Important Questions for CBSE Class 11 Chemistry Chapter 6 – Thermodynamics
Important Questions Class 11 Chemistry Chapter 6 – Thermodynamics
Chemistry involves the study of various experiments and their inference. For one to be good in Chemistry, one should have a clear understanding of every concept covered in the chapter. So, we provide the Important Questions Class 11 Chemistry Chapter 6 for students to understand every aspect thoroughly.
Thermodynamics is the study of heat and its behaviour. The behaviour varies in terms of Physics and Chemistry. This chapter includes vital topics like Thermodynamic Terms, Applications, Measurement Of ∆U And ∆H: Calorimetry, Enthalpy Change, ∆Rh Of Reaction – Reaction Enthalpy, Enthalpies For Different Types Of Reactions, Spontaneity And Gibbs Energy Change and Equilibrium.
The detailing of this chapter, along with lots of important questions, is covered in the Important Questions Class 11 Chemistry Chapter 6 for students to have a deep insight into the chapter and all the fundamentals related to it. Hence, it proves to be a vital resource for the students.
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|CBSE Class 11 Chemistry Important Questions|
|Sr No||Chapters||Chapter Name|
|1||Chapter 1||Some Basic Concepts of Chemistry|
|2||Chapter 2||Structure of Atom|
|3||Chapter 3||Classification of Elements and Periodicity in Properties|
|4||Chapter 4||Chemical Bonding and Molecular Structure|
|5||Chapter 5||States of Matter|
|8||Chapter 8||Redox Reactions|
|10||Chapter 10||The s-Block Elements|
|11||Chapter 11||The p block Elements|
|12||Chapter 12||Organic Chemistry – Some Basic Principles and Techniques|
|14||Chapter 14||Environmental Chemistry|
Thermodynamics Class 11 Important Questions with Answers
The following Important Questions and their solutions are included in the Chemistry Class 11 Chapter 6 Important Questions
Question 1: The thermodynamic state function is the quantity
(i) used to determine the heat changes
(ii) whose value is independent of the path
(iii) used to determine the pressure, volume and work
(iv) whose value is dependent on the temperature only
Answer 1: (ii) A quantity that is independent of the path.
Explanation: Functions like the pressure, volume and temperature are dependent on the state of the system only, not on the path.
Question 2: For the process to happen under the adiabatic conditions, the correct condition can be given by:
(i) ∆T = 0 (ii) ∆p = 0
(iii) q = 0 (iv) w = 0
Answer 2: (iii) q = 0
Explanation: For the adiabatic process, the heat transfer is zero, that is, q = 0.
Question 3: The enthalpies for all the elements in their standard states are given as:
(iii) < 0
(iv) Different from each of the element
Answer 3: (ii) Zero
Question 4. 18.0 g for the water completely evaporates at 100°C as well as 1 bar pressure, as well as the enthalpy change for the process is 40.79 kJ mol-1. What would be the enthalpy change for vaporising the two moles of the water under the given conditions? What is the standard enthalpy for the vaporisation of the water?
Answer 4. Given condition,
The amount of the water is 18.0 g, as well as the pressure, is 1 bar.
Now, we know that 18.0 g H2O = 1 mole H2O.
Enthalpy change of vaporising the 1 mole of H2O = 40.79 kJ mol-1
So, Enthalpy change of vaporising the 2 moles of H2O = 2 × 40.79 kJ = 81.358kJ
Standard enthalpy for evaporation at 100℃ and 1 bar pressure is given as,
Δvap H2O = + 40.79 k J mol-1.
Question 5. One mole of the acetone needs less heat to evaporate as compared to the 1 mol of the water. Which of the following two liquids has higher enthalpy for the vapourisatIon?
Answer 5. Acetone needs less heat to vaporise because of the weak force of attraction between the molecules.
As a result, the water has a higher enthalpy of evaporation.
It can be given as:
Δvap(water) > Δvap(acetone)
Question 6. Standard molar enthalpy for the formation, ΔfHӨ, is just the special case of the enthalpy of the reaction, ΔrHӨ. Is the ΔrHӨ for the following reaction equal to ΔfHΘ? Give proper reasons for your answer.
CaO(s) + CO2(g) → CaCO3(s); ΔfHӨ = –178.3 kJ mol-1
Answer 6. No, the ΔrHӨ for the given reactions is not equal to ΔfHӨ.
The standard enthalpy change to the formation of the one mole of the compound from the elements in their most stable states is known as the standard molar enthalpy for formation, ΔrHӨ.
CaO(s) + C(s) + 3/2 O2(g) → CaCO3(s)
This reaction is not the same as the given reaction.
Thus, ΔrHӨ ≠ ΔfHӨ.
Question 7. The value for the ΔfHӨ of the NH3 is
– 91.8 KJ mol-1. Calculate the enthalpy change for the given reaction:
2NH3 (g) → N2 (g) +3H2 (g)
Answer 7. ΔfHӨ is given for the formation. For the given reverse reaction, ΔfHӨ changes sign due to the opposite of the exothermic reaction in the endothermic reaction.
Thus, for the given one mole, ΔfHӨ for the decomposition is – (-91.8) = 91.8.
But the two moles are decomposing here.
ΔfHӨ = 2 × 91.8 = 183.6 KJ mol-1.
Question 8. Enthalpy is an extensive property. In general, if the enthalpy for the overall reaction A→B along one route is considered as ΔrH and ΔrH1, ΔrH2, ΔrH3 ….. represent enthalpies of the intermediate reactions leading to the formation of the product B. What will be the relation between the ΔrH for the overall reaction as well as the ΔrH1, ΔrH2 ….. etc., for the intermediate reactions?
Answer 8. In general, if the enthalpy of the overall reaction A→B along one route is considered as ΔrH and ΔrH1, ΔrH2, ΔrH3 ….. represent enthalpies of reaction leading to the formation of the same product B through the other route,
So, we get, ΔrH =ΔrH1 + ΔrH2 + ΔrH3 +…..
For the general reaction, Hess’s law of the constant heat addition can be given as.
Question 9. The enthalpy of the atomisation for the reaction CH4(g)→ C(g) + 4H (g) is 1665 kJ mol-1. What is the bond of energy of the C–H bond?
Answer 9. Enthalpy of the atomisation for the 4 moles of the C−H bonds = 1665 kJ mol-1
Hence, C−H bond energy for given per mole = 1665 kJ mol-1/4 = 416.2 kJmol-1.
Question 10. Use the given data to calculate the lattice Δ HӨ for NaBr.
Δsub HӨ for the sodium metal = 108.4 kJ mol-1
Ionisation enthalpy of the sodium = 496 kJ mol-1
Electron gain enthalpy for the bromine = – 325 kJ mol-1
Bond dissociation enthalpy for the bromine = 192 kJ mol-1
ΔfHӨ of the NaBr (s) = – 360.1 kJ mol-1
Answer 10. According to Hess’s Law,
ΔfHӨ = Δsub HӨ + ΔIEHӨ + ΔdissHӨ + ΔtgHӨ + U
ΔsubHӨ of the Na metal = 108.4 kJ/mol
Ionisation Enthalpy of Na = 496 k/mol
ΔtgHӨ for Br = – 325 kJ mol–1
ΔdissHӨ for Br = 192 kJ mol–1
ΔfHӨ of NaBr = – 360.1 kJ mol–1
ΔfHӨ = ΔsubHӨ +Ionisation Enthalpy of Na + ΔdissHӨ + ΔtgHӨ + U
-360.1 = 108.4 +496 + 96 + (-325) – U
U = +735.5 kJ/mol
Question 11. Given the condition, ΔH = 0 for the mixing of the two gases. Explain if the diffusion of these gases into each other in the closed container is a spontaneous process or not.
Answer 11. The diffusion will be a spontaneous process. As the change in the enthalpy is zero, the change in the randomness or disorder that is ΔS increases. As a result, for the equation ΔG = ΔH – TΔS, the term TΔS will be given as negative. Thus, ΔG will be negative. Hence the process will be spontaneous.
Question 12. Heat has a randomising influence on the system as well as temperature is the measure of the average chaotic motion of the particles in the system. Write the mathematical relation which relates to these three parameters.
Answer 12. Heat has a randomising influence on the system as well as temperature is the measure of the average chaotic motion of the particles in the system.
The mathematical relation that relates these three parameters is given as
ΔS = qrev / T
Where ΔS = change in the entropy
qrev = heat of the reversible reaction
T = temperature
Question 13. An increase In the enthalpy of the surroundings is the same as the decrease In the enthalpy of the system. Will the temperature of the system and its surroundings be equal if they are in thermal equilibrium?
Answer 13. Yes, if the system, as well as the surroundings, are in thermal equilibrium, their temperatures are equal. Also, an increase in the enthalpy of the surroundings is equal to a decrease in the enthalpy of the system.
Question 14. At the temperature 298 K, the Kp of the reaction is given as N2O4 (g) ⇄ 2NO2 (g) is 0.98. Predict when the reaction is spontaneous or not.
Answer 14. For the spontaneous reaction,
ΔrG° is given as -ve.
ΔrG° = – RT ln Kp = – RT ln (0.98)
In (0.98) is – 0.02
ΔrG° = – RT × –0.02
Δ will be positive.
Thus, the reaction will be considered as non-spontaneous.
Question 15. The sample for 1.0 mol of the monatomic ideal gas is taken through the cyclic process of expansion as well as compression, as shown in Fig. 6.1. What will be the value for the ΔH of the cycle as the whole?
Answer 15. The change in the internal energy during the cyclic process is zero.
ΔU = 0
When the system returns to the initial state, no work is said to have taken place.
Enthalpy for the steady-state cyclic process is one single value at any given stage, but it is different at the different stages.
ΔH = 0
Question 16. The standard molar entropy for H2O (l) is 70 J K-1 mol-1. Will the standard molar entropy for H2O (s) be more or less as compared to 70 J K-1 mol-1?
Answer 16. The solid form for the H2O is ice. In ice, the molecules are less random as compared with liquid water.
So, the molar entropy of H2O (s) < the molar entropy of H2O (l).
The standard molar entropy for H2O (s) is less compared with 70 J K-1 mol-1.
Question 17. Identify the state functions as well as path functions for the following: enthalpy, entropy, heat, temperature, work, and free energy.
Answer 17. State Functions are Enthalpy, Entropy, Temperature, and Free energy.
Path Functions are Heat and Work.
Question 18. The molar enthalpy for the vaporisation of acetone is less as compared with that of the water. Why?
Answer 18. As the acetone lacks a hydrogen bond, the intermolecular forces are weaker, causing it to boil/evaporate quickly, causing lower the molar enthalpy for the vaporisation. Moreover, as acetone lacks the polar O-H bond, it has a low enthalpy. Water has both the non-polar region as well as a strong hydrogen bond.
Question 19. Which quantity out of the following ΔrG and ΔrGӨ will be zero at equilibrium?
Answer 19. ΔrG will always be zero.
ΔrGӨ is zero for K = 1 as ΔrGӨ = – RT ln K
ΔrGӨ will be considered non-zero for the other values of K.
Question 20. Predict the change in the internal energy for the isolated system at the constant volume.
Answer 20. There is no energy transfer as the heat or the work in the isolated system,
Thus, w=0 and q=0.
According to the first law of thermodynamics-
Δ U = q + w = 0 + 0 = 0
ΔU = 0
Question 21. Although heat is the path function, heat absorbed by the system under certain specific conditions is independent of the path. What are the conditions for it? Explain.
Answer 21. (i) At the constant volume-
By the first law of thermodynamics:
q = ΔU + (–w)
(–w) = pΔV
q = ΔU + pΔV
ΔV = 0,
As the volume is constant.
qv = ΔU + 0 ⇒ qv = ΔU = change in the internal energy
(ii) At the constant pressure
qp = ΔU + pΔV
However, ΔU + pΔV = ΔH
qp = ΔH = change in the enthalpy.
Hence, at the constant volume as well as at the constant pressure, the heat change is the state function as it is equal to the change in the internal energy as well as the change in the enthalpy respectively that are in the state functions.
Question 22. Expansion of the gas in the vacuum is called free expansion. Calculate the work done as well as the change in the internal energy if 1 litre of the ideal gas expands isothermally into a vacuum until the total volume is 5 litres?
(–w) = pext (V2 – V1) = 0 × (5 –1) = 0
For the isothermal expansion,
q = 0
By the first law of thermodynamics-
q = ΔU + (–w)
0 = ΔU + 0
Hence, ΔU = 0
Question 23. Heat capacity (Cp) is the extensive property; however, the specific heat (c) is the intensive property. What will be the relation between the Cp and c for 1 mol of the water?
For the water, the heat capacity is = 18 × specific heat
And Cp = 18 × c
Specific heat is
c = 4.18 Jg-1 K-1
Heat capacity is
Cp = 18 × 4.18 JK-1 = 75.3 J K-1.
Question 24. The difference between the Cp and Cv can be derived as the empirical relation H = U + pv. Calculate the difference between the Cp and Cv for the ten moles of the ideal gas.
For the 1 mole of the gas,
Cp – Cv = R
For n moles of the gas, the relation is given as
Cp – Cv = nR = 10 × 4.184 J
Cp – Cv = 41.84 J.
Question 25. When the combustion of 1g of the graphite produces 20.7 kJ of heat, what will be the molar enthalpy change? Give the significance of the sign also.
The molar enthalpy change of the graphite = enthalpy change for the 1g carbon × molar mass of the carbon
ΔH = – 20.7 kJ g-1 × 12 g mol-1
ΔH = – 2.48 × 102 kJ mol-1.
The exothermic nature of the reaction is indicated with the negative sign for ΔH.
Question 26. The net enthalpy change for the reaction is the amount of energy needed to break all the bonds from the reactant molecules minus the amount of energy needed to form all the bonds in the product molecules. What will be the enthalpy change for the given reaction?
H2(g) + Br2(g) → 2HBr(g)
Given that the Bond energy of the H2, Br2 and HBr is 435 kJ mol-1, 192 kJ mol-1, as well as 368 kJ mol-1, respectively.
Answer 26. ΔrHӨ = Bond energy for H2 + Bond energy for Br2 – 2 × Bond energy for HBr
ΔrHӨ = 435 + 192 – (2 × 368) kJ mol-1
ΔrHӨ = –109 kJ mol-1.
Question 27. What will be the work done of an ideal gas enclosed in the cylinder if it is compressed by the constant external pressure, text in the single step as shown in Fig. Explain it graphically.
Answer 27. Work done on the ideal gas can be calculated based on the area covered by the P-V graph (shaded region) is the actual value of the work don = length × breadth = pext ∆V = AVI (or BVII ) × (VI-VII )
Question 28. How will you calculate the work done on the ideal gas in the compression if a change in the pressure Is carried out in infinite steps?
Answer 28. The process or change is said to be reversible when it is carried out in such a way that the process can be reversed at any time by the infinitesimal change.
When pressure is not constant and changes occur in it at the infinite steps (reversible conditions) during the compression from the initial volume, Vi, to the final volume, Vf, work done could be calculated using the pV-plot. The shaded area shows the work done on the gas.
Question 29. Represent the potential energy/enthalpy change in the given processes graphically.
(a) Throwing a stone from the ground to the roof.
(b) ½ H2(g) + ½ Cl2(g) ⇄ HCl(g); ΔrHΘ = –92.32 kJ mol-1
In which of the processes potential energy/enthalpy change is the contributing factor to the spontaneity?
Answer 29. (a) When throwing the stone from the ground to the roof, we must give energy to the stone.
(b) When the heat is produced in the reaction, it indicates that the decrease follows the process in the energy.
Energy increases in (a) part while decreases in (b). As a result, in process (b), the enthalpy change is the factor that contributes to spontaneity.
Question 30. Enthalpy diagram for the particular reaction is given in Fig. Is it possible to decide the reaction’s spontaneity from the given diagram? Explain.
Answer 30. No, enthalpy is not the only criterion for determining spontaneity; we must also consider entropy.
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Q.1 Calculate the free energy of the reaction
Also predict whether the reaction would be feasible or not
Q.2 Answer the following questions (any 2):
(a) What do you understand by the term standard enthalpy of sublimation
(b) Describe one application of Hess law.
(c) Predict the entropy change for the reaction:
(a) Standard enthalpy of sublimation is defined as the enthalpy change associated with one mole of a solid substance getting converted to its vapour state at constant temperature and standard pressure (1bar). It is represented as Hsub°.
(b) An application of Hesss law is that it is used to determine the enthalpy of formation of substances that cannot be measured experimentally.
(c) Both hydrogen and oxygen in the reactants side are in gaseous state and have high entropy. They combine to form a molecule of water in its liquid state which has lower entropy. Therefore, during the reaction, the entropy decreases.
Q.3 (i) Calculate the change in internal energy of a system it if it absorbs 20 kJ of heat and does 10 kJ of work
(ii) Write the third law of thermodynamics.
ii) The entropy of a perfect crystalline substance is zero at zero Kelvin.
Q.4 Pick the correct conditions for an endothermic reaction.
(a) rH is negative and qv is positive
(b) rH is positive and qp is negative
(c) Both rH and qp are negative
(d) Both rH and qv are positive
The heat absorbed or evolved (qp) at constant pressure is also called the heat of reaction or enthalpy of reaction rH.
rH = qp
In an endothermic reaction, heat is absorbed, therefore, qp is positive and rH will also be positive. Whereas, in an exothermic reaction, heat is evolved and system loses heat to the surroundings. Therefore, qp and rH will be negative for an exothermic reaction.
Q.5 A quantity which is determined by bomb calorimeter is
Bomb calorimeter is made up of a steel container with rigid walls to avoid its bursting when pressure is made to rise at a constant volume.
If a process is associated with the heat absorbed at constant volume then it is measured in a bomb calorimeter.
FAQs (Frequently Asked Questions)
1. Where can I get the important Questions Class 11 Chemistry Chapter 6?
You can get the Chapter 6 Class 11 Chemistry Important Questions from the Extramarks website, which covers all the crucial questions from the chapter and their solutions, ensuring students’ success in school as well as competitive examinations.
2. What are the essential topics covered in the chapter 'Thermodynamics'?
The essential topics covered in the chapter ‘Thermodynamics’ includes Thermodynamic Terms, Applications, Measurement Of ∆U As well as ∆H: Calorimetry, Enthalpy Change, ∆rh Of A Reaction – Reaction Enthalpy, Enthalpies for Different Types Of Reactions, Spontaneity and Gibbs Energy Change and Equilibrium.