements, all substances being in their standard states (1 bar pressure and 298 K).

(b) It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of oxygen or air.

or

(i) (a) Hess's law of constant Heat Summation:

According to Hess's law, the standard enthalpy of reaction is the sum of algebraic sum of the standard enthalpies of reaction into which the overall reaction may be splitted or divided.

(b) According to First law of Thermodynamics, energy can neither be created nor be destroyed although it may be changed from one form to another.

(ii) ∆G = ∆H - T∆S

= 91.18 × 10³ − 298 × 197.67

= 91180 − 58905.66 = 32274.34

For reaction to be spontaneous, ΔG should be negative. Temperature should be greater than 298 K for reaction to be spontaneous.

Given : 

ΔH⁰ = -219 kJ; 

ΔS⁰ = -21 JK^-1 = 0.021 kJ K^-1

ΔG⁰ = ?

For standard conditions : 

Pressure = 1 atm 

Temperature = T = 298 K

ΔG⁰ = ΔH⁰ - TΔS⁰

= -219 – 298 × (-0.021) 

= -219 + 6.258 

= -212.742 kJ

Since ΔG < 0, 

∴ The reaction is spontaneous.

Option : (d) Cl_2(g)

(a) Gibbs Energy: It is defined as the maximum amount of energy available to a system, during a process that can be converted into useful work.

Mathematical Expression of Gibbs Energy:

Gibbs energy is mathematically expressed as:

G = H – TS

Where G = Gibbs energy

H = enthalpy of system

S = entropy of system

T = absolute temperature

Change in free energy (ΔG) for the system:

ΔG = ΔΗ – TΔS

Where ΔΗ = change in enthalpy

ΔS = change in entropy

Gibbs Energy Criteria of Spontaneity:

For a spontaneous reaction, ΔG is negative i.e. ΔG < 0

(b) (i) Freezing of Water:

Here, entropy decreases (i.e. ΔS = −ve) because when liquid (water) freezes to form solid (ice), the molecules attain an ordered state.

(ii) C_(graphite) → C_(diamond)

Here, Entropy decreases (i.e. ΔS = −ve) because when C(graphite) is converted into C(diamond) , it attains more ordered state due to presence of covalent bonding in diamond.

Given : 

Equilibrium constant = KP = 1.64 × 10³ atm²

Temperature = T = 273 + 200 = 473 K

ΔG⁰ = ?

ΔG⁰ = -2.303 RTlog₁₀K_p

= – 2.303 × 8.314 × 473 × log₁₀1.64 × 10³

= – 2.303 × 8.314 × 473 × (3.2148)

= -29115 J

= -29.115 kJ

∴ ΔG⁰ = -29.115 kJ

Option : (c) \(\frac{RTInK}{ΔG^0}\) = -1

Option : (b) 42.875 JK^-1

ΔG⁰ = - 2.303 RT log K 

= - 2.303X8.314 J K^-1 mol^-1 x 300K X log 10 

=-5744.1 J.

Option : (c) ΔG < 0

(a) Enthalpy of vaporisation (ΔΗ) - 26 k/mol.

Boiling point (T) = 35°C = (35 + 273)

K = 308 K

Now, at constant pressure

Entropy change ∆S = \(\frac{-\Delta H}{T}\)

∆S = \(\frac{26K/mol}{308K}\)

∆S = 0.0844 kJ mol⁻¹ K⁻¹

(b) ∆_fH° = ∑∆H°_Products − ∑∆H°_Reacants

−1323 = {2(−393.5) + 2(−249.0)}

−{∆H°C₂H₄ + 0}

−1323 = −1285 − ∆H°C₂H₄ + 0}

∆H°C₂H₄ = −1285 + 1323 = 38 kJ mol⁻¹

(c) Path functions: These are the functions whose magnitude depend on the path followed during a process as well as on the end states. E.g. work (w), heat (Q) are path functions.

Option : (a) ΔS_total = 0

Option : (c) -122.6 JK^-1

Option : (a) ΔG = 0

Option : (d) 0.5 J·K^-1 mol^-1

Let quantity of heat from the reaction mixture = q

Heat capacity of calorimeter = C_v 

\(\therefore\) Quantity of heat absorbed by the calorimeter

q = C_v.∆T

q has opposite sign because the heat is lost by the system which is equal to the heat gained by the calorimeter.

\(\therefore\) q = −C_v × ∆T = −20.7kJ/K × (299 − 298)K

= -20.7 kJ

Here,−ve sign indicates the exothermic nature of the reaction

\(\therefore\) ∆U for the combustion of the 1 g of graphite

= −20.7 kJ K⁻¹

For combustion of 1 mole (12 g) of graphite 

= \(\frac{12g\,mol^{-1}\times(-20.7kJ)}{1g}\)

= −2.48 × 10² kJ mol⁻¹  (∴ ∆n_g = 0)

\(\therefore\) ∆H = ∆U = −2.48 × 10² kJ mol⁻¹

(i) q+w =∆U. As ∆U is a state function, hence q+w is a state function. 

(ii) On dissociation ,entropy increases, i.e. ,∆S is +ve .Though ∆His +ve but if T∆S>∆H, then according to the equation, ∆G =∆H - T∆S,∆G will be –ve. Hence, the process is spontaneous. 

(iii) A real crystal has some disorder due to presence of defects whereas ideal crystal has no disorder . Hence ,a real crystal has more entropy then ideal crystal. 

ΔU = q + w and ‘∆U’ is a state function, which depends upon initial and final states of the system and not upon its path. 

Both q (heat) and W (work done) are not state functions. However, q + w is equal to ∆U according to first law of thermodynamics, which is a state function.

(1) ∆G = ∆H − T∆S

+ 1.193 = 41.1 − 363 × ∆S

363 × ∆S = 41.1 − 1.93 = 39.907

\(\therefore\) ∆S = \(\frac{39.907s}{363}\)

= 0.1099 kJ mol⁻¹ K⁻¹

= 110 J mol⁻¹ K⁻¹

(2) T∆S = 383 × 109

= 41747 = 41.75 kJ mol⁻¹

∆G = ∆H − T∆S

= 40.7 − 41.75

(3) ∆G = ∆H − T∆S

− 0.979 = 40.1 − T∆S

\(\therefore\) T∆S = 40.1 + 0.979

= 41.079 kJ mol⁻¹

∆S = \(\frac{T\Delta S}{T}\) ⁼ \(\frac{41.079}{383}\) 

= 0.1072 kJ mol⁻¹ K⁻¹

= 107.2 J mol⁻¹ K⁻¹ .

(A) Expansion of a gas :

(1) When a gas expands against a constant pressure, P_ex changing the volume from initial volume V₁ to final volume V₂,

Change in volume, 

ΔV = V₂ - V₁

The mechanical work = W = -P_ex × ΔV

= -P_ex (V₂ – V₁)

(2) During expansion V₂ > V₁. The work is said to be performed by the system on the surroundings. This results in the decrease in the (work) energy of the system. Hence the work is negative, i.e. W is -ve.

(B) Compression of a gas : 

During compression, V₂ < V₁. The work is said to be performed on the system by the surroundings. This results in the increase in the (work) energy of the system. Hence the work is positive, i.e. W is + ve.

∆G =0 because the reaction is in equilibrium. 

∆G⁰ =- 2.303RT log 

K = - 2.303X8.314JK^-1 mol^-1 x300K log 10² 

= -11488J mol^-1 

= -11.488kJmol^-1 .

∆G = ∆H − T∆S

When ∆G = 0

∆H = T∆S

T = \(\frac{\Delta H}{\Delta S}\)

= \(\frac{-95.4\times1000\,J\,mol^{-1}}{-198.3\,J\,K}\)

= 481 K

At this temperature, the reaction would be in equilibrium. With increase in temperature, the opposing factor T∆S would become more and hence ∆G would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K.

Given, T = 298 K

K_p = 2.47 × 10⁻²⁹

∆_rG^Θ = ?

\(\because\) ∆_rG^Θ = −2.303 RT log K_p

= −2.303 × (8.314 J K⁻¹ mol⁻¹ ) × (298 K) × log(2.47 × 10⁻²⁹)

= 163000 J mol⁻¹

= 163 kJ mol⁻¹

N₂O₄(g) ⇌ 2NO₂(g)

If N₂O₄ is 50% dissociated, the mole fraction of both the substances is given by,

\(X_{N_2O_4}\) =  \(\frac{1-0.5}{1+0.5}\)

\(X_{NO_2}\) = \(\frac{2\times0.5}{1+0.5}\)

\(P_{N_2O_4}\) = \(\frac{0.5}{0.5}\)x 1 atm

\(P_{NO_2}\) = \(\frac{1}{1.5}\) x 1 atm

The equilibrium constant K_p is given by,

K_p = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\)

= \(\frac{\big(\frac{1}{1.5}\big)^2}{\frac{0.5}{1.5}}\)

= 1.33

\(\therefore\) ∆_fG^Θ = −RT In K_p

= −2.303 RT logK_p

= −2.303 × (8.314 J K⁻¹ mol⁻¹) × (333K) log 1.33

= −2.303 × 8.14 JK⁻¹ mol⁻¹ × 333K × 0.1239

= −763.8 kJ mol⁻¹

One decides the spontaneity of a reaction by considering

= ∆S_total = ∆S_sys + ∆S_surr

For calculating ∆S_surr, we have to consider the heat absorbed by the surroundings which is equal to −∆_rH^Θ. 

At temperature T, entropy change of the surroundings is:

∆S_surr = - \(\frac{\Delta_rH^\ominus}{T}\) (at constant pressure)

= \(\frac{-(-1648\times10^3\,J\,mol^{-1})}{298\,K}\)

= 5530 JK^-1 mol^-1

So, total entropy change for this reaction

Δ_rS_total = 5530JK⁻¹mol⁻¹ + (−549.4 JK⁻¹ mol⁻¹)

= 4980.6 J K⁻¹ mol⁻¹

This Show that the above reaction is spontaneous.

(i) Entropy decreases because molecules attain an ordered state. 

(ii) Entropy increases because from 0K to 115K average kinetic of molecules increases.