NCERT Solutions for Class 11 Chemistry Chapter 6

Q.1 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.

Ans.

Correct answer is (ii)

Explanation:

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.

Q.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) ∆T = 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0

Ans.

Correct answer is (iii)

Explanation:

When there is no exchange of heat between the system and its surroundings then the system is said to be under adiabatic conditions. Hence, under adiabatic conditions, q = 0.

Q.3 The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) Different for each element

Ans.

Correct answer is (ii)

Explanation:

The enthalpy of all elements in their standard state is zero.

Q.4 ∆U^θof combustion of methane is – X kJ mol^–1. The value of ∆H^θ is

(i) = ∆U^θ

(ii) > ∆U^θ

(iii) < ∆U^θ

(iv) = 0

Ans.

Correct answer is (iii)

Explanation:

Since ∆H^θ = ∆U^θ + ∆n_gRT and ∆U^θ = –X kJ mol^–1,

∆H^θ = (–X) + ∆n_gRT.

⇒ ∆H^θ < ∆U^θ

Q.5 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.

Ans.

Correct answer is (i)

Explanation:

Given: The enthalpy of combustion of,

1. CH_4(g) + 2O_2(g) → CO_2(g) +2H₂O_(g) ∆H=-890.3 kJ mol ^-1
2. C_(s)+O_2(g) → CO_{2(g) ∆H=-393.5kJ mol ^-1}
3. 2H_2(g)+ O_2(g) → 2H₂O_{(g) ∆H=-285.8 kJ mol ^-1}

Thus, the desired equation is the one that represents the formation of CH₄

C_(s)+2H_2(g) → CH_4(g)

∆_fH_CH4 =∆_cH_c+2∆_cH_H2 -∆_CH_CO2

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

=-74.8 kJ mol^-1

Enthalpy of formation of CH_4(g) = –74.8 kJ mol^–1

Hence, alternative (i) is correct.

Q.6 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

Ans.

For a reaction to be spontaneous, ∆G should be negative.

∆G = ∆H – T∆S

According to the question, for the given reaction,

∆S = positive

∆H = negative (since heat is evolved)

⇒ ∆G = negative

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Q.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Ans.

According to the first law of thermodynamics,

∆U = q + W (i)

Where,

∆U = 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

∆U = 701 J + (–394 J)

∆U = 307 J

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

Q.8 The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol^–1 at298 K. Calculate enthalpy change for the reaction at 298 K.

NH₂CN_(g)+ O_2(g) → N_2(g) +CO_2(g) +H₂O_(l)

Ans.

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

∆H = ∆U + ∆n_gRT

Where,

∆U = change in internal energy

∆n_g = change in number of moles

For the given reaction,

∆ng = ∑ng (products) – ∑ng (reactants)

= (2 – 1.5) moles

∆ng = 0.5 moles

And,

∆U = –742.7 kJ mol^–1

T = 298 K

R = 8.314 × 10^–3 kJ mol^–1 K^–1

Substituting the values in the expression of ∆H:

∆H = (–742.7 kJ mol^–1) + (0.5 mol) (298 K) (8.314 × 10^–3 kJ mol^–1 K^–1)

= 742.7 – 1.2

∆H = 741.5 kJ mol^–1

Q.9 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.

Ans.

From the expression of heat (q),

q = m. c. ∆T

Where,

c = molar heat capacity

m = mass of substance

∆T = change in temperature

Substituting the values in the expression of q:

q=[60/27 mole](24 j mole^-1 K^-1)(20K)

q = 1066.7 J

q = 1.07 kJ

Q.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.

∆_fusH = 6.03 kJ mol^–1 at 0°C.

C_p[H ₂O(l)] = 75.3 J mol^–1 K^–1

C_p[H ₂O(s)] = 36.8 J mol^–1 K^–1

Ans.

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 ∆H=C_p[H₂OCl]∆T+∆H _freezing +C_p [H₂O_(S)]∆T

= (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.11 Enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^–1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Ans.

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

C_{(s )+}O_2(g) →CO_2(g) ∆_∫H=-393.5 kJ mol^-1

(1 mole = 44 g)

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

Heat released on formation of 35.2 g CO₂

= 314.8 kJ mol^–1

Q.12 Enthalpies of formation of CO(g), CO₂(g), N ₂ O(g) and N ₂ O₄(g) are –110 kJ mol^–1, – 393 kJ

mol^–1, 81 kJ mol^–1 and 9.7 kJ mol^–1 respectively. Find the value of ∆rH for the reaction:

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

Ans.

∆_rH for a reaction is defined as the difference between ∆_fH value of products and ∆_fH

value of reactants.

Δ _rH = ∑ Δ_∫H (Products) – ∑ Δ_∫H (reactants)

For the given reaction,

N₂O₄(g) + 3CO(g) N₂O(g) + 3CO₂(g)

Δ_rH =[{ Δ_∫H(N₂O)+3 Δ_∫H (CO₂)}-{ Δ_∫H(N₂O₄)+3 Δ_∫H(CO)}]

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

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

Δ_rH =-777.7 kJ mol^-1

Hence, the value of ∆rH for the reaction is -777.7 kJ mol^-1

Q.13 Given

N_2(g)+3H_2(g) → 2NH_3(g) ∆rHθ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ gas?

Ans.

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 NH_3(g),

\frac{1}{2}{N}_{2\left(g\right)}+\frac{3}{2}{H}_{2\left(g\right)}\to N{H}_{3\left(g\right)}

.^..Standard enthalpy of formation of NH_3(g)

= ½ ∆_rH^θ

= ½ (–92.4 kJ mol^–1)

= –46.2 kJ mol^–1

Q.14 Calculate the standard enthalpy of formation of CH₃OH(_l) from the following data:

\begin{array}{l}C{H}_{3}O{H}_{\left(I\right)}+\frac{2}{3}{O}_{2\left(g\right)}\to C{O}_2{O}_{\left(I\right)};{?}_r{H}^{\theta }=-726{\text{kJ mol}}^{-1}\\ C_{\left(g\right)}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)};{}_{}\end{array}

Δ

c H θ =−393 kJ mol −1 H 2( g ) + 1 2 O 2( g ) → H 2 O ( I ) ;

Δ

f H θ =−286 kJ mol −1

Ans.

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

C\left(s\right)+2H_{2}O\left(g\right)+\frac{1}{2}{O}_{2\left(g\right)}\to C{H}_{3}O{H}_{\left(I\right)}\left(1\right)

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

calculations as:

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

∆_fH^θ [CH₃OH_(l)] = ∆cH^θ + 2∆_fH^θ [H₂O_(l)] – ∆_rH^θ

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

= (–393 – 572 + 726) kJ mol^–1

∆_fH^θ [CH₃OH_(l)] = –239 kJ mol^–1

Q.15 Calculate the enthalpy change for the process

CCl_4(g) → C_(g) + 4Cl_(g)

and calculate bond enthalpy of C–Cl in CCl_4(g).

∆_vapH^θ (CCl₄) = 30.5 kJ mol^–1.

∆_fH^θ (CCl₄) = –135.5 kJ mol^–1.

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

∆_aH^θ (Cl₂) = 242 kJ mol^–1

Ans.

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

CCI_4(l) → CCI_4(g) ∆_vapH^θ = 30.5 kJ mol^–1

C_{(s) →} C_(g) ∆_aH^θ = 715.0 kJ mol^–1

Cl_{2(g) →} 2Cl_(g) ∆_aH^θ = 242 kJ mol^–1

C_(g)+Cl_(g) → CCI_4(g) ∆_fH = –135.5 kJ mol^–1

Enthalpy change for the given processCCI_{4(g) →} C_(g)+4CI_(g)’ can be calculated using the following algebraic calculations as:

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

∆H = ∆_aH^θ(C) + 2∆_aH^θ (Cl₂) – ∆_vapH^θ – ∆_fH

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

∆H = 1304 kJ mol^–1

Bond enthalpy of C–Cl bond in CCl_4(g)

=326 kJ mol^-1

Q.16 For an isolated system, ∆U = 0, what will be ∆S?

Ans.

∆S will be positive i.e., greater than zero

Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

Q.17 For the reaction at 298 K,

2A + B → C

∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be

constant over the temperature range?

Ans.

From the expression,

∆G = ∆H – T∆S

Assuming, the reaction at equilibrium, ∆T for the reaction would be:

\begin{array}{l}T=\frac{\Delta H}{\Delta S}\left(\Delta G=0\text{at equilibrium}\right)\\ \frac{400{\text{kJ mol}}^{-1}}{0.2{\text{kJ K}}^{-1}mo{l}^{-1}}\end{array}

T = 2000 K

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction

to be spontaneous, T should be greater than 2000 K.

Q.18 For the reaction,

2Cl_(g) → Cl_2(g), what are the signs of ∆H and ∆S ?

Ans.

∆H and ∆S 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, ∆H is negative.

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

Q.19 For the reaction

2A_(g) + B_(g) → 2D_(g)

∆U^θ = –10.5 kJ and ∆S^θ= –44.1 JK^–1.

Calculate ∆G^θ for the reaction, and predict whether the reaction may occur spontaneously.

Ans.

For the given reaction,

2 A_(g) + B_(g) → 2D_(g)

∆n_g = 2 – (3)

= –1 mole

Substituting the value of ∆U^θ in the expression of ∆H:

∆H^θ = ∆U^θ + ∆n_gRT

= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K^–1 mol^–1) (298 K)

= –10.5 kJ – 2.48 kJ

∆H^θ = –12.98 kJ

Substituting the values of ∆H^θ and ∆S^θ in the expression of ∆G^θ:

∆G^θ = ∆H^θ – T∆S^θ

= –12.98 kJ – (298 K) (–44.1 J K^–1)

= –12.98 kJ + 13.14 kJ

∆G^θ = + 0.16 kJ

Since ∆G^θ for the reaction is positive, the reaction will not occur spontaneously.

Q.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G_θ? R = 8.314

JK^–1 mol^–1, T = 300 K.

Ans.

From the expression,

∆G^θ = –2.303 RT logK_eq

∆G^θ for the reaction,

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

= –5744.14 Jmol^–1

= –5.744 kJ mol^–1

Q.21

\begin{array}{l}\text{Comment on the thermodynamic stability of NO}\left(\text{g}\right),\text{given}\\ \frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to N{O}_{\left(g\right)};{\Delta }_r{H}^{\theta }=90kJmo{l}^{-1}\\ \frac{1}{2}{N}_{\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to N{O}_{2\left(g\right)};{\Delta }_r{H}^{\theta }=-74kJmo{l}^{-1}\end{array}

Ans.

The positive value of ∆_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 ∆_rH 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 unstable NO_2(g).

Q.22 Calculate the entropy change in surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions. ∆_fH^θ = –286 kJ mol^–1.

Ans.

It is given that 286 kJ mol^–1 of 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 (∆S_surr) for the surroundings =

\frac{q_{surr}}{T}

\frac{286{\text{kJ mol}}^{-1}}{298\text{k}}

∆S_surr =959.73 J mol ^-1 K^-1