 # NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics)

## NCERT Solutions For Class 11 Chemistry Chapter 6 PDF Free Download

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

1. Thermodynamic Terms
• The System and the Surroundings
• Types of Thermodynamic Systems
• State of the System
• Internal Energy as a State Function
2. Applications
• Work
• Enthalpy, H
3. Measurement Of ΔU And ΔH: Calorimetry
4. Enthalpy Change and Reaction Enthalpy
5. Enthalpies For Different Types Of Reactions
6. Spontaneity
7. 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.

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.

Q-2: 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.

Q-3: 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

Q-4: 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 $kJ mol^{-1}$,

$\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }$

Q-5: For, Methane, di-hydrogen 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 CH4(g)  + 2O2(g) $\rightarrow$ CO2(g) + 2H2O(g)

$\Delta H = -890.3kJmol^{-1}$

à C(s) + O2(g)  $\rightarrow$ CO2(g)

$\Delta H = -393.5kJmol^{-1}$

à 2H2(g) + O2(g) $\rightarrow$ 2H2O(g)

$\Delta H = -285.8kJmol^{-1}$

à C(s) + 2H2(g) $\rightarrow$ CH4(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$mol^{-1}$

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

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 +ve

Therefore, $\Delta G$ is negative

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 $\Delta U$ for the given process.

Ans:

As per Thermodynamics 1st law,

$\Delta U$ = q + W(i);

$\Delta U$ internal energy = heat

W = 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 J

Q-8: 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.

NH2CN(g) +3/2O2(g)à N2(g) + CO2(g) + H2O(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 energy

Here,

$\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)$

= (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 $kJmol^{-1}$

Q-9: 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 temperature

c = molar heat capacity

m = mass of substance

From (a)

$q = ( \frac{50}{27}mol )(24mol^{-1}K^{-1})(20K)$

q = 888.88 J q

Q-10: 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 $J mol^{-1}$(-10-0)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$kJ mol^{-1}$.

Q-11: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393$kJ mol^{-1}$,-110$kJ mol^{-1}$, 9.7$kJ mol^{-1}$ and 81$kJ mol^{-1}$ respectively. Then, $\Delta _{r}H$ = _____.

N2O4(g) + 3CO(g)à  N2O(g) + 3 CO2(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

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(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 $kJ mol^{-1}$

Q-12 Enthalpy of combustion of C to CO2 is -393.5 $kJ mol^{-1}$. Determine the heat released on the formation of 37.2g of CO2 from dioxygen and carbon.

Ans:

Formation of carbon dioxide from di-oxygen and carbon gas is given as:

C(s) + O2(g) à CO2(g); $\Delta _{f}H$ = -393.5 $kJ mol^{-1}$

1 mole CO2 = 44g

Heat released during formation of 44g CO2 = -393.5 $kJ mol^{-1}$

Therefore, heat released during formation of 37.2g of CO2  can be calculated as

= $\frac{-393.5kJmol^{-1}}{44g}\times 37.2g$

= -332.69 $kJ mol^{-1}$

Q-13: N2(g) + 3H2(g)à 2NH3(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)N2(g) + (1.5)H2(g) à 2NH3(g)

Therefore, Standard Enthalpy for formation of ammonia gas

= (0.5) $\Delta _{r}H^{\Theta }$

= (0.5)(-92.4$kJ mol^{-1}$)

= -46.2$kJ mol^{-1}$

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); $\Delta _{r}H^{\Theta }$ = -726 $kJ mol^{-1}$

C(g) + O2(g)à CO2(g); $\Delta _{c}H_{\Theta }$ = -393 $kJ mol^{-1}$

H2(g) + (1/2)O2(g)à H2O(l); $\Delta _{f}H^{\Theta }$ = -286 $kJ mol^{-1}$

Ans:

C(s) + 2H2O(g) + (1/2)O2(g) à CH3OH(l) …………………………(i)

CH3OH(l) can be obtained as follows,

$\Delta _{f}H_{\Theta }$ [CH3OH(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, $\Delta _{f}H_{\Theta }$ [CH3OH(l)] = -239$kJ mol^{-1}$

Q-15: Calculate $\Delta H$ for the following process

CCl4(g)à C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).

$\Delta _{vap}H^{\Theta }$ (CCl4) = 30.5 $kJ mol^{-1}$.

$\Delta _{f}H^{\Theta }$ (CCl4) = -135.5 $kJ mol^{-1}$.

$\Delta _{a}H^{\Theta }$ (C) = 715 $kJ mol^{-1}$,

$\Delta _{a}H^{\Theta }$  is a enthalpy of atomisation

$\Delta _{a}H^{\Theta }$ (Cl2) = 242 $kJ mol^{-1}$.

Ans:

“ The chemical equations implying to the given values of enthalpies” are:

(1) CCl4(l) à CCl4(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) Cl2(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 CCl4(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 C-Cl in CCl4(g)

= $\frac{1304}{4}kJmol^{-1}$

= 326 $kJ mol^{-1}$

Q-16: $\Delta U$ = 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.

Q-17:

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.

Q-18: 2Cl(g)à Cl2(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.

Q-19: 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

$\Delta H^{\Theta }$ = -12.98kJ

Putting 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

$\Delta G^{\Theta }$ = 0.16kJ

As, $\Delta G^{\Theta }$ is positive, the reaction won’t occur spontaneously.

Q-20: 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$kJmol^{-1}$

Q-21: What can be said about the thermodynamic stability of NO(g), given

(1/2)N2(g) + (1/2)O2(g); $\Delta _{r}H^{\Theta } = 90kJmol^{-1}$

NO(g) + (1/2)O2(g)à NO2(g) ; $\Delta _{r}H^{\Theta } = -74kJmol^{-1}$

Ans:

The +ve value of $\Delta _{r}H$ represents that during NO(g) formation from O2 and N2,  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 NO2(g) formation from O2(g) and NO(g),  heat is evolved. The obtained product, NO2(g) gets stabilized with minimum energy.

Thus, unstable NO(g) converts into unstable  NO2(g).

Q-22: 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 H2O(l).

Thus, the same heat will be absorbed by surrounding. Qsurr = +286$kJmol^{-1}$.

Now, $\Delta S_{surr}$ = Qsurr/7

= $\frac{286kJmol^{-1}}{298K}$

Therefore, $\Delta S_{surr} = 959.73Jmol^{-1}K^{-1}$

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 Also Access NCERT Exemplar for class 11 chemistry Chapter 6 CBSE Notes for class 11 chemistry Chapter 6 