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 $6{-1}$