CBSE · NCERT · Class 11 Chemistry · Chapter 5

NCERT Solutions: Class 11 Chemistry Chapter 5 - Thermodynamics

22 textbook Q&A22 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Thermodynamics, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercises 22
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1Exercises22 questions
Q.5.1Choose 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.v
  1. i. used to determine heat changes
  2. ii. whose value is independent of path
  3. iii. used to determine pressure volume work
  4. iv. whose value depends on temperature only.
Solution

A state function depends only on the initial and final states of the system, not on the path used to reach the state.

Answer:

(ii) whose value is independent of path.

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Q.5.2For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT = 0 (ii) Δp = 0 (iii) q = 0 (iv) w = 0v
  1. i. ΔT = 0
  2. ii. Δp = 0
  3. iii. q = 0
  4. iv. w = 0
Solution

An adiabatic process is one in which no heat is exchanged between system and surroundings. Therefore q = 0.

Answer:

(iii) q = 0.

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Q.5.3The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each elementv
  1. i. unity
  2. ii. zero
  3. iii. < 0
  4. iv. different for each element
Solution

By convention, the standard enthalpy of formation of an element in its standard state is taken as zero.

Answer:

(ii) zero.

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Q.5.4ΔU° of combustion of methane is - X kJ mol^-1. The value of ΔH° is (i) = ΔU° (ii) > ΔU° (iii) < ΔU° (iv) = 0v
  1. i. = ΔU°
  2. ii. > ΔU°
  3. iii. < ΔU°
  4. iv. = 0
Solution

For CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), Δng = 1 - 3 = -2. Since ΔH = ΔU + ΔngRT, ΔH = ΔU - 2RT. Therefore ΔH is more negative than ΔU, i.e. ΔH < ΔU.

Answer:

(iii) < ΔU°.

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Q.5.5The 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.v
  1. i. -74.8 kJ mol^-1
  2. ii. -52.27 kJ mol^-1
  3. iii. +74.8 kJ mol^-1
  4. iv. +52.26 kJ mol^-1.
Solution

For CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔcH = [ΔfH(CO2) + 2ΔfH(H2O)] - ΔfH(CH4). Thus -890.3 = [-393.5 + 2(-285.8)] - ΔfH(CH4) = -965.1 - ΔfH(CH4). Hence ΔfH(CH4) = -74.8 kJ mol^-1.

Answer:

(i) -74.8 kJ mol^-1.

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Q.5.6A 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 (v) possible at any temperaturev
  1. i. possible at high temperature
  2. ii. possible only at low temperature
  3. iii. not possible at any temperature
  4. v. possible at any temperature
Solution

Heat q appears on the product side, so the reaction is exothermic and ΔH is negative. Since ΔS is positive, ΔG = ΔH - TΔS is negative at all temperatures. Therefore the reaction is spontaneous at any temperature.

Answer:

(v) possible at any temperature.

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Q.5.7In 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?v
Solution

Heat absorbed by the system is q = +701 J. Work done by the system is w = -394 J. From the first law, ΔU = q + w = 701 - 394 = 307 J.

Answer:

+307 J.

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Q.5.8The 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(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)v
Solution

For NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l), use ΔH = ΔU + ΔngRT. Gaseous product moles = 1 mol N2 + 1 mol CO2 = 2 mol. Gaseous reactant moles = 1.5 mol O2; solid cyanamide and liquid water do not count. Thus Δng = 0.5. ΔH = -742.7 + (0.5)(8.314)(298)/1000 = -742.7 + 1.24 = -741.5 kJ mol^-1.

Answer:

-741.5 kJ mol^-1.

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Q.5.9Calculate 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.v
Solution

Moles of Al = 60.0/27.0 = 2.22 mol. ΔT = 20 K. Heat required q = nCΔT = (2.22)(24)(20) = 1.07 x 10^3 J = 1.07 kJ.

Answer:

1.07 kJ.

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Q.5.10Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol^-1 at 0°C. Cp [H2O(l)] = 75.3 J mol^-1 K^-1 Cp [H2O(s)] = 36.8 J mol^-1 K^-1v
Solution

Cool liquid water from 10°C to 0°C: q1 = (1)(75.3)(0 - 10) = -753 J. Freeze at 0°C: q2 = -6.03 kJ = -6030 J. Cool ice from 0°C to -10°C: q3 = (1)(36.8)(-10 - 0) = -368 J. Total q = -753 - 6030 - 368 = -7151 J = -7.15 kJ.

Answer:

-7.15 kJ mol^-1.

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Q.5.11Enthalpy 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.v
Solution

Moles of CO2 = 35.2/44.0 = 0.800 mol. Heat released = 0.800 x 393.5 = 314.8 kJ. The enthalpy change is -314.8 kJ for this amount.

Answer:

314.8 kJ of heat is released.

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Q.5.12Enthalpies 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 ΔrH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)v
Solution

ΔrH = ΣΔfH(products) - ΣΔfH(reactants). Products: 81 + 3(-393) = -1098 kJ mol^-1. Reactants: 9.7 + 3(-110) = -320.3 kJ mol^-1. Therefore ΔrH = -1098 - (-320.3) = -777.7 kJ mol^-1.

Answer:

-777.7 kJ mol^-1.

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Q.5.13Given N2(g) + 3H2(g) → 2NH3(g); ΔrH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH3 gas?v
Solution

The given reaction forms 2 mol NH3 from elements in their standard states. Therefore ΔfH° per mole of NH3 = -92.4/2 = -46.2 kJ mol^-1.

Answer:

-46.2 kJ mol^-1.

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Q.5.14Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ; ΔrH° = -726 kJ mol^-1 C(graphite) + O2(g) → CO2(g) ; ΔcH° = -393 kJ mol^-1 H2(g) + 1/2 O2(g) → H2O(l); ΔfH° = -286 kJ mol^-1.v
Solution

For CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l), -726 = [ΔfH(CO2) + 2ΔfH(H2O)] - ΔfH(CH3OH). Substitute values: -726 = [-393 + 2(-286)] - x = -965 - x. Hence x = -239 kJ mol^-1.

Answer:

-239 kJ mol^-1.

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Q.5.15Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C-Cl in CCl4(g). ΔvapH°(CCl4) = 30.5 kJ mol^-1. ΔfH° (CCl4) = -135.5 kJ mol^-1. ΔaH° (C) = 715.0 kJ mol^-1, where ΔaH° is enthalpy of atomisation ΔaH° (Cl2) = 242 kJ mol^-1v
Solution

The formation enthalpy of CCl4(g) is ΔfH°[CCl4(l)] + ΔvapH° = -135.5 + 30.5 = -105.0 kJ mol^-1. Atomising elements to C(g) + 4Cl(g) needs 715.0 + 2(242) = 1199 kJ mol^-1. Therefore CCl4(g) → C(g) + 4Cl(g) has ΔH = 1199 - (-105.0) = 1304 kJ mol^-1. Average C-Cl bond enthalpy = 1304/4 = 326 kJ mol^-1.

Answer:

ΔH = 1304 kJ mol^-1; average C-Cl bond enthalpy = 326 kJ mol^-1.

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Q.5.16For an isolated system, ΔU = 0, what will be ΔS ?v
Solution

The second law states that entropy of an isolated system increases for a spontaneous process and becomes maximum at equilibrium. Therefore ΔU = 0 alone does not make ΔS zero; the natural direction has ΔS positive until equilibrium is reached.

Answer:

For a spontaneous process in an isolated system, ΔS > 0; at equilibrium, ΔS = 0.

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Q.5.17For the reaction at 298 K, 2A + B → C ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.v
Solution

For spontaneity, ΔG = ΔH - TΔS < 0. Thus T > ΔH/ΔS = 400/0.2 = 2000 K. At 2000 K the reaction is at the threshold; above 2000 K it becomes spontaneous.

Answer:

Above 2000 K.

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Q.5.18For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?v
Solution

Formation of a Cl-Cl bond from chlorine atoms releases energy, so ΔH < 0. Two gaseous atoms combine to form one gaseous molecule, decreasing disorder, so ΔS < 0.

Answer:

ΔH is negative and ΔS is negative.

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Q.5.19For the reaction 2 A(g) + B(g) → 2D(g) ΔU° = -10.5 kJ and ΔS° = -44.1 JK^-1. Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.v
Solution

Here Δng = 2 - 3 = -1. At 298 K, ΔH° = ΔU° + ΔngRT = -10.5 - (8.314 x 298/1000) = -12.98 kJ. Convert entropy: ΔS° = -44.1 J K^-1 = -0.0441 kJ K^-1. ΔG° = ΔH° - TΔS° = -12.98 - 298(-0.0441) = +0.16 kJ. Since ΔG° is positive, the reaction is not spontaneous under standard conditions.

Answer:

ΔG° is about +0.16 kJ mol^-1 at 298 K; the reaction is not spontaneous under standard conditions.

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Q.5.20The equilibrium constant for a reaction is 10. What will be the value of ΔG° ? R = 8.314 JK^-1 mol^-1, T = 300 K.v
Solution

Use ΔG° = -RT ln K. With K = 10, T = 300 K and R = 8.314 J K^-1 mol^-1, ΔG° = -(8.314)(300)ln10 = -5744 J mol^-1 = -5.74 kJ mol^-1.

Answer:

-5.74 kJ mol^-1.

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Q.5.21Comment on the thermodynamic stability of NO(g), given 1/2 N2(g) + 1/2 O2(g) → NO(g); ΔrH° = 90 kJ mol^-1 NO(g) + 1/2 O2(g) → NO2(g): ΔrH°= -74 kJ mol^-1v
Solution

Formation of NO from N2 and O2 is endothermic, with ΔrH° = +90 kJ mol^-1, so NO is not very stable thermodynamically relative to the elements. Its oxidation to NO2 is exothermic, ΔrH° = -74 kJ mol^-1, showing that NO tends to form the more stable oxide NO2 in oxygen.

Answer:

NO(g) is thermodynamically unstable with respect to its elements and can be oxidised exothermically to NO2.

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Q.5.22Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfH° = -286 kJ mol^-1.v
Solution

The system releases 286 kJ mol^-1 to the surroundings. At 298 K, ΔSsurr = -ΔHsys/T = -(-286000 J mol^-1)/298 K = +960 J K^-1 mol^-1.

Answer:

+960 J K^-1 mol^-1.

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