Students should go through the Thermodynamics Classification Class 11 Chemistry notes provided below. These notes have been designed based on the latest NCERT Book for Class 11 Chemistry. These revision notes cover all important topics in your CBSE books. Students should revise all Class 11 Chemistry Notes as they will help the students to understand all topics given in your books. This will help you to get good marks in the Class 11 Chemistry exams.
CBSE Class 11 Chemistry Chapter 6 Thermodynamics Notes
1. Thermodynamics: Science which deals with study of different forms of energy and quantitative relationship.
2. System & Surroundings: The part of universe for study is called system and remaining portion is surroundings.
3. State of system & state function: State of system is described in terms of T, P, V etc. The property which depends only on state of system not upon path is called state function eg. P, V, T, E, H, S etc.
4. Extensive & Intensive Properties: Properties which depends on quantity of matter called extensive prop. eg. mass, volume, heat capacity, enthalpy, entropy etc. The properties which do not depends on matter present depends upon nature of substance called Intensive properties. eg. T,P, density, refractive index, viscosity, bp, pH, mole fraction etc.
5. Internal energy: The total energy with a system. i.e. U = Ee + En + Ec + Ep + Ek + —— ΔU = U2 – U1 or UP – UR & U is state function and extensive properly. If U1 > U2 energy is released.
6. Heat (q): It I a form of energy which is exchanged between system and surrounding due to difference of temperature. Unit is Joule (J) or Calorie (1 Calorie = 4.18 J).
7. First Law of Thermodynamics: It is law of conservation energy. Energy can neither be created not destroyed, it may be converted from one from into another. Mathematically U = q + w, w = – p. ΔV (work of expansion) ΔU = q – p. ΔV or q = ΔU + p. ΔV, q,w are not state function.
But U is state function.
8. Enthalpy (H): At constant volume ΔV = 0, → qv = ΔU
So H = U + p. ΔV, qp = H2 H1 = ΔH
⇒ ΔH = ΔU + P. ΔV.
9. Relationship between qp, qv i.e. H& U
It is ΔH = ΔU+ Δng.RT or qp = qv + Δng.RT
10. Exothermic and Endothermic reactions: H = –Ve for exothermic and H = +Ve for endothermic reaction i.e. evolution and absorption of heat.
Eg C+O2 → CO2 + 393.5 KJ, ΔH = –393.5 KJ (exothermic)
N2 + O2 → 2NO – 180.7 KJ, ΔH = 180.7 KJ (Endothermic)
11. Enthalpy of reaction ( ΔrH): The amount of heat evolved or absorbed when the reaction is completed.
12. Standard Enthalpy of reaction ( ΔrH0) at 1 bar pressure and specific temp.
(290K) i.e. standard state.
13. Different types of Enthalpies of reactions:
(i) Enthalpy of combustion ( ΔcH),
(ii) Enthalpy of formation ( ΔfH)
(iii) Enthalpy of neutralization
(iv) Enthalpy of solution
(v) Enthalpy of atomization( ΔaH),
(vi) Enthalpy of Ionisation (ΔiH)
(vii) Enthalpy of Hydration ( Δhyol.H)
(viii) Enthalpy of fusion ( Δfus.H)
(ix) Enthalpy of vaporization ( Δvap.H)
(x) Enthalpy of sublimation ( Δsub.H)
( Δsub.H) = Δfus.(H) – Δvap(H)
14. Hess’s Law of constant heat summation: The total amount of heat change is same whether the reaction takes place in one step or in several steps.
i.e. ΔH = ΔH1 + ΔH2 + ΔH3 + ——–
15. Bond enthalpy: It is amount of energy released when gaseous atoms combines to form one mole of bonds between them or heat absorbed when one mole of bonds between them are broken to give free gaseous atoms.
Further ΔrH = ΣB.E. (Reactants) – ΣB.E. (Products)
16. Spontaneous & Non Spontaneous Processes: A process which can take place by itself is called spontaneous process. A process which can neither take place by itself or by initiation is called non Spontaneous.
17. Driving forces for spontaneous process: (i) Tendency for minimum energy state. (ii) Tendency for maximum randomness.
18. Entropy (S): It is measure of randomness or disorder of system.
Entropy change ( S) = q(rev.)/T, j.k-1 .mol-1
19. Spontaneity in term of ( S)
ΔS(total) = ΔS(universe) = ΔS(system) + ΔS(surrounding)
If ΔS(total) is +ve, the process is spontaneous.
If ΔS(total) is –ve, the process is non spontaneous.
20. Second Law of thermodynamics: In any spontaneous process, the entropy of the universe always increases. A spontaneous process cannot be reversed.
21. Gibb’s free energy (G): defined as G = H – T.S & ΔG = ΔH – T. ΔS (Gibb’s Helmholts equation) it is equal useful work i.e. – G = W(useful) = W(max.)
If ΔG = ve, process is spontaneous.
22. Effects of T on spontaneity of a process: ΔG = ΔH – T. ΔS.
(i) For endothermic process may be non spontaneous at law temp.
23. Calculation of ( ΔrG0)
ΔrG0 = ΣΔfG0 (p) – ΣΔfG0 (r)
24. Relationship between ( ΔrG0) & equilibrium constant (k)
ΔG = ΔG0 + RTlnQ & ΔG0 = –2.303RT logk.
25. Calculation of entropy change:
ΔrS0 = ΣΔS0 (p) – ΣS0 (r)
Important Question Thermodynamics
Question. State First Law of thermodynamics.
Ans. Energy can neither be created nor destroyed. The energy of an isolated system is constant. ΔU = q + w.
Question. What is a thermodynamic state function?
Ans. A function whose value is independent of path. eg. P, V, E, H
Question. Give enthalpy (H) of all elements in their standard state.
Ans. In standard state enthalpies of all elements is zero.
Question. From thermodynamic point to which system the animals and plants belong?
Ans. Open system.
Question. Predict the sign of ΔS for the following reactions.
Ans. ΔS is positive (entropy increases)
Question. For the reaction 2Cl(g) → Cl2(g), What will be the sign of ΔH and ΔS?
Ans. ΔH: (–ve) b/c energy is released in bond formation and
ΔS: (–ve) b/c atoms combines to form molecules.
Question. State Hess’s Law for constant heat summation?
Ans. The change of enthalpy of reaction remains same, whether the reaction is carried out in one step or several steps.
Question. What is Gibb’s Helmhaltz equation?
Ans. ΔG = ΔH – T. ΔS
Question. Define extensive properties.
Ans. Properties which depends upon amount of substance called extensive properties. Volume, enthalpy, entropy.
Question. Give relationship between ΔH, ΔU for a reaction in gaseous state.
Ans. ΔH = ΔU + ng. RT.
Question. In a process, 701J heat is absorbed and 394J work is done by system. What is change in Internal energy for process?
Ans. q = 701J, w = 394J, so ΔU = q + w = 701 – 394 = 307J.
Question. Given: N2(g) + 3H2(g) → 2NH3(g), ΔrH0 = –92.4KJ.mol–1. What is the standard enthalpy of formation of NH3(g).
Ans. ΔfH NH3(g) = – 92.4/2 = 46.2KJ.mol-1
Question. Calculate entropy change in surroundings when 1.0 mol of H2O(l) is formed under standard conditions? Given ΔH0 = –286KJmol–1.
Ans. q(rev.) = – ΔH0 = –286 KJmol-1 = 286000Jmol-1
Question. Give relationship between entropy change and heat absorbed/evolved in a reversible reaction at temperature T.
Ans. Δs = q(rev)/T
Question. What is spontaneous change? Give one example.
Ans. A process which can take place of its own or initiate under some condition.
eg. Common salt dissolve in water of its own.
Question. A real crystal has more entropy than an Ideal Crystal. Why?
Ans. A real crystal has some disorder due to presence of defects in their6. A real crystal has some disorder due to presence of defects in their structural arrangement, and Ideal crystal does not have any disorder.
Question. Under what condition, the heat evolved/absorbed in a reaction is equal to its free energy change?
Ans. In ΔG = ΔH – T. ΔS, when reaction is carried out at OK or ΔS = 0, then ΔG = ΔH.
Question. Predict the entropy change in-
(i) A liquid crystallizes into solid
(ii) Temperature of a crystallize solid raised from OK to 115K
Ans. (i) Entropy decreases b/c molecules attain an ordered state.8. (i) Entropy decreases b/c molecules attain an ordered state.(ii) entropy increase b/c from OK to 115K particles begin to move.
Question. What is bond energy? Why is it called enthalpy of atomization?
Ans. It is the amount of energy required to dissociate one mole of bonds present b/w atoms in gas phase. As molecules dissociates into atoms in gas phase so bond energy of diatomic molecules is called enthalpy of atomization.
Question. Calculate entropy change for the following process.
H2O(s) ⇔ H2O(l), is 6.0 KJ mol-1 at 00C.
Ans. H2O(s) ⇔ H2O(l) at 00C, ΔfusH = 6KJ mol-1
= 6000J mol-1
ΔTf = 00C = 0 + 273 = 273K
Question. For oxidation of iron, 4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔS is –549.4J.K-1 mol-1, at 298K. Inspite of –ve entropy change of this reaction, Why the reaction is spontaneous? (ΔrH0 = –1648×103 J.mol-1)
& ΔS(system) = -549.4QJK-1 mol-1.
ΔrS(total) = 5530 – 5494 = 4980.6 J.K-1mol-1
Since ΔrS(total) is +ve, so the reaction is spontaneous
Question. Using the bond energy of Hr = 435 KJ mol-1, Br2 = 192 KJ mol-1, HBr = 368 KJmol-1. Calculate enthalpy change for the reaction H2(g) + Br2(g) → 2HBr(g)
Ans. ΔrH0 = Σbond enthalpies(rect.) – Σbond enthalpies(prod.)
= [435 – 192] – [2 x 368] KJ mol-1
= 627 – 736 = –109KJ. Mol-1
Question. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) and –110, –393, 81 and 9.7 KJ mol-1 respectively. Find value ΔrH for the reaction N2O4(g) + 3CO(g)
Ans. ΔrH = ΔfH0(prod.) – ΔrH0(rect)
= [81 + 1179] – [9.7 – 330] = –777.7KJ
Question. For the reaction at 298K, 2A+B → C, ΔH = 400 KJ mol-1, ΔS = 0.2 KJ mol-1 K-
1. At what temperature will the reaction become spontaneous, considering ΔH, S be constant at the temp.
Ans. ΔH = 400 KJ mol-1, ΔS = 0.2 KJK-1 mol-1.
ΔG = ΔH – T. ΔS
O = 400 – 0.2 x T ( G = 0 at equilibrium)
T = 400/0.2 = 2000K, so reaction will be spontaneous above 2000K
Question. The equilibrium constant for a reaction is 10. What will be the value of ΔG0? R = 8.314J.K-1 mol-1 T = 300K.
Ans. ΔrG0 = –2.303 RT logK
= –2.303 x 8.314 x 300 x log10
= –19.147 x 300 x 1 = –5744.1J
ΔrG0 = –5.7441KJ.mol-1
Question. What do you understand by state function? Neither q nor w is a state function but q + w is a state function? Explain.
Ans.6 The property whose value depends upon state of system and is independent of path. q + w = ΔU, which is a state function as value of ΔU does not depends upon path.
Question.7 Justify the following statements:
(i) An endothermic reaction is always thermodynamically spontaneous.
(ii) The entropy always increases on going from liquid to vapour state at any temperature T.
Ans.7 (a) It is false, exothermic reaction is not always spontaneous. If ΔS = +ve and T. ΔS> H. The process will be non spontaneous even it. It is endothermic.
(b) The entropy of vapour is more than that of liquid, so entropy increases during vaporization
Question.8 Calculate the temperature above which the reduction reaction becomes spontaneous:
PbO(s) + C(s) Pb(s) + CO(g), given [ ΔH = 108.4 KJ mol-1, ΔS = 190J.K-1 mol-1].
Ans. ΔG = H – T. ΔS, at equilibrium ΔG = 0, ⇒ H = T. ΔS
So the reaction will be spontaneous above 570.52K, as above this temperature ΔG will be –ve.
Question. From the data given below at 298K for the reaction:
CH4(g) + 2O2(g) CO2(g) +2H2O(l) Calculate enthalpy of formation of CH4(g) at 298K. Given:[ ΔrH = -890.5KJ,
[ΔfH (CO2) = – 393.5KJ.mol-1 , ΔfH(H2O) = – 286.0KJ.mol-1]
Ans. ΔrH = ΔfH(CO2 ) +2 ΔfH CH4(g) – ΔfH( (O2 )
–890.5KJ = – 393.5KJ + 2v – 286 – ΔHf(CH4) – O
= ΔHf(CH4) = –75.0 KJ.mol-1.
ΔHf(CH) = – 75.0KJ.mol-1
Question.10 For the reaction NH4Cl(s) NH3(g) + HCl(g) at 250C enthalpy change ΔH = 177KJ.mol-1 and ΔS = 285J.K-1 mol-1. Calculate free energy change ΔG at 250C and predict whether the reaction is spontaneous or not.
Ans. ΔH = 177 KJ mol-1, ΔS = 285 JK-1 mol-1
Question. What is entropy? Why is the entropy of a substance taken as zero at 0K? Calculate the ΔrG for the reaction? N2(g) + 3H2(g) ⇒ 2NH3(g) at 298K The value of equilibrium constant (K) is 6.6×105, R = 8.314JK-1mol-1.
Ans. It is measure of randomness or disorder of system.
Because at O K there is complete order in the system.
ΔG0 = –2.303 RT logK = –2.303 x 8.314 x 298 x log6.6 x 105
= -5705.8[log6.6 + log 105]
= –5705.8[0.8195 + 5.0] = –5705.8 + 5.8195J
ΔG0 = –33.205 KJ mol-1.
Question. (i) What are extensive property and intensive properties?
(ii) Calculate the value of equilibrium constant (K) at 400K for 2 NOCl(g) →2NO(g) + Cl2(g).
ΔH0 = 77.2KJ.mol-1, ΔS0 = 122J.K-1 mol-1 at 400K, R = 8.314 J.K-1mol-1
Ans. (i) An extensive property is a property whose value depends on the quantity or size of matter present in the system Those properties which do not depend on the quantity or size of matter present are known as intensive properties
Question. Define standard enthalpy of formation. Calculate the enthalpy of formation of benzene from data
Ans. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation.
Question. Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
Ans. A substance has a perfectly ordered arrangement only at absolute zero.
Hence , entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from it means no heat change.
Question. The equilibrium constant for a reaction is one or more if ΔG° for it is less than zero. Explain
Ans. ―ΔG° = RT ln K, thus if ΔG° is less than zero. i.e., it is negative, then ln K will be positive and hence K will be greater than one.
Question. Many thermodynamically feasible reactions do not occur under ordinary conditions. Why?
Ans. Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.