NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics)

NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics) can be found on this page. Here, students can access detailed, explanative solutions to all the intext and exercise questions listed in Chapter 6 of the NCERT Class 11 Chemistry textbook. Furthermore, the NCERT Solutions provided on this page can be downloaded as a PDF for free by clicking the download button provided above.

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NCERT Solutions for Chemistry – Class 11, Chapter 6: Thermodynamics

“Thermodynamics” is the sixth chapter in the NCERT Class 11 Chemistry textbook. Thermodynamics is a branch of science that deals with the relationship between heat and other forms of energy. A part of the universe where observations are made is called system. The surrounding of a system is the part of the universe that does not contain the system. Based on the exchange of energy and matter, thermodynamic systems can be classified into 3 types: closed system, open system and an isolated system.

In a closed system, only energy can be exchanged with the surrounding. In open system energy as well as matter can be exchanged with the surrounding. In an isolated system, both energy and matter cannot be exchanged with the surrounding.

Subtopics included in Class 11 Chemistry Chapter 6 Chemical Thermodynamics

1. Thermodynamic Terms
   • The System and the Surroundings
   • Types of Thermodynamic Systems
   • State of the System
   • Internal Energy as a State Function
2. Applications
   • Work
   • Enthalpy, H
3. Measurement Of ΔU And ΔH: Calorimetry
4. Enthalpy Change and Reaction Enthalpy
5. Enthalpies For Different Types Of Reactions
6. Spontaneity
7. Gibbs Energy Change And Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 6

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Q-1: 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

Ans:

(ii) A quantity which is independent of path.

Reason:

Functions like pressure, volume and temperature depends on the state of the system only 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:

(iii) q = 0

Reason:

For an adiabatic process heat transfer is zero, i.e. q = 0.

 

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

(i) Unity   (ii) Zero

(iii) < 0     (iv) Different for every element

Ans:

(ii) Zero

 

Q-4: ∆U⁰ of combustion of methane is – X kJ mol^–1. The value of ∆H⁰ is

(i) = $Delta U^{Theta }$

(ii) > $Delta U^{Theta }$

(iii) < $Delta U^{Theta }$

(iv) 0

Ans:

(iii) < $Delta U^{Theta }$

Reason:

$Delta H^{Theta } = Delta U^{Theta } + Delta n_{g}RT$ ; $Delta U^{Theta }$ = – Y $kJ mol^{-1}$,

$Delta H^{Theta } = ( – Y) + Delta n_{g}RT Rightarrow Delta H^{Theta } < Delta U^{Theta }$

 

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 CH_4(g) will be

(i) -74.8 kJ$mol^{-1}$

(ii) -52.27 kJ$mol^{-1}$

(iii) +74.8 kJ$mol^{-1}$

(iv) +52 kJ$mol^{-1}$

Ans:

(i) -74.8kJ$mol^{-1}$

1. CH_4(g)  + 2O_2(g) $ightarrow$ CO_2(g) + 2H₂O_(g)

$Delta H = -890.3kJmol^{-1}$

2. C_(s) + O_2(g)  $ightarrow$ CO_2(g)

$Delta H = -393.5kJmol^{-1}$

3. 2H_2(g) + O_2(g) $ightarrow$ 2H₂O_(g)

$Delta H = -285.8kJmol^{-1}$

C_(s) + 2H_2(g) $ightarrow$ CH_4(g)

$Delta _{f}H_{CH_{4}}$  = $Delta _{c}H_{c}$ + $2Delta _{f}H_{H_{2}}$ – $Delta _{f}H_{CO_{2}}$

= [ -393.5 +2(-285.8) – (-890.3)] kJ$mol^{-1}$

= -74.8kJ$mol^{-1}$

 

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:

(iv) possible at any temperature

$Delta G$ should be –ve, for spontaneous reaction to occur

$Delta G$ = $Delta H$ – T$Delta S$

As per given in question,

$Delta H$ is –ve ( as heat is evolved)

$Delta S$ is +ve

Therefore, $Delta G$ is negative

So, the reaction will be possible at any temperature.

 

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:

As per Thermodynamics 1^st law,

$Delta U$ = q + W(i);

$Delta U$ internal energy = heat

W = work done

W = -594 J (work done by system)

q = +801 J (+ve as heat is absorbed)

Now,

$Delta U$ = 801 + (-594)

$Delta U$ = 207 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 at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH₂CN_(g) + 3/2 O_2(g) → N_2(g) + CO_2(g) + H₂O_(l)

Ans:

$Delta H$ is given by,

$Delta H = Delta U + Delta n_{g}RT$………………(1)

$Delta n_{g}$ = change in number of moles

$Delta U$ = change in internal energy

Here,

$Delta n_{g} = sum n_{g}(product) – sum n_{g}(reactant)$

  = (2 – 1.5) moles

 $Delta n_{g}$ = 0.5 moles

Here,

T =298K

$Delta U$ = -742.7 $kJmol^{-1}$

R  = $8.314 imes 10^{-3}kJmol^{-1}K^{-1}$

Now, from (1)

$Delta H = (-742.7 kJmol^{-1}) + (0.5mol)(298K)( 8.314 imes 10^{-3}kJmol^{-1}K^{-1})$

= -742.7 + 1.2

$Delta H$ = -741.5 $kJmol^{-1}$

 

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:

Expression of heat(q),

$q = mCPDelta T$;………………….(a)

$Delta T$ = Change in temperature

c = molar heat capacity

m = mass of substance

From (a)

$q = ( frac{60}{27}mol )(24mol^{-1}K^{-1})(20K)$

q = 1066.67 J = 1.067 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.

$C_{p}[H_{2}O_{(l)}] = 75.3 J;mol^{-1}K^{-1}$

$C_{p}[H_{2}O_{(s)}] = 36.8 J;mol^{-1}K^{-1}$

 Ans:

$Delta H_{total}$ = sum of the changes given below:

(a) Energy change that occurs during transformation of 1 mole of water from $10^{circ}C;to;0^{circ}C$.

(b) Energy change that occurs during transformation of 1 mole of water at $0^{circ}C$  to  1 mole of ice at $0^{circ}C$.

(c) Energy change that occurs during transformation of 1 mole of ice from $0^{circ}C;to;(-10)^{circ}C$.

$Delta H_{total} = C_{p}[H_{2}OCl]Delta T + Delta H_{freezing} C_{p}[H_{2}O_{s}]Delta T$

= (75.3 $J mol^{-1}K^{-1}$)(0 – 10)K + (-6.03*1000 $J mol^{-1}$(-10-0)K

= -753 $J mol^{-1}$ – 6030$J mol^{-1}$ – 368$J mol^{-1}$

= -7151 $J mol^{-1}$

= -7.151$kJ mol^{-1}$

Thus, the required change in enthalpy for given 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 carbon dioxide from di-oxygen and carbon gas is given as:

C_(s) + O_2(g) → CO_2(g); $Delta _{f}H$ = -393.5 $kJ mol^{-1}$

1 mole CO₂ = 44g

Heat released during formation of 44g CO₂ = -393.5 $kJ mol^{-1}$

Therefore, heat released during formation of 35.2g of CO₂  can be calculated as

= $frac{-393.5kJmol^{-1}}{44g} imes 35.2g$

= -314.8 $kJ mol^{-1}$

 

Q-12: Enthalpies of formation of CO _(g), CO_{2 (g)}, N₂O _(g) and N₂O_4(g) are –110, – 393, 81 and 9.7 kJ mol^–1 respectively. Find the value of ∆_rH for the reaction:

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

Ans:

“$Delta _{r}H$ for any reaction is defined as the fifference between $Delta _{f}H$ value of products and $Delta _{f}H$ value of reactants.”

$Delta _{r}H = sum Delta _{f}H (products) – sum Delta _{f}H (reactants)$

Now, for

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

$Delta _{r}H = [(Delta _{f}H(N_{2}O) + (3Delta _{f}H(CO_{2})) – (Delta _{f}H(N_{2}O_{4}) + 3Delta _{f}H(CO))]$

Now, substituting the given values in the above equation, we get:

$Delta _{r}H$ = [{81$kJ mol^{-1}$ + 3(-393) $kJ mol^{-1}$} – {9.7$kJ mol^{-1}$ + 3(-110) $kJ mol^{-1}$}] $Delta _{r}H$ = -777.7 $kJ mol^{-1}$

 

Q-13: Given N₂ (g) + 3H₂ (g) → 2NH₃ (g) ; ∆_rH⁰= –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 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.”

Dividing the chemical equation given in the question by 2, we get

(0.5)N_2(g) + (1.5)H_2(g) → 2NH_3(g)

Therefore, Standard Enthalpy for formation of ammonia gas

= (0.5) $Delta _{r}H^{Theta }$

= (0.5)(-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:

CH₃OH_(l) + 3/2 O_2(g)→ CO_2(g) + 2H₂O_(l); $Delta _{r}H^{Theta }$ = -726 $kJ mol^{-1}$

C_(g) + O_2(g)→ CO_2(g); $Delta _{c}H_{Theta }$ = -393 $kJ mol^{-1}$

H_2(g) + 1/2 O_2(g)→ H₂O_(l); $Delta _{f}H^{Theta }$ = -286 $kJ mol^{-1}$

Ans:

C_(s) + 2H₂O_(g) + (1/2)O_2(g) → CH₃OH_(l) …………………………(i)

CH₃OH_(l) can be obtained as follows,

$Delta _{f}H_{Theta }$ [CH₃OH_(l)] = $Delta _{c}H_{Theta }$

2$Delta _{f}H_{Theta }$ – $Delta _{r}H_{Theta }$

= (-393 $kJ mol^{-1}$) +2(-286$kJ mol^{-1}$) – (-726$kJ mol^{-1}$)

= (-393 – 572 + 726) $kJ mol^{-1}$

= -239$kJ mol^{-1}$

Thus, $Delta _{f}H_{Theta }$ [CH₃OH_(l)] = -239$kJ mol^{-1}$

 

Q-15: Calculate the enthalpy change for the process

CCl_4(g)→ C_(g) + 4Cl_(g) and determine the value of bond enthalpy for C-Cl in CCl_4(g).

$Delta _{vap}H^{Theta }$ (CCl₄) = 30.5 $kJ mol^{-1}$.

$Delta _{f}H^{Theta }$ (CCl₄) = -135.5 $kJ mol^{-1}$.

$Delta _{a}H^{Theta }$ (C) = 715 $kJ mol^{-1}$,

$Delta _{a}H^{Theta }$  is a enthalpy of atomisation

$Delta _{a}H^{Theta }$ (Cl₂) = 242 $kJ mol^{-1}$.

Ans:

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

(1) CCl_4(l) à CCl_4(g) ; $Delta _{vap}H^{Theta }$ = 30.5 $kJ mol^{-1}$

(2) C_(s) à C_(g) $Delta _{a}H^{Theta }$ = 715 $kJ mol^{-1}$

(3) Cl_2(g) à 2Cl_(g) ; $Delta _{a}H^{Theta }$ = 242 $kJ mol^{-1}$

(4) C_(g) + 4Cl_(g) à CCl4(g); $Delta _{f}H^{Theta }$ = -135.5 $kJ mol^{-1}$ $Delta H$ for the process CCl_4(g) à C_(g) + 4Cl_(g) can be measured as:

$Delta H = Delta _{a}H^{Theta }(C) + 2Delta _{a}H^{Theta }(Cl_{2}) – Delta _{vap}H^{Theta } – Delta _{f}H$

= (715$kJ mol^{-1}$) + 2($kJ mol^{-1}$) – (30.5$kJ mol^{-1}$) – (-135.5$kJ mol^{-1}$)

Therefore, $H = 1304kJmol^{-1}$

The value of bond enthalpy for C-Cl in CCl_4(g)

= $frac{1304}{4}kJmol^{-1}$

= 326 $kJ mol^{-1}$

 

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

Ans:

$Delta U$ is positive ; $Delta U$ > 0.

As, $Delta U$ = 0 then$Delta S$ will be +ve, as a result reaction will be spontaneous.

 

Q-17: For the reaction at 298K,

2A + B → C

$Delta H$ = 400 $kJ mol^{-1}$

$Delta H$ = 0.2 $kJ mol^{-1}K^{-1}$

At what temperature will the reaction become spontaneous considering $Delta S$ and $Delta H$ to be constant over the temperature range?

Ans:

Now,

$Delta G = Delta H – TDelta S$

Let, the given reaction is at equilibrium, then $Delta T$ will be:

T = $(Delta H – Delta G)frac{1}{Delta S}$ $frac{Delta H}{Delta S}$; ($Delta G$ = 0 at equilibrium)

= 400$kJ mol^{-1}$/0.2$kJ mol^{-1}K^{-1}$

Therefore, T = 2000K

Thus, for the spontaneous, $Delta G$ must be –ve and T > 2000K.

 

Q-18: For the reaction

2Cl_(g)→ Cl_2(g)

What are the signs of $Delta S$ and $Delta H$?

Ans:

$Delta S$ and $Delta H$ are having negative sign.

The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So,  $Delta H$ is negative.

Also, 2 moles of Chlorine atoms are having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, $Delta S$ is negative.

 

Q-19: For the reaction

2A_(g) + B_(g)→ 2D_(g)

$Delta U^{Theta }$ = -10.5 kJ and $Delta S^{Theta }$ = -44.1$JK^{-1}$

Calculate $Delta G^{Theta }$ for the reaction, and predict whether the reaction may occur spontaneously.

Ans:

2A_(g) + B_(g) → 2D_(g)

$Delta n_{g}$ = 2 – 3

= -1 mole

Putting value of $Delta U^{Theta }$ in expression of $Delta H$:

$Delta H^{Theta } = Delta U^{Theta } + Delta n_{g}RT$

= (-10.5KJ) – (-1)( $8.314 imes 10^{-3}kJK^{-1}mol^{-1}$)(298K)

= -10.5kJ -2.48kJ

$Delta H^{Theta }$ = -12.98kJ

Putting value of $Delta S^{Theta }$ and $Delta H^{Theta }$ in expression of $Delta G^{Theta }$:

$Delta G^{Theta } = Delta H^{Theta } – TDelta S^{Theta }$

= -12.98kJ –(298K)(-44.1$JK^{-1}$)

= -12.98kJ +13.14kJ

$Delta G^{Theta }$ = 0.16kJ

As, $Delta G^{Theta }$ is positive, the reaction won’t occur spontaneously.

 

Q-20: The equilibrium constant for a reaction is 10. What will be the value of ∆G₀? R = 8.314 JK^–1 mol^–1, T = 300 K.

Ans:

Now,

$Delta G^{Theta }$ = $-2.303RTlog eq$

= (2.303)( $8.314 imes kJK^{-1}mol^{-1}$)(300K) $log 10$

= -5744.14$Jmol^{-1}$

= -5.744$kJmol^{-1}$

 

Q-21: Comment on the thermodynamic stability of NO(g), given,

(1/2)N_2(g) + (1/2)O_2(g) → NO_(g); $Delta _{r}H^{Theta } = 90kJmol^{-1}$

NO_(g) + (1/2)O_2(g) → NO_2(g); $Delta _{r}H^{Theta } = -74kJmol^{-1}$

Ans:

The +ve value of $Delta _{r}H$ represents that during NO_(g) formation from O₂ and N₂,  heat is absorbed. The obtained product, NO_(g) is having more energy than reactants. Thus, NO_(g) is unstable.

The -ve value of $Delta _{r}H$ represents that during NO_2(g) formation from O_2(g) and NO_(g),  heat is evolved. The obtained product, NO_2(g) gets stabilized with minimum energy.

Thus, unstable NO_(g) converts into unstable  NO_2(g).

 

Q-22: Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. ∆_f H⁰= –286 kJ mol^–1.

 Ans:

$Delta _{r}H^{Theta } = -286kJmol^{-1}$ is given so that amount of heat is evolved during the formation of 1 mole of H₂O_(l).

Thus, the same heat will be absorbed by surrounding Q_surr = +286$kJmol^{-1}$.

Now, $Delta S_{surr}$ = Q_surr/7

= $frac{286kJmol^{-1}}{298K}$

Therefore, $Delta S_{surr} = 959.73Jmol^{-1}K^{-1}$

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