Thermodynamics Question Answers: NCERT Class 11 Chemistry

In thermodynamics, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state. A state function describes the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, T etc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C.

Total ΔH = C_p [H₂OCI] ΔT  + ΔH_freezing + C_p[H₂O_(s)] ΔH

= (75.3 J mol^-1 K^-1) (0 - 10)K + (-6.03 × 10³ J mol^-1) + (36.8 J mol^-1 K^-1) (-10 - 0)K

= -753 J mol^-1 - 6030 J mol^-1 - 368 J mol^-1

= -7151 J mol^-1

= -7.151 kJ mol^-1

Hence, the enthalpy change involved in the transformation is -7.151 kJ mol^-1.

Formation of CO₂ from carbon and dioxygen gas can be represented as:

C_(s) + O_2(g)  → CO_2(g)           Δ_fH = -393.5 kJ mol^-1

(1 mole = 44 g)

Heat released on formation of 44g CO₂ = -393.5 kJ mol^-1

Heat released on formation of 35.2 g CO₂

= [ -393.5kJ mol^-1  /  44g  ]  x  35.2g

= -314.8 kJ mol^-1

Δ_rH for a reaction is defined as the difference between Δ_fH value of products and Δ_fH value of reactants.

Δ_rH =  Δ_fH (products) -  Δ_fH (reactants)

For the given reaction,

N₂O_4(g)+ 3CO_(g)  →  N₂O_(g) + 3CO_2(g)

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

Substituting the values of Δ_fH for N₂O, CO₂, N₂O₄,and CO from the question, we get:

Δ_rH = [{81 KJ mol^-1 + 3(-393) KJ mol^-1} - {9.7KJ mol^-1 + 3(-110)KJ mol^-1}]

Δ_rH = -777.7 KJ mol^-1

Hence, the value of Δ_rH for the reaction is -777.7 KJ mol^-1.

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.

Re-writing the given equation for 1 mole of NH₃(g),

½ N_2(g) + 3/2 H_2(g) → NH_3(g)

Standard enthalpy of formation of NH_3(g)

= ½ Δ_rH^θ

= ½ (-92.4 kJ mol^-1)

= - 46.2 kJ mol^-1

The reaction that takes place during the formation of CH₃OH(l) can be written as:

C_(s) + 2H₂O_(g) + ½O_2(g)  → CH₃OH_(l)     (1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) - equation (i)

Δ_fH^θ [CH₃OH_(l)]  =  Δ_cH^θ  +  2Δ_fH^θ [H₂O_(l)]  - Δ_rH^θ

= (-393 kJ mol^-1) + 2 (-286 kJ mol^-1) - (-726 kJ mol^-1)

= (-393 - 572 + 726) kJ mol^-1

Δ_fH^θ[CH₃OH_(l)]

= -239 kJ mol^-1

The chemical equations implying to the given values of enthalpies are:

(i) CCl_4(l)    →    CCL_4(g)      Δ_vapH^θ = 30.5 kJ mol^-1

(ii) C_(s)    →   C_(g)               Δ_aH^θ = 715.0 kJ mol^-1

(iii) Cl_2(g)  →   2Cl_(g)          Δ_aH^θ = 242 kJ mol^-1

(iv) C_(g)  + 4Cl_(g)  →  CCl_4(g)  Δ_fH = -135.5 kJ mol^-1

Enthalpy change for the given process  C_(g)  + 4Cl_(g)  →  CCl_4(g)   can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)

ΔH = Δ_aH^θ(C)  +  2Δ_aH^θ (Cl₂) -  Δ_vapH^θ - Δ_fH

= (715.0 kJ mol^-1) + 2(242 kJ mol^-1) - (30.5 kJ mol^-1) - (-135.5 kJ mol^-1)

∴ΔH = 1304 kJ mol^-1

Bond enthalpy of C-Cl bond in CCl_4(g) = 326 kJ mol^-1

ΔS will be positive i.e., greater than zero

Since ΔU= 0, ΔS will be positive and the reaction will be spontaneous.

From the expression,

ΔG = ΔH - TΔS

Assuming the reaction at equilibrium, ΔTfor the reaction would be:

T = (ΔH - ΔG) / ΔS

T = ΔH / ΔS  (ΔG = 0 at equilibrium)

= 400 kJ mol^-1 /  0.2 kJ K^-1mol^-1

T= 2000 K

For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

ΔH and ΔS are negative The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.

For the given reaction,

2 A_(g) + B_(g) → 2D_(g)

Δn_g = 2 - (3) = -1 mole

Substituting the value of ΔU⁰¸ in the expression of ΔH:

ΔHº  = ΔUº + Δn_gRT

= (-10.5 kJ) - (-1) (8.314 x 10^-3 kJ K^-1 mol^-1) (298 K)

= -10.5 kJ - 2.48 kJ

ΔHº  =  -8.02 kJ

Substituting the values of ΔHº and ΔSº¸ in the expression of ΔH:

ΔGº  =  ΔHº  - TΔSº

= -8.-02 kJ - (298 K) (-44.1 J K^-1)

= -8.02 kJ + 13.14 kJ

ΔGº = + 5.12 kJ

Since ΔGº for the reaction is positive, the reaction will not occur spontaneously.

An adiabatic process is one that occurs without transfer of heat or matter between a system and its surroundings & it helps in explaining first law of thermodynamics

Hence, under adiabatic conditions, q = 0.

Therefore, alternative (iii) is correct.

From the expression, ΔG⁰ = -2.303 RTlogK_eq

ΔG⁰ for the reaction,

= (2.303) (8.314 JK^-1mol^-1) (300 K) log10

= -5744.14 Jmol^-1

= -5.744 kJ mol^-1

The positive value of Δ_rH indicates that heat is absorbed during the formation of NO_(g). This means that NO_(g) has higher energy than the reactants (N₂ and O₂). Hence, NO_(g) is unstable. The negative value of ΔrH indicates that heat is evolved during the formation of NO_2(g) from NO_(g) and O_2(g). The product, NO_2(g) is stabilized with minimum energy. Hence, unstable NO_(g) changes to stable NO_2(g).

It is given that 286 kJ mol^-1of heat is evolved on the formation of 1 mol of H₂O_(l). Thus, an equal amount of heat will be absorbed by the surroundings.

q_surr = +286 kJ mol^-1

Entropy change (ΔS_surr) for the surroundings =  q_surr / 7

= 286 kJ mol^-1  / 298k

 ΔS_surr = 959.73 J mol^-1K^-1

(ii) zero

(iii) < ΔU⁰

According to the question,

Thus, the desired equation is the one that represents the formation of CH_4(g) i.e.,

[-393.5 + 2(-285.8) - (-890.3)] kJ Mol^-1

= -74.8 kJ Mol^-1

So,Enthalpy of formation of CH_4(g) is -74.8 kJ Mol^-1

That means answer is (i).

For a reaction to be spontaneous, ΔG should be negative.

ΔG = ΔH- TΔS

According to the question, for the given reaction,

ΔS = positive

ΔH= negative (since heat is evolved)

⇒ ΔG= negative

Therefore, the reaction is spontaneous at any temperature. Hence, answer is (iv).

According to the first law of thermodynamics,

ΔU= q + W    (i)

Where, ΔU = change in internal energy for a process

q = heat

W = work

Given, q = + 701 J (Since heat is absorbed)

W = -394 J (Since work is done by the system)

Substituting the values in expression (i), we get

ΔU= 701 J + (-394 J) ΔU = 307 J

Hence, the change in internal energy for the given process is 307 J.

Enthalpy change for a reaction (ΔH) is given by the expression,

ΔH = ΔU + Δn_gRT

Where, ΔU = change in internal energy

Δng = change in number of moles

For the given reaction,

Δn_g =  ng (products) -  n_g (reactants)

= (2 - 1.5) moles

Δn_g = 0.5 moles

And,

ΔU = -742.7 kJ mol^-1

T = 298 K

R = 8.314 x 10^-3 kJ mol^-1 K^-1

Substituting the values in the expression of ΔH:

ΔH = (-742.7 kJ mol^-1) + (0.5 mol) (298 K) (8.314 x 10^-3 kJ mol^-1 K^-1)

= -742.7 + 1.2

ΔH = -741.5 kJ mol^-1

From the expression of heat (q),

q = m. c. ΔT

Where,

c = molar heat capacity

m = mass of substance

ΔT = change in temperature

Substituting the values in the expression of q:

q = (60/27 mol) (24 Jmol^–1 K^–1) (20K)

q = 1066.7 J

q = 1.07 kJ