Ncert Solution of Thermodynamics: As we know that most of the physical and chemical changes are accompanied by energy changes. a These energy changes may take place in the form of heat, light, work, electrical energy, etc. All these forms of energy can be converted into one another. We shall learn about their changes in this chapter. This conversion is the main theme of this chapter. Hence, thermodynamics is defined as the branch of science which deals with the study of different forms of energy and the quantitative relation ship between them.
The study of thermodynamics (Ncert Solution of Thermodynamics) is based on three generalizations called first, second and third Laws of thermodynamics. These laws apply only when the system is in equilibrium or move from one equilibrium state to another equilibrium state. These laws have been arrived at purely on the basis of human experience and there is no theoretical proof for any of these laws. However, the validity of these laws is supported by the fact that nothing contrary to these laws has been found so far and nothing contrary is expected.
Ncert Solution of Thermodynamics: Class 11 Chemistry free Pdf Download
Ncert Solution of Thermodynamics is very important for all the students who are preparing for Neet, Jee(mains), IIT advance and other board exams. Here we have solved all the questions given in the exercise of the ncert chemistry for class 11th and make available free download pdf. Let’s start to solve the questions.
Questions and Answers of Ncert Solution of Thermodynamics
Q.No. 1. Choose the correct answer. A thermodynamic state function is a quantity (Ncert Solution of Thermodynamics)
(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:- A Thermodynamic state function are those properties whose value depends on the initial state and final state of the system and independent of the path hence answer will be (ii).
Q.No. 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:- In adiabatic condition, there is no change in of heat between system and surrounding hence q = 0. Therefore the answer will be (iii).
Q.No.3. The enthalpies of all elements in their standard states are:(Ncert Solution of Thermodynamics)
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Ans:- The enthalpies of all elements in their standard state is zero. Therefore, the answer of this question will be (ii).
Q.No.4. ΔU⊖ of combustion of methane is −X kJ mol−1. The value of ΔH⊖ is(Ncert Solution of Thermodynamics)
(i)=ΔU⊖
(ii) >ΔU⊖
(iii) < ΔU⊖
(iv) = 0
Ans:- As we know that
ΔHθ = ΔUθ + ΔngRT and given ΔUθ=−Xkmol−1 ∴ ΔHθ=(−X) + ΔngRT
Or ΔHθ< ΔUθ Therefore, the answer will be (iii).
Q.No.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3kJmol−1 −393.5kJmol−1, and −285.8kJmol−1 respectively. Enthalpy of formation of CH4 will be (Ncert Solution of Thermodynamics)
(i)−74.8kJmol−1
(ii)−52.27kJmol−1
(iii)+74.8kJmol−1
(iv)+52.26kJmol−1
Ans:- Given:
(i) CH4(g)+2O2(g)⟶CO2(g)+2H2O(l) ΔH=−890.3 kJmol−1
(ii)C(s)+O2(g)⟶CO2(g) ΔH=−393.5kJmol−1
(iii)2H2(g)+O2(g)⟶2H2O(l) ΔH=−285.8kJmol−1
Aim: c(s) + 2H2(g) → CH4(g), ΔH = ? Now Eqn. (ii) + 2 × Eqn. (iii) – Eqn. (i) gives the required equation with ΔH = – 393.5 + 2(– 285.8) – (–890.3) KJmol-1 = – 74.8 KJmol-1. Hence, (i) is the correct answer.
Q.No.6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be(Ncert Solution of Thermodynamics)
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Ans:- Here ΔH = –ve and ΔS = +ve. ΔG = ΔH – TΔS. For the reaction to be spontaneous, ΔG should be negative which will be so at any temperature. Hence the Option (iv) is correct.
Q.No.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? (Ncert Solution of Thermodynamics)
Ans:- As given, q = + 70 J, w = – 394 J, ΔU = ? By first law of Thermodynamics ΔU = q + w = + 70 J + (– 394 J) = + 307 J ∴ internal energy of the system increases by 307 J.
Q.No.8. The reaction of cyanamide,NH2CN(s) with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7kJmol−1 at 298 K. Calculate enthalpy change for the reaction at 298 K. (Ncert Solution of Thermodynamics)
NH2CN(g) + 3/2O2(g)→N2(g)+CO2(g)+H2O(l)
Ans:- For the given reaction, Δng = (np – nr)g = 2 – 3/2 = +1/2 mol and ΔH = ΔU + ΔngRT. Hence ΔH = – 742.7 KJmol-1 + (+1/2 mol)(8.314 × 10–3 KJmol-1 K–1) (298 K) = – 742.7 + 1.2 KJmol-1 = – 741.5 KJmol-1 .
Q.No.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 Jmol−1K−1.(Ncert Solution of Thermodynamics)
Ans:- According to the question, n = 60/27 mol, C (molar heat capacity) = 24 Jmol−1K−1 and ΔT = (55 – 35) K = 20 K. q= n × C × ΔT = (60/27) mol × (24 Jmol−1K−1 ) × 20 K = 1066.7 J = 1.07 kJ.
Q.No.10. Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at
–10.0°C.ΔfusH=6.03kJmol−1 at 0∘C
Cp[H2O(I)]=75.3Jmol−1K−1
Cρ[H2O(s)]=36.8Jmol−1K−1
Ans:- ΔH for 1 mol water at 10oC → 1 mol of water at 0oC = Cp[H2O(I)] × ΔT = 75.3 Jmol−1K−1 × (– 10 K) = – 753 Jmol–1
ΔH for 1 mol water at 0oC → 1 mol of ice at 0oC = ΔfreezH= – 6.03 kJmol−1
ΔH for 1 mol ice at 0oC → 1 mol of ice at –10oC = Cρ[H2O(s)] × ΔT =36.8Jmol−1K−1 × (— 10 K) = =368Jmol−1K−1
Now Total ΔH = — 0.753 KJmol-1 – 6.03 kJmol−1 – 0.368Jmol−1K−1 = – 7.151 kJmol−1
Q.No.11. Enthalpy of combustion of carbon to CO2 is –393.5−7.151kJmol−1 . Calculate the heat released upon formation of 35.2 g ofCO2 from carbon and dioxygen gas. (Ncert Solution of Thermodynamics)
Ans:- Chemical equation for the above reaction can be written as:
C(s) + O2(g)⟶CO2(g), ΔfH = −393.5 kJmol−1 and 1 mol CO2(g) = 44 gram. Heat released when 44 g CO2(g) is formed = 393.5 KJ. ∴ Heat released when 35.2 gram CO2(g) is formed = (393.5/44) × 35.2 kJ = 314 .8 kJ.
Q.No.12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are −110,−393,81 and 9.7kJmol−1 respectively.
Find the value of ∆H for the reaction:
N2O4(g)+3CO(g)→N2O(g+3CO2(g]
Ans:- Enthalpy of a reaction (ΔrH) is the difference value of the enthalpy change of the formation of products and the enthalpy change of the formation of reactants. Hence formula can be written as
Δ,H=∑Δ,H( products )−∑ΔfH( reactants )
For the given reaction,
N2O4(g) + 3CO(g)⟶ N2O(g) + 3CO2(g)
ΔrH=[ {ΔfH(N2O) + 3ΔJH(CO2)}−{ΔfH (N2O4) + 3ΔjH(CO)} ] = [81+ 3(–393)] + [9.7 + 3(—110)] kJ = – 777.7 KJ.
Q.No.13. Given
N2(g) + 3H2(g) ⟶ 2NH3(g);
ΔrHθ=−92.4kJmol−1
What is the standard enthalpy of formation ofNH3 gas? (Ncert Solution of Thermodynamics)
Ans:- Reaction for the formation of NH3 (g) is 1/2 N2(g) + 3/2H2(g) ⟶ NH3(g) ∴ ΔfHo of NH3= – 92.4/2 = – 46.2 KJmol-1
Q.No.14. Calculate the standard enthalpy of formation ofCH3OH(l) from the following data:(Ncert Solution of Thermodynamics)
CH3OH(l)+3/2O2(g]→CO2(g)+2H2O(l): Δ,H∘=−726kJmol−1
C(graphite) +O2(g)→CO2(g]: ΔcHo=−393kJmol−1
H2(g)+1/2O2(g)→H2O(1); Δ,H=−286kJmol−1
Ans:- Our aim is C(s) + 2H2(g) + 1/2 O2(g) → CH3OH (l), ΔfHo = ?
Eqn. (ii) + 2 × Eqn. (iii) – Eqn. (i) gives the required equation with. ΔH = —393 + 2 × (– 286) – (– 726) KJmol-1 = — 239 KJmol-1.
Q.No.15. Calculate the enthalpy change for the process CCl4(g)→C(g)+4Cl(g) and calculated bond enthalpy of C−Cl in CCl4(g) (Ncert Solution of Thermodynamics)
ΔvapHθ (CCl4) = 30.5kJmol−1 ΔfHθ(CCl4) =−135.5kJmol−1
ΔaHθ(C) = 715.0kJmol−1, where ΔaHθ is enthalpy of atomisation
Δ2Hθ(Cl2) = 242kJmol−1
Ans:- The given data can be expressed as (1) CCl4(l) → CCl4(g) ; ΔHo = 30.5 kJmol−1
(2) C(s) → C(g) ΔHo = 715 kJmol−1
(3) Cl2(g) → 2Cl(g) ; ΔHo = 242 kJmol−1
(4) C(s) + 2Cl2(g) → CCl4(l); ΔHo = –135 kJmol−1
Our Aim: CCl4(g) → C(g) + 4 Cl(g), ΔH = ? Eqn. (2) + 2 × Eqn. (3) – Eqn. (1) – Eqn.(4) gives the required equation with. ΔH = 715.0 + 2 (242) – 30.5 – (– 135.5) kJmol−1 = 1304 kJmol−1 ∴ Bond enthalpy of C — Cl in CCl4 (average value) = 1304/4 = 326 kJmol−1
Q.No.16. For an isolated system, ∆U = 0, what will be ∆S ? (Ncert Solution of Thermodynamics)
Ans:- ΔU = 0 means that energy factor has no role to play. Hence, for the process to be spontaneous, entropy factor should favour the process hence ΔS must be +ve. This process can be understood by the mixing of gases kept in two bulbs connected by a stop cock and isolated from the surroundings where ΔU = 0.
Q.No.17. For the reaction at 298 K,
2A + B → C
∆H = 400kJmol−1
and ∆S = 0.2kJmol−1
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.
Ans:- In this solution, let’s first calculate the temperature at which the reaction will be in equilibrium that means ΔG = 0. Now, ΔG = ΔH – TΔS ∴ 0 = ΔH – TΔS or T = ΔH/ΔS = 400 KJmol-1/ 0.2 kJK–1mol–1= 2000 K . Now it can be concluded that for reaction to be spontaneous, for ΔG to be –ve, T should be greater than 2000 K.
Q.No.18. For the reaction,2Cl(g)→Cl2(g) , what are the signs of ∆H and ∆S ? (Ncert Solution of Thermodynamics)
Ans:- The given reaction represents the formation of bonds. Hence, energy is released that means ΔH is negative. Further, 2 moles of atoms have greater randomness than 1 mole of molecule. Hence, randomness decreases that means ΔS is –ve.
Q.No.19. For the reaction (Ncert Solution of Thermodynamics)
2A(g)+B(g)→2D(g)ΔUe=−10.5kJ and ΔS∘=−44.1JK−1
Calculate ΔG⊖ for the reaction, and predict whether the reaction may occur spontaneously
Ans:- For the given reaction, Δng = 2 – 3 = – 1 and ΔH = ΔU + ΔngRT. Hence ΔH = – 10.5 KJ + (–1) (8.314 × 10–3 kJ)(298) = –10.5 – 2.48 = – 12.98 kJ.
Now ΔGo = ΔHo – TΔSo = –12.98 kJ – 298(– 44.1 × 10–3 kJ ) = –12.98 kJ + 13.14 kJ = 0.16 kJ.
Q.No.20. The equilibrium constant for a reaction is 10. What will be the value of ∆G⊖ ? R = 8.314JK−1 mol−1
T = 300 K.
Ans:- ΔGo = – 2.303 RT log K = – 2.303 × 8.314 JK-1mol-1 × 300 K × log 10 = –5744.1 J.
Q.No.21. Comment on the thermodynamic stability of NO(g) , given
12N2(g)+12O2(g)→NO(g);
ΔrH⊖= 90kJmol−1NO(g) + 12O2(g)→NO2(g):
ΔrHe= −74kJmol−1
Ans:- As energy is absorbed in the first reaction, NO (g) is unstable. As energy is released in the second reaction, NO2 (g) is stable. Therefore, unstable NO(g) changes into the stable NO2(g).
Q.No.22. Calculate the entropy change in surroundings when 1.00 mol ofH2O(l) is formed under standard conditions.ΔHθ=−286kJ mol−1 (Ncert Solution of Thermodynamics)
Ans:- H2(g) + 1/2 O2(g) → H2O (l) ΔfHo = – 286 KJmol-1. This means that when 1 mol of H2O(l) is formed, 286 kJ of heat is released and this heat is absorbed by the surroundings, i.e., qsurr = + 286 KJmol-1 ∴ ΔS = qsurr /T = 286 KJmol-1/298 K = 0.9597 kJK–1mol–1= 959.7 JK-1mol-1.
Conclusion of Ncert Solution of Thermodynamics
Here, we have published the solution of all the questions given in the exercise of NCERT text book of chemistry for 11th. This is the fifth chapter of chemistry according to the new revised Syllabus for cbse board exam and also ncert Syllabus. There are total 22 questions in this exercise. Answers of all the questions have been written in proper manner using the maximum concepts like all the laws of Thermodynamics, state functions and Hess law.
We hope that Ncert Solution of Thermodynamics you all have liked very much. If you have any doubts or suggestions, please contact us through my email. You are kindly requested to share this article among your friends and favourites. Thanks for coming this website.