NCERT Solutions for Class 11 Chemistry Chapter 6 Chemical Thermodynamics is provided here. This topic is an extremely important topic in chemistry and is important for class 11, Class 12 and competitive exams like JEE and NEET.
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Introduction to Chemical Thermodynamics
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. Surrounding is part of universe excluding system. Based on the exchange of energy and matter, the system is 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. This was brief on Chemical Thermodynamics.
Subtopics included in Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Thermodynamic Terms
 The System And The Surroundings
 Types Of The System
 The State Of The System
 Internal Energy As A State Function

Applications
 Work
 Enthalpy, H
 Measurement Of Du And Dh: Calorimetry
 Enthalpy Change, Drh Of A Reaction – Reaction Enthalpy
 Enthalpies For Different Types Of Reactions
 Spontaneity
 Gibbs Energy Change And Equilibrium.
Class 11 Chemistry Chapter 6 Chemical Thermodynamics Important Questions
Q1: A thermodynamic state function is _______.
1. A quantity which depends upon temperature only.
2. A quantity which determines pressurevolume work.
3. A quantity which is independent of path.
4. A quantity which determines heat changes.
Ans:
(3) A quantity which is independent of path.
Reason:
Functions like pressure, volume and temperature depends on the state of the system only and not on the path.
Q2: Which of the following is a correct conditions for adiabatic condition to occur.
1. q = 0
2. w = 0
3. \(\Delta p = 0\)
4. \(\Delta T = 0\)
Ans:
1: q = 0
Reason:
For an adiabatic process heat transfer is zero, i.e. q = 0.
Q3: The value of enthalpy for all elements in standard state is _____.
(1) Zero
(2) < 0
(3) Different for every element
(4) Unity
Ans:
(1) Zero
Q4: For combustion of methane \(\Delta U^{\Theta }\) is – Y \(kJ mol^{1}\). Then value of \(\Delta H^{\Theta }\) is ____.
(a) > \(\Delta U^{\Theta }\)
(b) = \(\Delta U^{\Theta }\)
(c) = 0
(d) < \(\Delta U^{\Theta }\)
Ans:
(d) < \(\Delta U^{\Theta }\)
Reason:
\(\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT\) ; \(\Delta U^{\Theta }\) = – Y \(kJ mol^{1}\), \(\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }\)Q5: For, Methane, dihydrogen and and graphite the enthalpy of combustion at 298K are given 890.3kJ \(mol^{1}\), 285.8kJ\(mol^{1}\) and 393.5kJ\(mol^{1}\) respectively. Find the enthalpy of formation of Methane gas?
(a) 52.27kJ\(mol^{1}\)
(b) 52kJ\(mol^{1}\)
(c) +74.8kJ\(mol^{1}\)
(d) 74.8kJ\(mol^{1}\)
Ans:
(d) 74.8kJ\(mol^{1}\)
à CH_{4(g) } + 2O_{2(g)} \(\rightarrow\) CO_{2(g)} + 2H_{2}O_{(g)}
\(\Delta H = 890.3kJmol^{1}\)à C_{(s)} + O_{2(g) } \(\rightarrow\) CO_{2(g)}
\(\Delta H = 393.5kJmol^{1}\)à 2H_{2(g)} + O_{2(g) }\(\rightarrow\) 2H_{2}O_{(g)}
\(\Delta H = 285.8kJmol^{1}\)à C_{(s)} + 2H_{2(g) }\(\rightarrow\) CH_{4(g)}
\(\Delta _{f}H_{CH_{4}}\) = \(\Delta _{c}H_{c}\) + \(2\Delta _{f}H_{H_{2}}\) – \(\Delta _{f}H_{CO_{2}}\)= [ 393.5 +2(285.8) – (890.3)] kJ\(mol^{1}\)
= 74.8kJ\(mol^{1}\)
Q6: 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
Ans:
(c) will be possible at any temperature
\(\Delta G\) should be –ve, for spontaneous reaction to occur \(\Delta G\) = \(\Delta H\) – T\(\Delta S\)As per given in question,
\(\Delta H\) is –ve ( as heat is evolved) \(\Delta S\) is +veTherefore, \(\Delta G\) is negative
So, the reaction will be possible at any temperature.
Q7: In the process, system absorbs 801 J and work done by the system is 594 J. Find \(\Delta U\) for the given process.
Ans:
As per Thermodynamics 1^{st} law,
\(\Delta U\) = q + W(i); \(\Delta U\) internal energy = heatW = work done
W = 594 J (work done by system)
q = +801 J (+ve as heat is absorbed)
Now,
\(\Delta U\) = 801 + (594) \(\Delta U\) = 207 JQ8: The reaction given below was done in bomb calorimeter, and at 298K we get, \(\Delta U\) = 753.7 kJ \(mol^{1}\). Find \(\Delta H\) at 298K.
NH_{2}CN_{(g)} +3/2O_{2(g)}à N_{2(g)} + CO_{2(g)} + H_{2}O_{(l)}
Ans:
\(\Delta H\) is given by, \(\Delta H = \Delta U + \Delta n_{g}RT\)………………(1) \(\Delta n_{g}\) = change in number of moles \(\Delta U\) = change in internal energyHere,
\(\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)\)_{ }= (2 – 2.5) moles
_{ }\(\Delta n_{g}\) = 0.5 moles
Here,
T =298K
\(\Delta U\) = 753.7 \(kJmol^{1}\)R = \(8.314\times 10^{3}kJmol^{1}K^{1}\)
Now, from (1)
\(\Delta H = (753.7 kJmol^{1}) + (0.5mol)(298K)( 8.314\times 10^{3}kJmol^{1}K^{1})\)= 753.7 – 1.2
\(\Delta H\) = 754.9 \(kJmol^{1}\)Q9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from \(45^{\circ}C\; to\; 65^{\circ}C\). For aluminium molar haet capacity is 24 \(J mol^{1}K^{1}\)
Ans:
Expression of heat(q),
\(q = mCP\Delta T\);………………….(a) \(\Delta T\) = Change in temperaturec = molar heat capacity
m = mass of substance
From (a)
\(q = ( \frac{50}{27}mol )(24mol^{1}K^{1})(20K)\)q = 888.88 J q
Q10: Calculate \(\Delta H\) for transformation of 1 mole of water into ice from\(10^{\circ}C\;to\;(10)^{\circ}C\).. \(\Delta _{fus}H = 6.03 kJmol^{1} \;at\;10^{\circ}C\).
\(C_{p}[H_{2}O_{(l)}] = 75.3 J\;mol^{1}K^{1}\)
\(C_{p}[H_{2}O_{(s)}] = 36.8 J\;mol^{1}K^{1}\)
Ans:
\(\Delta H_{total}\) = sum of the changes given below:(a) Energy change that occurs during transformation of 1 mole of water from \(10^{\circ}C\;to\;0^{\circ}C\).
(b) Energy change that occurs during transformation of 1 mole of water at \(0^{\circ}C\) to 1 mole of ice at \(0^{\circ}C\).
(c) Energy change that occurs during transformation of 1 mole of ice from \(0^{\circ}C\;to\;(10)^{\circ}C\).
\(\Delta H_{total} = C_{p}[H_{2}OCl]\Delta T + \Delta H_{freezing} C_{p}[H_{2}O_{s}]\Delta T\)= (75.3 \(J mol^{1}K^{1}\))(0 – 10)K + (6.03*1000 \(J mol^{1}\)(100)K
= 753 \(J mol^{1}\) – 6030\(J mol^{1}\) – 368\(J mol^{1}\)
= 7151 \(J mol^{1}\)
= 7.151\(kJ mol^{1}\)
Thus, the required change in enthalpy for given transformation is 7.151\(kJ mol^{1}\).
Q11: Enthalpies of formation for CO_{2(g)}, CO_{(g), }N_{2}O_{4(g)}, N_{2}O_{(g)} are 393\(kJ mol^{1}\),110\(kJ mol^{1}\), 9.7\(kJ mol^{1}\) and 81\(kJ mol^{1}\) respectively. Then, \(\Delta _{r}H\) = _____.
N_{2}O_{4(g)} + 3CO_{(g)}à _{ }N_{2}O_{(g)} + 3 CO_{2(g)}
Ans:
“\(\Delta _{r}H\) for any reaction is defined as the fifference between \(\Delta _{f}H\) value of products and \(\Delta _{f}H\) value of reactants.”
\(\Delta _{r}H = \sum \Delta _{f}H (products) – \sum \Delta _{f}H (reactants)\)Now, for
N_{2}O_{4(g)} + 3CO_{(g)} à _{ }N_{2}O_{(g)} + 3 CO_{2(g)}
\(\Delta _{r}H = [(\Delta _{f}H(N_{2}O) + (3\Delta _{f}H(CO_{2})) – (\Delta _{f}H(N_{2}O_{4}) + 3\Delta _{f}H(CO))]\)Now, substituting the given values in the above equation, we get:
\(\Delta _{r}H\) = [{81\(kJ mol^{1}\) + 3(393) \(kJ mol^{1}\)} – {9.7\(kJ mol^{1}\) + 3(110) \(kJ mol^{1}\)}] \(\Delta _{r}H\) = 777.7 \(kJ mol^{1}\)
Q12 Enthalpy of combustion of C to CO_{2} is 393.5 \(kJ mol^{1}\). Determine the heat released on the formation of 37.2g of CO_{2} from dioxygen and carbon.
Ans:
Formation of carbon dioxide from dioxygen and carbon gas is given as:
C_{(s)} + O_{2(g)} à CO_{2(g)}; \(\Delta _{f}H\) = 393.5 \(kJ mol^{1}\)
1 mole CO_{2} = 44g
Heat released during formation of 44g CO_{2} = 393.5 \(kJ mol^{1}\)
Therefore, heat released during formation of 37.2g of CO_{2} can be calculated as
= \(\frac{393.5kJmol^{1}}{44g}\times 37.2g\)
= 332.69 \(kJ mol^{1}\)
Q13: N_{2(g)} + 3H_{2(g)}à 2NH_{3(g)} ; \(\Delta _{r}H^{\Theta }\) = 92.4 \(kJ mol^{1}\)
Standard Enthalpy for formation of ammonia gas is _____.
Ans:
“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)N_{2(g)} + (1.5)H_{2(g)} à 2NH_{3(g)}
Therefore, Standard Enthalpy for formation of ammonia gas
= (0.5) \(\Delta _{r}H^{\Theta }\)
= (0.5)(92.4\(kJ mol^{1}\))
= 46.2\(kJ mol^{1}\)
Q14: Determine Standard Enthalpy of formation for CH_{3}OH_{(l)} from the data given below:
CH_{3}OH_{(l)} + (3/2)O_{2(g)}à CO_{2(g)} + 2H_{2}O_{(l)}; \(\Delta _{r}H^{\Theta }\) = 726 \(kJ mol^{1}\)
C_{(g)} + O_{2(g)}à CO_{2(g)}; \(\Delta _{c}H_{\Theta }\) = 393 \(kJ mol^{1}\)
H2_{(g)} + (1/2)O_{2(g)}à H_{2}O_{(l)}; \(\Delta _{f}H^{\Theta }\) = 286 \(kJ mol^{1}\)
Ans:
C_{(s)} + 2H_{2}O_{(g)} + (1/2)O_{2(g)} à CH_{3}OH_{(l)} …………………………(i)
CH_{3}OH_{(l)} can be obtained as follows,
\(\Delta _{f}H_{\Theta }\) [CH_{3}OH_{(l)}] = \(\Delta _{c}H_{\Theta }\)2\(\Delta _{f}H_{\Theta }\) – \(\Delta _{r}H_{\Theta }\)
= (393 \(kJ mol^{1}\)) +2(286\(kJ mol^{1}\)) – (726\(kJ mol^{1}\))
= (393 – 572 + 726) \(kJ mol^{1}\)
= 239\(kJ mol^{1}\)
Thus, \(\Delta _{f}H_{\Theta }\) [CH_{3}OH_{(l)}] = 239\(kJ mol^{1}\)
Q15: Calculate \(\Delta H\) for the following process
CCl_{4(g)}à C_{(g)} + 4Cl_{(g)} and determine the value of bond enthalpy for CCl in CCl_{4(g)}.
\(\Delta _{vap}H^{\Theta }\) (CCl_{4}) = 30.5 \(kJ mol^{1}\).
\(\Delta _{f}H^{\Theta }\) (CCl_{4}) = 135.5 \(kJ mol^{1}\).
\(\Delta _{a}H^{\Theta }\) (C) = 715 \(kJ mol^{1}\),
\(\Delta _{a}H^{\Theta }\) is a enthalpy of atomisation
\(\Delta _{a}H^{\Theta }\) (Cl_{2}) = 242 \(kJ mol^{1}\).
Ans:
“ The chemical equations implying to the given values of enthalpies” are:
(1) CCl_{4(l)} à CCl_{4(g)} ; \(\Delta _{vap}H^{\Theta }\) = 30.5 \(kJ mol^{1}\)
(2) C_{(s)} à C_{(g)} \(\Delta _{a}H^{\Theta }\) = 715 \(kJ mol^{1}\)
(3) Cl_{2(g)} à 2Cl_{(g)} ; \(\Delta _{a}H^{\Theta }\) = 242 \(kJ mol^{1}\)
(4) C_{(g)} + 4Cl_{(g)} à CCl4(g); \(\Delta _{f}H^{\Theta }\) = 135.5 \(kJ mol^{1}\) \(\Delta H\) for the process CCl_{4(g)} à C_{(g)} + 4Cl_{(g) }can be measured as:
\(\Delta H = \Delta _{a}H^{\Theta }(C) + 2\Delta _{a}H^{\Theta }(Cl_{2}) – \Delta _{vap}H^{\Theta } – \Delta _{f}H\)= (715\(kJ mol^{1}\)) + 2(\(kJ mol^{1}\)) – (30.5\(kJ mol^{1}\)) – (135.5\(kJ mol^{1}\))
Therefore, \(H = 1304kJmol^{1}\)
The value of bond enthalpy for CCl in CCl_{4(g)}
= \(\frac{1304}{4}kJmol^{1}\)
= 326 \(kJ mol^{1}\)
Q16: \(\Delta U\) = 0 for isolated system, then what will be \(\Delta U\)?
Ans:
\(\Delta U\) is positive ; \(\Delta U\) > 0.As, \(\Delta U\) = 0 then\(\Delta S\) will be +ve, as a result reaction will be spontaneous.
Q17:
Following reaction takes place at 298K,
2X + Y à Z
\(\Delta H\) = 400 \(kJ mol^{1}\)
\(\Delta H\) = 0.2\(kJ mol^{1}K^{1}\)
Find the temperature at which the reaction become spontaneous considering \(\Delta S\) and \(\Delta H\) to be constant over the entire temperature range?
Ans:
Now,
\(\Delta G = \Delta H – T\Delta S\)Let, the given reaction is at equilibrium, then \(\Delta T\) will be:
T = \((\Delta H – \Delta G)\frac{1}{\Delta S}\) \(\frac{\Delta H}{\Delta S}\); (\(\Delta G\) = 0 at equilibrium)
= 400\(kJ mol^{1}\)/0.2\(kJ mol^{1}K^{1}\)
Therefore, T = 2000K
Thus, for the spontaneous, \(\Delta G\) must be –ve and T > 2000K.
Q18: 2Cl_{(g)}à Cl_{2(g)}
In above reaction what can be the sign for \(\Delta S\) and \(\Delta H\)?
Ans:
\(\Delta S\) and \(\Delta H\) are having negative sign.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, \(\Delta H\) is negative.
Also, 2 moles of Chlorine atoms is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, \(\Delta S\) is negative.
Q19: 2X_{(g)} + Y_{(g)}à 2D_{(g)}
\(\Delta U^{\Theta }\) = 10.5 kJ and \(\Delta S^{\Theta }\) = 44.1\(JK^{1}\)
Determine \(\Delta G^{\Theta }\) for the given reaction, and predict that whether given reaction can occur spontaneously or not.
Ans:
2X_{(g)} + Y_{(g)} à 2D_{(g)}
\(\Delta n_{g}\) = 2 – 3= 1 mole
Putting value of \(\Delta U^{\Theta }\) in expression of \(\Delta H\):
\(\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT\)= (10.5KJ) – (1)( \(8.314\times 10^{3}kJK^{1}mol^{1}\))(298K)
= 10.5kJ 2.48kJ
\(\Delta H^{\Theta }\) = 12.98kJPutting value of \(\Delta S^{\Theta }\) and \(\Delta H^{\Theta }\) in expression of \(\Delta G^{\Theta }\):
\(\Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }\)= 12.98kJ –(298K)(44.1\(JK^{1}\))
= 12.98kJ +13.14kJ
\(\Delta G^{\Theta }\) = 0.16kJAs, \(\Delta G^{\Theta }\) is positive, the reaction won’t occur spontaneously.
Q20: Find the value of \(\Delta G^{\Theta }\) for the reaction, if equilibrium is given 10.given that T = 300K and R = \(8.314\times 10^{3}kJK^{1}mol^{1}\).
Ans:
Now,
\(\Delta G^{\Theta }\) = \(2.303RT\log eq\)= (2.303)( \(8.314\times 10^{3}kJK^{1}mol^{1}\))(300K) \(\log 10\)
= 5744.14\(Jmol^{1}\)
5.744\(kJmol^{1}\)
Q21: What can be said about the thermodynamic stability of NO(g), given
(1/2)N_{2(g)} + (1/2)O_{2(g)}; \(\Delta _{r}H^{\Theta } = 90kJmol^{1}\)
NO_{(g)} + (1/2)O_{2(g)}à NO_{2(g)} ; \(\Delta _{r}H^{\Theta } = 74kJmol^{1}\)
Ans:
The +ve value of \(\Delta _{r}H\) represents that during NO_{(g)} formation from O_{2} and N_{2}, heat is absorbed. The obtained product, NO_{(g)} is having more energy than reactants. Thus, NO_{(g)} is unstable.
The ve value of \(\Delta _{r}H\) represents that during NO_{2(g)} formation from O_{2(g)} and NO_{(g)}, heat is evolved. The obtained product, NO_{2(g)} gets stabilized with minimum energy.
Thus, unstable NO_{(g) }converts into unstable NO_{2(g).}
Q22: Determine \(\Delta S\) in surrounding given that mole of H2O(l) is formed I standard condition. \(\Delta _{r}H^{\Theta } = 286kJmol^{1}\).
Ans:
\(\Delta _{r}H^{\Theta } = 286kJmol^{1}\) is given so that amount of heat is evolved during the formation of 1 mole of H_{2}O_{(l)}.Thus, the same heat will be absorbed by surrounding. Q_{surr} = +286\(kJmol^{1}\).
Now, \(\Delta S_{surr}\) = Q_{surr}/7
= \(\frac{286kJmol^{1}}{298K}\)
Therefore, \(\Delta S_{surr} = 959.73Jmol^{1}K^{1}\)
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