Comprehensive NCERT solutions, Q&A of Thermodynamics of Class 11 Chemistry. As On <DM>.

Q:

Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.

A:

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.

Q:

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. 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

A:

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.

Q:

Enthalpy of combustion of carbon to CO₂is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂from carbon and dioxygen gas.

A:

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

$\therefore$ Heat released on formation of 35.2 g CO₂

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

= -314.8 kJ mol^-1

Q:

Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of Δ_rH for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

A:

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

Δ_rH = $\sum$ Δ_fH (products) - $\sum$ 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.

Q:

Given

N₂(g) + 3H₂(g) → 2NH₃(g) ; Δ_rH⁰ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ gas?

A:

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)

$\therefore$ Standard enthalpy of formation of NH_3(g)

= ½ Δ_rH^θ

= ½ (-92.4 kJ mol^-1)

= - 46.2 kJ mol^-1

Q:

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:

CH₃OH (l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l) ; Δ_rH⁰= –726 kJ mol^–1

C(g) + O₂(g) → CO₂(g) ; Δ_cH⁰ = –393 kJ mol^–1

H₂(g) + 1/2 O₂(g) → H₂O(l) ; Δ_fH⁰ = –286 kJ mol^–1.

A:

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

$\therefore$ Δ_fH^θ[CH₃OH_(l)]

= -239 kJ mol^-1

Q:

Calculate the enthalpy change for the process

CCl₄(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl₄(g).

Δ_vapH⁰(CCl₄) = 30.5 kJ mol^–1.

Δ_fH⁰ (CCl₄) = –135.5 kJ mol^–1.

Δ_aH⁰ (C) = 715.0 kJ mol^–1 , where Δ_aH⁰ is enthalpy of atomisation

Δ_aH⁰ (Cl₂) = 242 kJ mol^–1

A:

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

Q:

For an isolated system, ΔU = 0, what will be ΔS?

A:

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

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

Q:

For the reaction at 298 K,

2A + B → C

ΔH = 400 kJ mol^-1and ΔS = 0.2 kJ K^-1mol^-1

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

A:

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.

Q:

For the reaction, 2Cl_(g) → Cl_2(g),what are the signs of ΔH and ΔS ?

A:

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.

Q:

For the reaction

2 A(g) + B(g) → 2D(g)

ΔU⁰ = –10.5 kJ and ΔS⁰ = –44.1 JK^–1.

Calculate ΔG⁰ for the reaction, and predict whether the reaction may occur spontaneously.

A:

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.

Q:

For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0

A:

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.

Q:

The equilibrium constant for a reaction is 10. What will be the value of ΔG⁰ ? R = 8.314 JK^–1 mol^–1, T = 300 K.

A:

From the expression, ΔG⁰= -2.303 RTlogK_eq

ΔG⁰for the reaction,

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

= -5744.14 Jmol^-1

= -5.744 kJ mol^-1

Q:

Comment on the thermodynamic stability of NO(g), given

1/2 N₂(g) + 1/2 O₂(g) → NO(g) ; Δ_rH⁰ = 90 kJ mol^–1

NO(g) + 1/2 O₂(g) → NO₂(g) : Δ_rH⁰= –74 kJ mol^–1

A:

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).

Q:

Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δ_fH⁰ = –286 kJ mol^–1.

A:

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

$\therefore$ ΔS_surr= 959.73 J mol^-1K^-1

Q:

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

A:

(ii) zero

Q:

ΔU⁰of combustion of methane is – X kJ mol^–1. The value of ΔH⁰ is

(i) = ΔU⁰

(ii) > ΔU⁰

(iii) < ΔU⁰

(iv) = 0

A:

(iii) < ΔU⁰

Q:

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol^–1 , –393.5 kJ mol^–1, and –285.8 kJ mol^–1 respectively. Enthalpy of formation of CH₄(g) will be

(i) –74.8 kJ mol^–1

(ii) –52.27 kJ mol^–1

(iii) +74.8 kJ mol^–1

(iv) +52.26 kJ mol^–1

A:

According to the question,

Thus, the desired equation is the one that represents the formation of CH₄_(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).

Q:

A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(iv) possible at any temperature

A:

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).

Q:

In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?

A:

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.

Q:

The reaction of cyanamide, NH₂CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol^–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

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

A:

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 = $\sum$ ng (products) - $\sum$ 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

Q:

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^–1 K^–1.

A:

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

Thermodynamics Question Answers: NCERT Class 11 Chemistry

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