Thermodynamics

Chemistry Class 11

NCERT

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.

(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.

Solution :

A thermodynamic state function is a quantity whose value is independent of a path. Functions like $p, V, T$ etc. depend only on the state of a system and not on the path. Hence, alternative (ii) is correct.

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$

Solution :

A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, $q = 0.$$\\$ Therefore, alternative (iii) is correct.

3   The enthalpies of all elements in their standard states are:$\\$

(i) unity$\\$ (ii) zero$\\$ (iii) < 0$\\$ (iv) different for each element

Solution :

The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.

4   $\Delta U^\theta$ of combustion of methane is $- X kJ mol{-1}$ . The value of $\Delta H^\theta$ is$\\$

Solution :

Since $\Delta H \theta = \Delta U \theta + \Delta n _g RT $and $\Delta U \theta = -X kJ mol^{-1},$$\\$ $\Delta H \theta = (-X) + \Delta n g RT.$$\\$ $\Rightarrow \Delta H \theta < \Delta U \theta $$\\$ Therefore, alternative (iii) is correct.

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 $CH_{4 (g)}$ will be

Solution :

According to the question,$\\$ (i)$CH_{4(g)}+2O_{2(g)} \to CO_{2(g)}+2H_2O(l);$$\\$ $\Delta _c H^{\Theta}=-890.3kJ mol^{-1}$$\\$ (ii)$C(s)+O_2(g)\to CO_2;$$\\$ $\Delta _c H^{\Theta} =-285.8k J mol^{-1}$$\\$ (iii)$2H_2(g)+O_2(g)\to 2H_2O(l);\Delta _c H^{\Theta}=-285.8kJ mol^{-1}$$\\$ Thus, the desired equation is the one that represents the formation of $CH _4$ ( g) i.e.,$\\$ $C(s)+2H_2(g)\to CH_4(g);$$\\$ $\Delta _f H_{Ch4}=\Delta _CH_C+2\Delta CH_{h2}-\Delta CH_{Co_2}$$\\$ $=(-393.5 )+ 2* (-285.8 )-(- 890.3 )=- 74.8kJmol ^{- 1}$$\\$ $\therefore $ Enthalpy of formation of $CH _4 (g) = -74.8 kJ mol^{-1}$$\\$ Hence, alternative (i) is correct

6   A reaction, $A + B \to C + D + q$ is found to have a positive entropy change. The reaction will be$\\$

Solution :

For a reaction to be spontaneous, $\Delta G$ should be negative.$\\$ $\Delta G = \Delta H - T\Delta S$ $\\$ According to the question, for the given reaction,$\\$ $\Delta S $= positive $\\$ $\Delta H$ = negative (since heat is evolved) $\Rightarrow \Delta G =$ negative$\\$ Therefore, the reaction is spontaneous at any temperature.$\\$ Hence, alternative (iv) is correct.

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?

Solution :

According to the first law of thermodynamics,$\\$ $\Delta U = q + W (i)$ $\\$ Where,$\\$ $\Delta U =$ change in internal energy for a process$\\$ $q =$ heat$\\$ $W =$ work$\\$ Given,$\\$ $q = + 701 J $(Since heat is absorbed)$\\$ $W = -394 J $(Since work is done by the system$\\$ Substituting the values in expression (i), we get$\\$ $\Delta U = 701 J + (-394 J)$ $\\$ $\Delta U = 307 J $ $\\$ Hence, the change in internal energy for the given process is $307 J.$

8   The reaction of cyanamide, $NH_ 2 CN(s)$ with dioxygen was carried out in a bomb calorimeter and $\Delta U$ was found to be $-742.7 KJ $ mol $^{-1 }$at $298 K.$ Calculate the enthalpy change for the reaction at $298 K$. $NH _2 CN (g) + \dfrac{3}{2}O _2 (g)\to N _2 ( g ) +CO _2 ( g ) +H _2 O (l )$

Solution :

Enthalpy change for a reaction $(\Delta H)$ is given by the expression,$\\$ $\Delta H =\Delta U +\Delta _{ ng} RT$ $\\$ Where,$\\$ $\Delta U $= change in internal energy$\\$ $\Delta n_{ g}$ = change in number of moles$\\$ For the given reaction,$\\$ $\Delta _{ n g} = \Sigma {n g}$ (products) $- \Sigma {n g}$ (reactants)$\\$ $= (2 - 1.5)$ moles $\\$ $\Sigma {n g} = +0.5$ moles$ \\$ And, $\Delta U = -742.7 kJ $ mol $ ^{-1}$ $\\$ $T = 298 K$ $\\$ $R = 8.314 × 10 ^{-3} kJ $ mol $^{-1} K ^{-1}$ $\\$ Substituting the values in the expression of $\Delta H:$ $\\$ $\delta H = (-742.7 kJ$ mol$^{ -1} ) + (+0.5 $mol$) (298 K) 8.314 × 10 ^{-3} kJ$ mol $^{ -1} K ^{-1}$ $\\$ $= -742.7 + 1.2$ $\\$ $\Delta H = -741.5kJ $ mol$^{ -1}$

9   Calculate the number of $kJ $of heat necessary to raise the temperature of $60 g$ of aluminium from $35^ o C$ to $55^ o C$. Molar heat capacity of $Al $ is $24 J$ mol $^{-1} K ^{-1 }.$ $\\$

Solution :

From the expression of heat $(q),$ $\\$ $q = m. c. \Delta T$ $\\$ Where,$\\$ c = molar heat capacity$\\$ m = mass of substance$\\$ $\Delta T$ = change in temperature$\\$ Substituting the values in the expression of$\\$ $q= (\dfrac{60}{27}mol)(24 J \ mol^{-1} K^{-1})(20 K)$ $\\$ $q = 1066.7J$ $\\$ $q= 1.07 kJ$ $\\$

10   Calculate the enthalpy change on freezing of $1.0 $ mol of water at $10.0^ o C$ to ice at $- 10.0^ o C$. ,$\Delta _{ fus} H = 6.03 KJ $mol $^{-1}$ at $0^ o C.$ $\\$ $C_ p [H _2 O(l) ] = 75.3 J $mol $^{-1} K ^{-1 };$ $\\$ $C_ p [H _2 O(s) ] = 36.8 J $ mol $^{-1 }K ^{-1} .$ $\\$

Solution :

Total enthalpy change involved in the transformation is the sum of the following changes:$\\$ (a) Energy change involved in the transformation of $1$ mol of water at $10^oC $ to $1$ mol of water at $0^oC.$ $\\$ (b) Energy change involved in the transformation of $1$ mol of water at $0^o$ to $1$ mol of ice at $0^oC.$ $\\$ (c) Energy change involved in the transformation of $1$ mol of ice at $0^oC $to$ 1$ mol of ice at $10^oC.$ $\\$ Total $\Delta H = C_ p [ H_ 2 O ( l ) ]\Delta T +\Delta H_{ freezing} + C _p [ H _2 O (s ) ]\Delta T$ $\\$ $= (75.3 J$ mol$^{ -1} K^{ -1} ) (0 - 10)K + (-6.03 × 10 ^3 J$ mol $^{-1} ) + (36.8 J$ mol$^{ -1} K^{ -1 }) (-10 - 0)K$ $\\$ $= -753 J$ mol $^{-1 }- 6030 J $ mol $^{-1 } - 368 J$ mol$^{ -1}$ $\\$ $= -7151 J$ mol$^{ -1}$ $\\$ $= -7.151 kJ$ mol$^{ -1}$ $\\$ Hence, the enthalpy change involved in the transformation is $-7.151 kJ $mol $6{-1}$

11   Enthalpy of combustion of carbon to carbon dioxide is $-393.5 kJ \ mol ^{-1}$ Calculate the heat released upon formation of $35.2 g\ of\ CO _2$ from carbon and dioxygen gas.

Solution :

Formation of $CO _2$ from carbon and dioxygen gas can be represented as$\\$ $C(s)+O_{2()g}\to CO_{2(g)};\Delta H=-393.5kJ \ mol^{-1}\\ (1 mole=44g)$ $\\$ Heat released in the formation of $44 g$ of $CO _2 = 393.5 KJ\ mol ^{-1}$$\\$ Heat released in the formation of $35.2 g$ of$\\$ $CO_ 2 =(393.5KJ)×(35.2g)/(44g)=314.8 KJ$

12   Enthalpies of formation of $CO(g), CO 2 (g), N 2 O(g)$and $N 2 O 4 (g)$ are $-110 ,-393, 81 kJ$ and $9.7 kJ\ mol^{-1}$ respectively. Find the value of $\Delta_ r H$ for the reaction:$\\$ $N _2 O _{4(g)} + 3CO(g) \to N _2 O(g) + 3CO _2(g)$

Solution :

$\Delta_ r H$ for a reaction is defined as the difference between $\Delta_ f H$ value of products and $\Delta_ f H$ value of reactants.$\\$ $\Delta_ r H= \sum \Delta_ f H ( product )-\sum \Delta_ f H( reac tan t )$$\\$ For the given reaction,$\\$ $N _2 O _4(g) + 3CO(g) \to N _2 O(g) + 3CO _2(g)\\ \Delta_ r H =[ { \Delta_ f H( N_ 2 O )+ 3 \Delta_ f H ( CO_ 2 ) } - { \Delta_ f H( N _2 O )+ 3\Delta_ f H( CO ) } ]$ Substituting the values of $\Delta_ f H$ for N 2 O, CO 2 , N 2 O 4 , and CO from the question, we get: $\Delta_ r H=[{ 81 kJ mol ^{- 1 }+ 3 (- 393 ) KJ mol ^{- 1 }} - {9.7 kJ mol mol ^{- 1 }+ 3 (- 110 ) KJ mol ^{- 1 } } ]\\ \Delta_ r H =- 777.7 kJmol ^{- 1}$ Hence, the value of $\Delta_ r H$ for the reaction is - 777.7kJmol$ ^{- 1} $.

13   Given$\\$ $N _2 ( g )+ 3 H_ 2 ( g )\to 2 NH _3 ( g ) ; \Delta_ r H ^{\theta} = -92.4 kJ\ mol^{ -1}$$\\$ What is the standard enthalpy of formation of NH 3 gas?

Solution :

Standard enthalpy of formation of a compound is the change in 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. Re-writing the given equation for 1 mole of NH 3 (g),$\\$ $\dfrac{1}{2}N_2(g)+\dfrac{3}{2}H_2(g)\to 2 NH_3(g)$ $\\$ $\therefore $ Standard enthalpy of formation of NH 3 (g)$\\$ $=1/2 \Delta_r H^{theta}\\ =1/2 (-92.4 kJ\ mol^{-1})\\ =-46.2 kJ\ mol^{-1}$

14   Calculate the standard enthalpy of formation of $CH _3 OH(l)$ from the following data:$\\$ $CH_3OH(l)+\dfrac{3}{2}O_2(g)\to CO_2(g)+2H_2O(l);\Delta_r H^{\theta}=-726kJ\ mol^{-1}\\ C(g)+O_2(g)\to CO_2(g); \Delta_c H^{\theta}=-393kJ\ mol^{-1}\\ H_2(g)+\dfrac{1}{2}O_2(g)\to H_2O(l);\Delta_f H^{\theta}=-286 kJ mol^{-1}.$

Solution :

The reaction that takes place during the formation of $CH _3 OH(l)$ can be written as:$\\$ $C(s) + 2H_ 2 O(g) + O_ 2 (g) \to CH_ 3 OH(l) (1)$$\\$ The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:$\\$ Equation (ii) + 2 × equation (iii) - equation (i)$\\$ $\Delta_f H^{\theta} [CH _3 OH(l)] = \Delta_c H^{\theta} + 2\Delta_f H^{\theta} [H2O(l)] - \Delta_r H^{\theta}\\ = (-393 kJ mol^{-1}) + 2(-286 kJ mol^{-1}) - (-726 kJ mol^{-1})\\ = (-393 - 572 + 726) kJ mol^{-1} \Delta_f H^{\theta} [CH 3 OH(l)] = -239 kJ mol^{-1}$

15   Calculate the enthalpy change for the process$\\$ $CCl _4 (g) \to C (g) + 4Cl(g)$$\\$ and calculate bond enthalpy of $C-Cl$ in $ CCl_ 4 (g).$$\\$ $\Delta_{ vap} H^{\theta} (CCl ) = 30.5 kJ\ mol^{-1}.\\ \Delta_ f H^{\theta}(CCl 4 ) = -135.5 kJ\ mol^{-1}.\\ \Delta_ a H^{\theta} (C) = 715.0 kJ mol^{-1} $, where $\Delta_ aH^{\theta}$ is enthalpy of atomisation$\\$ $\Delta_ a H^{\theta} (Cl 2 ) = 242 kJ mol^{-1}$

Solution :

( i ) CCl _4 ( l )\to CCl_ 4 ( g ) \Delta_{ vap} H^{\theta} (CCl ) = 30.5 kJ\ mol^{-1}.\\ ( ii ) C (s)\to C(g)\Delta_ a H^{\theta} (C) = 715.0 kJ mol^{-1} \\ ( iii ) Cl _2 ( g)\to 2 Cl ( g )$\Delta_ a H^{\theta} (Cl 2 ) = 242 kJ mol^{-1}\\ (iii)C(g)+4Cl(g)\to CCl_4(g)\Delta_ f H^{\theta}(CCl 4 ) = -135.5 kJ\ mol^{-1}.$ $\\$ Enthalpy change for the given process $CCl_4(g) \to C(g)+4Cl(g)$can be calculated using the following algebraic calculations as:$\\$ Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)$\\$ $\Delta H=\Delta _a H^{\theta}(C)+2\Delta _a H^{\theta}(C12)-\Delta _{vap}H^{theta}-\Delta_f H\\ = (715.0 kJ mol^{-1}) + 2(242 kJ mol^{-1}) - (30.5 kJ mol^{-1}) - (-135.5 kJ mol^{-1})\\ \Delta H = 1304 kJ mol^{-1}$ $\\$ Bond enthalpy of C-Cl bond in CCl 4 (g)$\\$ $\dfrac{1304}{4}kJ \ \ mol^{-1}\\ =326 kJ\ mol^{-1}$

16   For an isolated system, $\Delta U = 0,$ what will be $\Delta S?$

Solution :

$\Delta S$ will be positive i.e., greater than zero.$\\$ Since $\Delta U = 0,\Delta S$ will be positive and the reaction will be spontaneous.

17   For the reaction at 298 K,$\\$ $2A + B \to C$ $\\$ $\Delta H = 400 kJ mol^{-1}$ and $\Delta S = 0.2 kJ K^{-1} mol^{-1}$$\\$ At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?

Solution :

From the expression,$\\$ $\Delta G = \Delta H - T\Delta S$ $\\$ Assuming the reaction at equilibrium, $\Delta T$ for the reaction would be:$\\$ $ T=(\Delta H-\Delta G)\dfrac{1}{\Delta S}\\ =\dfrac{\Delta H}{\Delta S}$ $\\$ ($\Delta G=0$at equilibrium)$\\$ $\dfrac{400 kJ\ mol ^{- 1}}{0.2 kJ\ K ^{- 1}\ mol ^{- 1}}\\ T=2000 K$ $\\$ For the reaction to be spontaneous, $\Delta G$ must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

18   For the reaction,$\\$ $2 Cl ( g ) \to Cl_ 2 ( g )$$\\$ what are the signs of $\Delta _H $ and $\Delta _S ?$

Solution :

$\Delta _H $ and $\Delta _S $are negative$\\$ The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released.$\\$ Hence, $\Delta _H $ is negative.$\\$ Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\Delta _S $ is negative for the given reaction.$\\$

19   For the reaction $2A(g) + B(g) \to 2D (g)$$\\$ $2 A ( g )+ B ( g )\to 2 D ( g )\\ \Delta U \theta = -10.5 kJ$ and $\Delta S ^{\theta}= -44.1 JK^{-1}.$ Calculate $\Delta G^{\theta}$ for the reaction, and predict whether there action may occur spontaneously.

Solution :

For the given reaction,$\\$ $2 A ( g )+ B ( g )\to 2 D ( g )\\ \Delta n g = 2 - (3)\\ = -1 mole$ $\\$ Substituting the value of $\Delta U ^{\theta}$ in the expression of $\Delta H:$$\\$ $\Delta H ^{\theta}= \Delta U ^{\theta} + \Delta n g RT\\ = (-10.5 kJ) - (-1) (8.314 × 10^{-3} kJ K^{-1} mol^{-1}) (298 K)\\ = -10.5 kJ - 2.48 kJ\\ \Delta H ^{\theta}= -12.98 kJ$$\\$ Substituting the values of $\Delta H ^{\theta}$ and $\Delta S ^{\theta}$ in the expression of $\Delta G ^{\theta}:$$\\$ $\Delta G ^{\theta} = \Delta H ^{\theta} - T\Delta S^{\theta} \\ = -12.98 kJ - (298 K) (-44.1 J K^{-1})\\ = -12.98 kJ + 13.14 kJ\\ \Delta G ^{\theta} = + 0.16 kJ$$\\$ Since$ \Delta G ^{\theta}$ for the reaction is positive, the reaction will not occur spontaneously.

20   The equilibrium constant for a reaction is 10. What will be the value of $\Delta G ^{\theta} ? R = 8.314 JK^{-1} mol^{-1}, T = 300 K.$

Solution :

From the expression,$\\$ $\Delta G ^{\theta} = -2.303 RT \log Keq$ $\Delta G ^{\theta}$ for the reaction,$\\$ $= (2.303) (8.314 JK^{-1} mol^{-1}) (300 K) \log10 = -5744.14 Jmol^{-1}\\ = -5.744 kJ mol^{-1}$

21   Comment on the thermodynamic stability of NO(g), given$\\$ $\dfrac{1}{2}NO(g)+\dfrac{1}{2}O_2(g)\to NO_2(g): \Delta_r H^{\theta}=90kJ\ mol^{-1}\\ NO(g)+\dfrac{1}{2}O_2(g)\to O_2(g):\Delta _r H^{\theta} =-74kJ\ mol^{-1}$

Solution :

The positive value of $\Delta_r H^{\theta}$ indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N 2 and O 2 ).$\\$ Hence, NO(g) is unstable.$\\$ The negative value of $ \Delta_r H^{\theta}$ indicates that heat is evolved during the formation of NO 2 (g) from NO(g) and O 2 (g). The product, NO 2 (g) is stabilized with minimum energy.$\\$ Hence, unstable NO(g) changes to unstable NO 2 (g).

22   Calculate the entropy change in surroundings when 1.00 mol of $H _2 O(l)$ is formed under standard conditions. $\Delta_ f H^{\theta} = -286 kJ mol^{-1.}$

Solution :

It is given that $286 kJ\ mol^{-1}$ of heat is evolved on the formation of 1 mol of $H_ 2 O(l).$ $\\$ Thus, an equal amount of heat will be absorbed by the surroundings. $\\$ $q _{surr }= +286 kJ\ mol^{-1}$ $\\$ Entropy change ($\Delta $Ssurr) for the surroundings = $\dfrac{q_{surr}}{7}\\ =\dfrac{286kJ\ mol^{-1}}{298k}\\ \Delta S_{surr}=959.73 J\ mol^{-1}K^{-1}$