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Thermodynamics is an important branch of Chemistry. It deals with energy, its transformation, and the laws governing the entire process of transformation of energy. Have you ever wondered why a hot cup of tea cools down after some time or why ice melts, so the only reason behind this kind of phenomenon is thermodynamics. NCERT Class 11 Chapter 5 plays a very important role in Chemistry as it is the basic chapter that builds the foundations of advanced chapters and it also helps to explain why changes occur around us.
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In Thermodynamics, there are 22 questions in the exercise. The solutions of thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you prepare for the class 11 final examination and competitive exams like JEE Mains, NEET, etc. If you are looking for an answer from any other chapter even from any other class then go with NCERT Solutions, there you will get all the answers of NCERT easily.
Also Read :
Thermodynamics Class 11 Chemistry NCERT Exemplar Solutions |
Thermodynamics Class 11 Chemistry Chapter 5 Notes |
Answer :
A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of the path.
Like- $p, V, T$ depends on the state of the system, not on the path.
So, the correct option is (ii)
Question 5.2 For the process to occur under adiabatic conditions, the correct condition is:
(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) q = 0
(iv) w = 0
Answer :
In adiabatic conditions, the heat exchange between the system and the surroundings through its boundary is not allowed.
So, the correct option is (iii) $q$ = 0
Answer :
The enthalpies of all elements in their standard states are Zero.
So, the correct option is (ii)
Answer :
Equation of combustion of methane-
$CH_4(g)+\:2O_2(g)\rightarrow CO_2(g)\:+\:2H_2O(l)$
Since, $\Delta H^o=\Delta U^o+\Delta n_gRT$ and $\Delta U^o=-X\:KJ\:mol^{-1}$
from the equation $\Delta n_g=(n_p-n_R)=1-3=-2$
$\Delta H^o=-X-2RT$
$\Delta H^o<\Delta U^o$
so the correct option is (iii).
(i) –74.8 $kJ mol^{-1}$
(ii) –52.27 $kJ mol^{-1}$
(iii) +74.8 $kJ mol^{-1}$
(iv) +52.26 $kJ mol^{-1}$
Answer :
$CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H$ ................. -890.3 kJ/mol
$C+2O_{2}\rightarrow CO_{2}$ $\Delta H$ ..................-393.5 kJ/mol
$2H_{2}+O_{2}\rightarrow 2H_{2}O$ $\Delta H$ ..................-258.8 kJ/mol
So, the required equation is to get the formation of $CH_4 (g)$ by combining these three equations-
Thus, $C+2H_{2}\rightarrow CH_{4}$
$\Delta _fH_{CH_{4}} = \Delta _cH_c+2\Delta _cH_{H_2}$
= [-393.5 + 2(-285.8) + 890.3]
= -74.8 kJ/mol
therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)
Answer :
For the reaction to be feasible the $\Delta G$ should be negative
$\Delta G = \Delta H - T\Delta S$
According to the question,
$\Delta S$ = positive unit
$\Delta H$ = negative (as heat is evolved in the reaction)
Overall the $\Delta G$ is negative.
Therefore the reaction is possible at any temperature. So, the correct option is (iv)
Answer :
The first law of Thermodynamics states that,
$\Delta U = q+W$
where, $\Delta U$ = change in internal energy for the process
q = heat and W = work
Given that,
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)
Substituting the value in the equation of the first law we get,
$\Delta U = 701+(-394) =307J$
So, the change in internal energy for the process is 307 J
$NH_2 CN ( g ) + \frac{3}{2} O_2 (g) \rightarrow N_2 + CO_2 ( g ) H_2 O ( l)$
Answer :
Given information,
$\Delta U = -742.7kJmol^{-1}$
T = 298 K
R = 8.314 $\times 10^{-3}$
$\Delta n_g =$ $n_g$ (products) $- n_g$ (reactants)
= (2 - 1.5)
= 0.5 moles
The enthalpy change for the reaction is expressed as;
$\Delta H = \Delta U+n_gRT$
where, $\Delta U$ = change in internal energy and
$\Delta n_g =$ change in no. of moles
By putting the values we get,
$\Delta H = (-742.7)+0.5(298)(8.314\times 10^{-3})$
$=-741.5 kJmol^{-1}$
Answer :
We know that
$q = mc\Delta T$
m = mass of the substance
c = heat capacity
$\Delta T$ = change in temperature
By putting all these values we get,
$q = (\frac{60}{27}mol)(24 J/mol/K)(20K)$
= 1066.7 J
= 1.07 kJ
Question 5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at $10.0 ^\circ C$ to ice at $- 10.0 ^\circ C$.
$\Delta _{ fus }H = 6.03 KJ mol ^{-1}$ at $0 ^\circ C$ .
$Cp [H_2O(l)] = 75.3 J mol^{-1} K^{-1}\\\\.\: \: \: \: Cp [H_2O(s)] = 36.8 J mol^ { -1}K^{-1}$
Answer :
Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from $10^o C$ to $0^o C$ of 1 mol of water (water $\rightarrow$ water )
(ii) from $0^o C$ to $0^o C$ of 1 mol of ice (water $\rightarrow$ ice)
(iii) from $0^o C$ to $-10^oC$ of 1 mole of ice (ice $\rightarrow$ ice)
So, the total enthalpy change
$\Delta H = C_p[H_2O(l)](\Delta T)+\Delta H_(freezing)+C_p[H_2O(s)]\Delta T$
$\\ = (75.3)(-10)+(-6.03\times 10^{3})+(36.8)(-10)\\ =-753-6030 - 368\\ =-7151 Jmol^{-1}\\ = 7.151kJ/mol$
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol
Answer :
Formation of carbon dioxide from carbon and dioxygen reaction is-
$C+O_{2}\rightarrow CO_{2}$ ............. $\Delta H_f$ = -393.5 kJ/mol
1 mole of $CO_2$ = 44 g
So, the heat released in the formation of 44g of $CO_2$ = -393.5 kJ/mol
therefore, In 35.5 g of $CO_2$ the amount of heat released =
$\frac{-393.5}{44}\times 35.2 = -314.8 kJ/mol$
$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$
Answer :
Given,
Enthalpies of formation of $CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g)$ are –110, – 393, 81 and 9.7 $KJ mol ^{-1}$
$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$
We know that the $\Delta _rH=\sum \Delta _fH$ (product) $-\sum \Delta _fH$ (reactants)
For the above reaction,
$\Delta _rH =[{\Delta _fH(N_{2}O)+3\Delta _fH(CO_{2})}]-[\Delta _fH(N_{2}O_{4})+3\Delta _fH(CO)]$
Substituting the given values we get,
$\\= [81+3(-393)]-[9.7+3(-110)]\\ =-777.7 kJ/mol$
Thus the value of $\Delta _ r H$ of the reaction is -777.7 kJ/mol
Answer :
the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.
Thus we can re-write the reaction as;
$\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightarrow NH_{3}$
Therefore, standard enthalpy of formation of ammonia is = $\frac{1}{2}\Delta _rH^\Theta$
= 1/2 (-92.4)
= -46.2 kJ/mol
Question 5.14 Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:
Answer :
for the formation of $CH_3OH$ the reaction is ,
$C+2H_{2}O+\frac{1}{2}O_{2}\rightarrow CH_3OH$
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)
$\Delta _fH[CH_{3}OH] = \Delta _cH^\Theta +2\Delta _fH[H_{2}O]-\Delta _rH^\Theta$
$=(-393) + 2(-286)-(-726)$
$=-239 kJmol^{-1}$
Question 5.15 Calculate the enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ and calculate bond enthalpy of $C-Cl$ in $CCl_4(g)$
$\Delta _{vap} H ^{\ominus }(CCl_4) = 30.5 kJ mol^{-1}.\\\\ .\: \: \: \: \: \Delta _f H ^\ominus (CCl_4) = -135.5 kJ mol^{-1}.\\\\.\: \: \: \: \Delta _ a H ^ \ominus (C) = 715.0 kJ mol ^{-1} , where \: \: \Delta_ a H ^\ominus \: \: is\: \: enthalpy \: \: of \: \: atomisation\\\\.\: \: \: \: \Delta _ a H ^\ominus (Cl_2) = 242 kJ mol ^{-1}$
Answer :
We have the following chemical reaction equations-
$(1)\ CCl_{4}(l)\rightarrow CCl_{4}(g)$ ................ $\Delta _{vap}H^o = 30.5kJ/mol$
$(2)\ C(s)\rightarrow C(g)$ ........................ $\Delta _{a}H^o =715.0kJ/mol$
$(3)\ Cl2(g)\rightarrow 2Cl(g)$ .................. $\Delta _{a}H^o =242kJ/mol$
$(4)\ C(g)+4Cl\rightarrow CCl_4(g)$ ............ $\Delta _{f}H =-135.5kJ/mol$
The enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ by the above reaction is calculated as;
$\Delta H =\Delta _aH^o(C)+2(\Delta _aH^o(Cl_{2}))-\Delta _{vap}H^o-\Delta _fH$
$= [(715)+2(242)-(30.5)-(-135.5)] kJ/mol$
$=1304 kJmol^{-1}$
And the bond enthalpy of C-Cl bond in $CCl_4$ (g) = 1304/4 = 326 kJ/mol
Question 5.16 For an isolated system, $\Delta U = 0$, what will be $\Delta S$?
Answer :
Since $\Delta U = 0$
So, the $\Delta S>0$ (positive) therefore the reaction will be feasible.
Answer :
From the equation,
$\Delta G = \Delta H-T\Delta S$
Suppose the reaction is at equilibrium, So the change in temperature is given as;
$T=\frac{\Delta H -\Delta G}{\Delta S}$
$= \frac{400}{0.2} =2000K$ ( $\Delta G$ at equilibrium is zero)
To reaction should be spontaneous, $\Delta G$ should be negative. So, for the given reaction T should be greater than 2000 K
Question 5.18 For the reaction, $2 Cl (g) \rightarrow Cl_2 ( g)$ , what are the signs of $\Delta H \: \: and \: \: \Delta S$?
Answer :
The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So, $\Delta H$ is negative.
Two moles of an atom have more randomness than one mole of chlorine. So, spontaneity is decreased. Hence $\Delta S$ is negative.
Answer :
Given reaction is $2 A (g) + B ( g ) \rightarrow 2 D ( g)$
We know that, $\Delta G^o = \Delta H^o-T\Delta S^o$
and $\Delta H^o = \Delta U^o+\Delta n_{g}RT$
here $\Delta n_{g}= 2 - (3) = -1$
and $\Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}$
substituting the given values in equations-
$\Delta G = (-10.5+8.314\times 10^{-3}\times 298)-298(-44.1)$
$\\\Delta G = (-12.98 +13.14 )kJ\\ \Delta G= +0.16 kJ$
Hence the reaction will not occur spontaneously because the $\Delta G ^ \ominus$ value is positive for this reaction.
Question 5.20 The equilibrium constant for a reaction is 10. What will be the value of
$\Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$
Answer :
Given values are,
$R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$
and equilibrium constant = 10
It is known that,
$\Delta G^o = -2.303\ RT \log K_{eq}$
$\\=-2.303(8.314)(300)\log 10\\ =-5744.14 Jmol^{-1}\\ =-5.744 kJmol^{-1}$
Hence the value of $\Delta G^o$ is -5.744 kJ/mol
Question 5.21 Comment on the thermodynamic stability of NO(g), given
Answer :
The formation of $NO$ is unstable because $\Delta _rH$ is positive which means heat is absorbed during the reaction. So, $NO$ (g) has higher energy than its reactants $N_{2}$ and $O_{2}$.
On the other hand, $NO_{2}$ is stable because $\Delta _rH$ is negative means heat is released during the formation of $NO_{2}$. It is established with minimum energy. Hence unstable $NO$ changes to stable $NO_{2}$.
Answer :
-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also.
$q_{surr}= +286$ kJ/mol
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298
= 959.73 J/mol/K
Thermodynamics enables us to study energy changes quantitatively and to make useful predictions. For these, we divide the universe into the system and the surroundings. In thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes its surroundings. After completing NCERT solutions for Thermodynamics you will be able to explain terms like system, surroundings, and boundary; In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically.
Question:
At standard conditions, if the change in the enthalpy for the following reaction is $-109 \mathrm{kJmol}^{-1}$.
$$
\mathrm{H}_{2(g)}+B r_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}
$$
Given that bond energy of $\mathrm{H}_2$ and $\mathrm{Br}_2$ is $435 \mathrm{kJmol}^{-1}$ and $192 \mathrm{kJmol}^{-1}$ respectively, What is the bond energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of HBr ?
1) 736
2) 518
3) 259
4) 368
Solution:
$$
\begin{aligned}
& \quad \mathrm{H}_2 \rightarrow 2 \mathrm{H} \quad \mathrm{~B} \cdot E_{H_2} \\
& \mathrm{Br}_2 \rightarrow 2 \mathrm{Br} \quad \mathrm{~B} \cdot E_{B r_2} \\
& 2 \mathrm{H}+2 \mathrm{Br} \rightarrow 2 \mathrm{HBr} \quad-2 \mathrm{~B} \cdot E_{H B r} \\
& \mathrm{H}_2+\mathrm{Br}_2 \rightarrow 2 \mathrm{HBr}
\end{aligned}
$$
From the Bond energy concept, we have
$$
\begin{aligned}
& B E_{H_2}+B E_{B r_2}-2 B E_{H B r}=\Delta H_{H B r} \\
\rightarrow & 435+192-2 x=-109 \\
\rightarrow & 2 x=736 \\
\rightarrow & x=368 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the answer is the option (4).
Thermodynamics comes under physical chemistry. This chapter is about learning the concepts and then applying them to solve the numerical. The approach to solving numerical should be simple yet time-saving. For this, you first need to understand the key concepts as a system, surroundings, internal energy, work, heat, enthalpy, etc. For solving any questions first give it a thorough reading and note down all the information given in the question. Begin by identifying what is given in the question and what needs to be found. Carefully note the units and convert them if necessary to ensure consistency. Use the appropriate formula based on the concept like $\Delta U=q+w$ (change in internal energy $=$ heat + work done) or $\Delta H=\Delta U+P \Delta V$ (enthalpy change).
Let us understand it by taking an example. Suppose the question says "A system absorbs 300 J of heat and does 100 J of work. Calculate the change in internal energy."
Solution- Here, heat absorbed $q=+300 J$ (positive as it is absorbed) and work done by the system $w=$ $-100 J$ (negative as energy is leaving the system).
Now use the formula:
$$
\Delta U=q+w=300+(-100)=200 J
$$
So, the change in internal energy is 200 J.
Practice helps in mastering thermodynamics numerical. Start with simpler problems and gradually move to ones involving concepts like enthalpy change, work in isothermal or adiabatic processes, and heat capacity.
5.1 Thermodynamic Terms |
5.2 Applications |
5.3 Measurement of $\Delta U$ and $\Delta H$: Calorimetry |
5.4 Enthalpy Change, $\Delta H$ of a Reaction – Reaction Enthalpy |
5.5 Enthalpies for Different Types of Reactions |
5.6 Spontaneity |
5.7 Gibbs Energy Change and Equilibrium |
Concept | JEE | NCERT |
Thermodynamics | ✅ | ✅ |
Thermodynamics: Properties Of System | ✅ | |
Path, State Function, Types Of Process | ✅ | ✅ |
Reversible, Irreversible, Polytropic Process | ✅ | |
Thermodynamic Equilibrium | ✅ | |
Zeroth Law Of Thermodynamics | ✅ | |
Heat And Work | ✅ | ✅ |
Internal Energy | ✅ | ✅ |
Isothermal Reversible And Isothermal Irreversible | ✅ | ✅ |
Graphical Representation Of Work Done In Thermodynamics | ✅ | |
Heat Capacity | ✅ | ✅ |
Relation Between Cp And Cv | ✅ | ✅ |
Thermochemistry And Enthalpy For Chemical Reaction | ✅ | |
Standard Enthalpy And Enthalpy Of Formation | ✅ | ✅ |
Enthalpy Of Combustion | ✅ | ✅ |
Enthalpy Of Dissociation, Atomisation And Phase Change | ✅ | |
Lattice Enthalpy, Hydration Enthalpy And Enthalpy Of Solution | ✅ | |
Enthalpy Of Neutralisation | ✅ | |
Ionization And Electron Gain Enthalpy | ✅ | |
Resonance Enthalpy | ✅ | |
Kirchoff’s Equation | ✅ | |
Born Habers Cycle | ✅ | |
Hess’s Law | ✅ | ✅ |
Bomb Calorimeter | ✅ | |
Calculation Of Changes In S For Different Process | ✅ | |
Spontaneity Criteria Through Entropy | ✅ | ✅ |
Spontaneity Criteria Through Enthalpy (H) And Entropy (S) | ✅ | ✅ |
Gibbs Energy And Change In Gibbs Energy | ✅ | ✅ |
Spontaneity Criteria With Gibbs Energy (G) | ✅ | ✅ |
Gibbs Energy At Equilibrium | ✅ | ✅ |
The Thermodynamics chapter in Class 11 NCERT Chemistry offers a bunch to students preparing for JEE and NEET:
Also Check NCERT Books and NCERT Syllabus here:
NCERT Books Class 11 Chemistry |
NCERT Syllabus Class 11 Chemistry |
NCERT Books Class 11 |
NCERT Syllabus Class 11 |
The enthalpy and heat capacity is related by the formula $C_p=\left(\frac{d H}{d T}\right)_{p^{\prime}}$ where $C_p$ is the heat capacity at constant pressure. It means that heat capacity is the rate of change of enthalpy with respect to temperature at constant pressure.
A system's degree of disorder or unpredictability is measured by its entropy. It helps determine the direction of spontaneous processes. The greater is the entropy of a system, higher will be the randomness.
Gibbs Free Energy Equation is $G=H-T S$; it's used to predict spontaneity. If $\Delta G<0$ then the process is spontaneous. The value of G, depends on the enthaply and entropy of the system.
Work in Isothermal Process is $W=n R T \ln \frac{V_f}{V_i}$; in Adiabatic Process, $W=\frac{P_i V_i-P_f V_f}{\gamma-1}$, where no heat exchange occurs.
According to Hess's Law, the overall enthalpy change of a reaction remains constant regardless of the number of steps involved. It allows us to compute unknown enthalpy changes using known reactions.
The first law of thermodynamics states that energy can neither be created nor be destroyed it can only transfer from one form to another form.
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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
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