NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics) can be found on this page. Here, students can access detailed, explanative solutions to all the intext and exercise questions listed in chapter 6 of the NCERT class 11 chemistry textbook. 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 Chemistry – Class 11, 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 which 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 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 Class 11 Chemistry Chapter 6 Chemical Thermodynamics
- 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
NCERT Solutions for Class 11 Chemistry Chapter 6
Q-1: A thermodynamic state function is _______.
1. A quantity which depends upon temperature only.
2. A quantity which determines pressure-volume work.
3. A quantity which is independent of path.
4. A quantity which determines heat changes.
(3) 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: Which of the following is a correct conditions for adiabatic condition to occur.
1. q = 0
2. w = 0
1: q = 0
For an adiabatic process heat transfer is zero, i.e. q = 0.
Q-3: The value of enthalpy for all elements in standard state is _____.
(2) < 0
(3) Different for every element
Q-4: For combustion of methane
(c) = 0
Q-5: For, Methane, di-hydrogen and and graphite the enthalpy of combustion at 298K are given -890.3kJ
à CH4(g) + 2O2(g)
à C(s) + O2(g)
à 2H2(g) + O2(g)
à C(s) + 2H2(g)
= [ -393.5 +2(-285.8) – (-890.3)] kJ
Q-6: A reaction, X + Y à U + V + q is having a +ve entropy change. Then the reaction ____.
(a) will be possible at low temperature only
(b) will be possible at high temperature only
(c) will be possible at any temperature
(d) won’t be possible at any temperature
(c) will be possible at any temperature
As per given in question,
So, the reaction will be possible at any temperature.
Q-7: In the process, system absorbs 801 J and work done by the system is 594 J. Find
As per Thermodynamics 1st law,
W = work done
W = -594 J (work done by system)
q = +801 J (+ve as heat is absorbed)
Q-8: The reaction given below was done in bomb calorimeter, and at 298K we get,
NH2CN(g) +3/2O2(g)à N2(g) + CO2(g) + H2O(l)
= (2 – 2.5) moles
Now, from (1)
= -753.7 – 1.2
Q-9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from
Expression of heat(q),
c = molar heat capacity
m = mass of substance
q = 888.88 J q
(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: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393
N2O4(g) + 3CO(g)à 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-12 Enthalpy of combustion of C to CO2 is -393.5
Formation of carbon dioxide from di-oxygen and carbon gas is given as:
C(s) + O2(g) à CO2(g);
1 mole CO2 = 44g
Heat released during formation of 44g CO2 = -393.5
Therefore, heat released during formation of 37.2g of CO2 can be calculated as
Q-13: N2(g) + 3H2(g)à 2NH3(g) ;
Standard Enthalpy for formation of ammonia gas is _____.
“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) à 2NH3(g)
Therefore, Standard Enthalpy for formation of ammonia gas
Q-14: Determine Standard Enthalpy of formation for CH3OH(l) from the data given below:
CH3OH(l) + (3/2)O2(g)à CO2(g) + 2H2O(l);
C(g) + O2(g)à CO2(g);
H2(g) + (1/2)O2(g)à H2O(l);
C(s) + 2H2O(g) + (1/2)O2(g) à CH3OH(l) …………………………(i)
CH3OH(l) can be obtained as follows,
= (-393 – 572 + 726)
CCl4(g)à 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)
Following reaction takes place at 298K,
2X + Y à Z
Find the temperature at which the reaction become spontaneous considering
Let, the given reaction is at equilibrium, then
Therefore, T = 2000K
Thus, for the spontaneous,
Q-18: 2Cl(g)à Cl2(g)
In above reaction what can be the sign for
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 is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus,
Q-19: 2X(g) + Y(g)à 2D(g)
2X(g) + Y(g) à 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: Find the value of
Q-21: What can be said about the thermodynamic stability of NO(g), given
(1/2)N2(g) + (1/2)O2(g);
NO(g) + (1/2)O2(g)à NO2(g) ;
The +ve value of
The -ve value of
Thus, unstable NO(g) converts into unstable NO2(g).
Thus, the same heat will be absorbed by surrounding. Qsurr = +286
Continue learning the Laws of Thermodynamics and its applications with Byju’s video lectures.
|NCERT Exemplar for class 11 chemistry Chapter 6|
|CBSE Notes for class 11 chemistry Chapter 6|
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