NCERT Solutions Class 11 Chemistry Chapter 6 – Free PDF Download
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics can be found on this page. Here, students can access detailed and explanative solutions according to the latest CBSE Syllabus 2022-23 for all the questions present in the NCERT textbook. This chapter contains various subdivisions like closed, open and isolated systems, isothermal and free expansion of gas, internal energy as a state function etc. NCERT Solutions for Class 11 Chemistry contain a brief description of all these important topics to enable a high conceptual knowledge among students.
Chemistry is a branch of Science which deals with molecules, atoms, states of matter, thermodynamics, elements etc. Students have to pay attention to each of these concepts and learn them completely. Many students do not identify the crucial concepts in each chapter. So, following the NCERT Solutions available at BYJU’S will help you attain remarkable grades in the exam. Furthermore, the NCERT Solutions provided on this page can be downloaded as a PDF for free by clicking the download button provided above.
NCERT Solutions for Class 11 Chemistry 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 NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics
- Thermodynamic Terms
- The System and the Surroundings
- Types of Thermodynamic Systems
- State of the System
- Internal Energy as a State Function
- Enthalpy, H
- Measurement Of ΔU And ΔH: Calorimetry
- Enthalpy Change and Reaction Enthalpy
- Enthalpies For Different Types Of Reactions
- Gibbs Energy Change And Equilibrium
Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 6
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
(ii) A quantity which is independent of path.
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
(iii) q = 0
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
Q-4: ∆U0 of combustion of methane is – X kJ mol–1. The value of ∆H0 is
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 CH4(g) will be
(i) -74.8 kJ
(ii) -52.27 kJ
(iii) +74.8 kJ
(iv) +52 kJ
1. CH4(g) + 2O2(g)
2. C(s) + O2(g)
3. 2H2(g) + O2(g)
C(s) + 2H2(g)
= [ -393.5 +2(-285.8) – (-890.3)] kJ
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
(iv) possible at any temperature
As per given in question,
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?
As per Thermodynamics 1st law,
W = work done
W = -594 J (work done by system)
q = +701 J (+ve as heat is absorbed)
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 O2(g) → N2(g) + CO2(g) + H2O(l)
= (2 – 1.5) moles
Now, from (1)
= -742.7 + 1.2
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.
Expression of heat(q),
c = molar heat capacity
m = mass of substance
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.
(a) Energy change that occurs during transformation of 1 mole of water from
(b) Energy change that occurs during transformation of 1 mole of water at
(c) Energy change that occurs during transformation of 1 mole of ice from
Thus, the required change in enthalpy for given transformation is -7.151
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.
Formation of carbon dioxide from di-oxygen and carbon gas is given as:
C(s) + O2(g) → CO2(g);
1 mole CO2 = 44g
Heat released during formation of 44g CO2 = -393.5
Therefore, heat released during formation of 35.2g of CO2 can be calculated as
Q-12: Enthalpies 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 ∆rH for the reaction:
N2O4(g) + 3CO(g)→ N2O(g) + 3 CO2(g)
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
Now, substituting the given values in the above equation, we get:
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?
“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)N2(g) + (1.5)H2(g) → 2NH3(g)
Therefore, Standard Enthalpy for formation of ammonia gas
Q-14: Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH(l) + 3/2 O2(g)→ CO2(g) + 2H2O(l);
C(g) + O2(g)→ CO2(g);
H2(g) + 1/2 O2(g)→ H2O(l);
C(s) + 2H2O(g) + (1/2)O2(g) → CH3OH(l) …………………………(i)
CH3OH(l) can be obtained as follows,
= (-393 – 572 + 726)
Q-15: Calculate the enthalpy change for the process
CCl4(g)→ C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).
“ The chemical equations implying to the given values of enthalpies” are:
(1) CCl4(l) à CCl4(g) ;
(2) C(s) à C(g)
(3) Cl2(g) à 2Cl(g) ;
(4) C(g) + 4Cl(g) à CCl4(g);
The value of bond enthalpy for C-Cl in CCl4(g)
Q-16: For an isolated system, ∆U = 0, what will be ∆S?
Q-17: For the reaction at 298K,
2A + B → C
At what temperature will the reaction become spontaneous considering
Let, the given reaction is at equilibrium, then
Therefore, T = 2000K
Thus, for the spontaneous,
Q-18: For the reaction
What are the signs of
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,
Also, 2 moles of Chlorine atoms are having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus,
Q-19: For the reaction
2A(g) + B(g)→ 2D(g)
2A(g) + B(g) → 2D(g)
= -1 mole
Putting value of
= (-10.5KJ) – (-1)(
= -10.5kJ -2.48kJ
Putting value of
= -12.98kJ –(298K)(-44.1
= -12.98kJ +13.14kJ
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.
Q-21: Comment on the thermodynamic stability of NO(g), given,
(1/2)N2(g) + (1/2)O2(g) → NO(g);
NO(g) + (1/2)O2(g) → NO2(g);
The +ve value of
The -ve value of
Thus, unstable NO(g) converts into stable NO2(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.
Thus, the same heat will be absorbed by surrounding Qsurr = +286
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|CBSE Notes for class 11 chemistry Chapter 6|
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The NCERT Solutions for Class 11 Chemistry (Chapter 6) provided on this page contains all the intext and exercise questions listed in Chapter 6 of the NCERT chemistry textbook for Class 11. It therefore covers many important questions that could be asked in the board examinations.
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