Chemistry NCERT Exemplar Solutions Class 11th Chapter Six: Overview, Questions, Preparation

This is a Long Answer Type Questions as classified in NCERT Exemplar

A contrast is drawn in thermodynamics between extensive and intense qualities. An extensive property is one whose value is proportional to the amount or size of matter in the system. Extensive properties include mass, volume, internal energy, enthalpy, and heat capacity, to name a few.

Properties that are independent of the amount or size of matter present are known.

As though they were intensive properties Temperature, density, and pressure, for example, are intense properties. A molar property? m, is the value of an extensive property of the system for 1 mol of the substance. If n is the amount of matter? m =? /n, is independent of the amount of matter. An intensive property is always the ratio of two extensive properties.

Extensive/Extensive = Intensive

E.g., Mole Fraction = Moles / Total number of moles = Extensive/Extensive

This is a Long Answer Type Questions as classified in NCERT Exemplar

We know,

ΔS_total = ΔS_sys + ΔS_surr

When a system is in thermal equilibrium with its surroundings, the surroundings' temperature is the same as the system's. Furthermore, a rise in the enthalpy of the surroundings equals a decrease in the system's enthalpy. As a result of the entropy shift in the environment,

ΔS_surr = ΔH_surr/T = -ΔH_sys/T

ΔS_total = ΔS_sys = (-ΔH_sys/T)

Rearranging the above equation:

ΔS_total = TΔS_sys - ΔH_sys

For spontaneous process,

ΔS_total > 0, so 

TΔS_sys - ΔH_sys> 0

⇒ ( - ΔH_sys - TΔS_sys) > 0

The above equation can be written as

- ΔG > 0

ΔG = ΔH - TΔS < 0

ΔH_sys is the enthalpy change of a reaction, TΔS_sys is the energy which is not available to do

useful work. As a result, G is a measure of 'free energy, ' as it is the net energy available to conduct beneficial work. As a result, it's also known as the reaction's free energy. At constant pressure and temperature, G gives a set of conditions for spontaneity,

(i) If ΔG is negative (<0), the process is spontaneous.

(ii) If ΔG is positive (>0), the process is nonspontaneous.

 Unit of ΔG is Joule.

 The reaction will be spontaneous at high temperature.

(i) Reversible work is represented by the combined areas

(ii) Work against constant pressure, P_f  is represented by the area

          

Work (i) > Work (ii). 

The Approach While Dealing With the Concept of Thermodynamics

Since the concept of Thermodynamics and the terminologies of Chemistry are a bit new to the students, they should first learn the names of different chemical components and how to write the chemical equations properly through NCERT Exemplars and Solutions books.

While writing the chemical equations, they might make some mistakes. To avoid making errors, they should practice writing the equations at home.

Solving previous years' question papers will help them to manage time and quickly analyze the answers in their minds.

Comparing their own answers with the solutions provided in the NCERT Exemplar Solutions will help them to be sure of the fact that how efficient they are at answering the questions of Chemistry. It will help them to boost their confidence as well.

Revising is extremely important when it comes to answering Chemistry questions since a lack of practice and habits of writing can make the students forget some important information that they should include in their answers. The mock papers are beneficial for them to serve this purpose

Rational thinking plays an important part in answering the questions in Chemistry. The students need to have a clear idea of the basic physical characteristics of different chemical compounds and how they will react depending on the presence of other external factors. Practicing is the only way to sharpen their thinking ability and analysis power.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Enthalpy of a reaction is the energy change per mole for the process.

18 g of H₂O = 1 mole ΔHvap = 40.79 kJ/ mol

 Enthalpy change for vapourising 2 moles of H2O = 2 x 40.79 = 81.58 kJ ΔH°vap = 40.79 kJ mol-1.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Due to weak force of attraction between molecules, acetone requires less heat to vaporise. Hence, water has higher enthalpy of vaporization.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Standard molar enthalpy of formation,  _rH₁^- is just a special case of _fH₂^-, where

one mole of a compound is formed from its constituent elements. In the above equation, enthalpy of formation and enthalpy of reaction is not the same.

This is a Short Answer Type Questions as classified in NCERT Exemplar

ΔH for formation is given. For the reverse reaction, ΔHchanges sign as the reverse of exothermic reaction will be endothermic. So, ΔH for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,

ΔH=2×91.8=183.6KJ

This is a Short Answer Type Questions as classified in NCERT Exemplar

According to Hess’s law, Δ_rH = Δ_rH₁+ Δ_rH₂+Δ_rH₃

This is so because during the reaction A→ B, B’s formation undergoes various intermediate reactions, with the overall value of the enthalpy being Δ_rH.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The reaction presented in the question is

CH₄ (g)→C (g) + 4H (g)

Now, Δ_aH = 1665  kJ/mol

The mean bond enthalpy of the C-H bond should be used here. For the atomisation of 4 moles of C-H bonds, the value is 1665 kJ/mol. So, per mole energy = 1665/4 = 416.2 kJ / mol

This is a Short Answer Type Questions as classified in NCERT Exemplar

In order to calculate the lattice enthalpy of NaBr,

(i) Na (s) →Na (g) ; Δ_subH? =108.4 kJ/mol

(ii) Na→Na⁺ + e^-   Δ_iH? = 496kJ/ mol

(iii) $\mathrm{}\frac{1}{2}$ Br→ Br, $\mathrm{}\frac{1}{2}$ Δ_diss H? = 96kJ/ mol

(iv) Br+e^-Br^-  Δ_egH? = - 325 kJmol^-1

? _fH? =? _subH? + Δ_diss H + Δ_i H? + Δ_i H? + Δ_eg H? +? _lattice H?

= -360.1 -108.4-96-496+325 = -735.5KJ/ mol

This is a Short Answer Type Questions as classified in NCERT Exemplar

It is a spontaneous process. Although enthalpy change is zero, randomness or disorder (ΔS) increases and ΔS is positive. Therefore, in the equation, ΔG = ΔH – TΔS, the term TΔS will be negative. Hence ΔG will be negative.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Heat has a randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is? S = q_rev/ T

Here? S = change in entropy  ^

q_rcv = heat of reversible reaction '

T = temperature

This is a Short Answer Type Questions as classified in NCERT Exemplar

Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are the same.

This is a Short Answer Type Questions as classified in NCERT Exemplar

ΔrG° = -RT ln Kp

= -RT ln (0.98)

Since In (0.98) is negative

.’. ΔrG° is positive the reaction is non spontaneous

This is a Short Answer Type Questions as classified in NCERT Exemplar

During cyclic process, change in internal energy is zero.

ΔU = 0

and no work is said to be done, as system returns to the initial state.
For a steady state cyclic process at any given stage enthalpy is one single value however, at different stage it would vary.

ΔH = 0

This is a Short Answer Type Questions as classified in NCERT Exemplar

The standard molar entropy of H₂0 (1) is 70 J K^-1 mol^-1. The solid form of H₂0 is ice. In ice, molecules of H₂0 are less random than in liquid water. Thus, molar entropy of H₂0 (s) < molar entropy of H₂0 (1). The standard molar entropy of H₂0 (s) is less than 70 J K^-1 mol^-1.

This is a Short Answer Type Questions as classified in NCERT Exemplar

State functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Molar enthalpy of vaporization is more for water due to hydrogen bonding between water molecules.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Gibbs energy for a reaction in which all reactants and products are in standard state. ΔrG° is related to the equilibrium constant of the reaction as follows

ΔrG = ArG° + RT In K

At equilibrium, 0 = ΔrG° + RT InA– ( {ΔrG = 0) or  ΔrG° =-RT lnK

ΔrG° = 0 when K= 1

For all other values of K, ArG° will be non-zero.

This is a Short Answer Type Questions as classified in NCERT Exemplar

For an isolated system w = 0, q = 0

Since ΔU= q + w = 0 + 0 = 0, ΔU= 0

This is a Short Answer Type Questions as classified in NCERT Exemplar

At constant volume

q = ΔU + (-w)

-w = pΔ q = AU + pΔV

ΔV = 0 (at constant volume)

Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV

Since ΔU + pΔV=ΔH

=> qp = ΔH change in enthalpy

Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.

This is a Short Answer Type Questions as classified in NCERT Exemplar

During free expansion, external pressure is zero, so Work done, w = -pextΔV

= -0(5 – 1) = 0

Since the gas is expanding isothermally, therefore, q = 0

ΔU = q + w =0+0=0

This is a Short Answer Type Questions as classified in NCERT Exemplar

For water, molar heat capacity = 18 x Specific heat or Cp = 18 x c

But, specific heat,

C = 4.18 J g^-1 K^-1 Heat capacity,

Cp = 18 x 4.18 JK^-1  = 75.24 JK^-1

This is a Short Answer Type Questions as classified in NCERT Exemplar

Given that, Cv = heat capacity at constant volume,

Cp = heat capacity at constant pressure

 Difference between Cp and Cv is equal to gas constant (R).

.’. Cp – Cv = nR                                (where, n = no. of moles)

= 10 x 8.314 = 83.14J

This is a Short Answer Type Questions as classified in NCERT Exemplar

Molar enthalpy change for graphite (ΔH)

= enthalpy change for 1 g x molar mass of C = -20.7×12 = -2.48 x 102 kJ mol-1

Since the sign of ΔH = -ve, it is an exothermic reaction.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Δ_rH? =ΣB.E _(reactant)-ΣB.E _(product)

 =B.E._H2 + BE_BR2 -2 ×B.E>_HBr

= 435+192-2 (×368)

= 109KJmol^-1

This is a Short Answer Type Questions as classified in NCERT Exemplar

As per the information provided in the question, for one mole of CCl₄ (154 g), the heat of vaporisation required is 30.5 kJ/mol .

Hence for the vaporisation of 284 g of CCl₄, we require:

$\frac{284}{154}*30.5$  = 56.2 kJ

Chemistry NCERT Exemplar Solutions Class 11th Chapter six 
Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol^–1

Now the enthalpy of formation for H₂O will be half the enthalpy of the value in the given equation. So now we can calculate that

? _fH? = $\frac{-572\mathrm{}\mathrm{K}\mathrm{J}}{2}$ = -286KJ/mol

This is a Short Answer Type Questions as classified in NCERT Exemplar

Assumption: Cylinder is filled with one mole gas, and the piston is frictionless. Let the pressure of gas inside be p and the volume of gas be V_ {I}.

Piston is moved towards the inside to make the external pressure (P_ {ext}) equal to p. Now, let us assume that this change takes place in a single step, hence, V is the final volume. The work done by the piston is depicted in the graph shown below by shading the area.

P_extΔV= AV₁   (V₁-V₂)

This is a Short Answer Type Questions as classified in NCERT Exemplar

When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesimal amounts as compression happens from initial volume V_i to the final volume V_f. The below graph depicts the work done with the shaded area.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Throwing a stone from ground to roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:In process b), potential energy/enthalpy change is a contributing factor to the spontaneity.

This is a Short Answer Type Questions as classified in NCERT Exemplar

No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

This is a Short Answer Type Questions as classified in NCERT Exemplar

We can conclude from the figure that this change is a reversible change.

Now,

W= -2.303nRT log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

But,  p₁V₁ = p₂V_{2 = $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$} = $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$ = $\frac{2}{1}$ =2

 W= -2.303nRT log $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

 = -2.303 *8.314*1*298*log2

= -2.303 *8.314*298*0.3010J

= -1717.46J

This is a Short Answer Type Questions as classified in NCERT Exemplar

We know that the amount of work done =-p_ext? V

On substituting the values in the formula, we get,

-2bar* (50-10)L=-80Lbar

According to the described problem,1 LBar = 100J

Therefore, -80 L bar= (-80*100)= -8000J

= -8kJ, which is the amount of work done

The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i)

Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or 1 Kelvin). That is why it is an intensive property which does not depend on mass.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)

Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

The correct option is B <ΔU $°$
The balanced equation for the combustion of methane is:

CH₄ (g)+2O₂ (g)→CO₂ (g)+2H₂O (l)

Here,  Δng=1−3=−2

ΔH $°$ =ΔU $°$ +ΔngRT

ΔH $°$ =−393−2RT

∴ΔH $°$ <ΔU $°$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

In free expansion, w=0 because volume is constant, as the process is adiabatic q = 0 and from first law of thermodynamics.

U=q+w

This means that internal energy remains constant.

Kindly go through the solution

option (ii)

 

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)? S (system) decreases, but? S (surroundings) increases.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)

When we add equations (b) and (c) we will get equation (a); hence, the Algebraic sum of y and z will give x, x =y + z.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)

Same bonds are formed in reaction (a) and (b) but bonds between the reactant molecules are broken only in reaction (b)

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound as the reaction could be exothermic or endothermic.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

Enthalpy of sublimation can be shown as

Solid | Liquid | Vapour

Hence, Enthalpy of sublimation of a substance is equal to enthalpy of fusion + enthalpy of vaporisation.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (ii)

positive for a spontaneous reaction

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i)

The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i) and (ii)

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and? _rH will also be negative. Hence, option (i) and (ii) is the correct answer.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii) and (iv)

Gas expands to fill the available space spontaneously, and burning of carbon to carbon dioxide is spontaneous

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iv)

For
isothermal reversible change

q= -w = nRTln $\frac{V_f}{V_i}$  =2.303nRTlog

$\frac{{\mathrm{W}}_{600\mathrm{k}}}{W_{300\mathrm{k}}}$ = $\frac{1\times \mathrm{R}\times 600\mathrm{K}\mathrm{l}\mathrm{n}\frac{10}{1}\mathrm{}\mathrm{}}{1\times \mathrm{R}\times 300\mathrm{K}\mathrm{l}\mathrm{n}\frac{10}{1}}$ = $\frac{600}{300}$  =2

For
isothermal expansion of ideal gases,  ? U = 0

Since,
temperature is constant this means there is no change in internal energy.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i) and (iii)

Since, the above reaction is an exothermic reaction, the enthalpy of reactants is more than the enthalpy of products

Kindly go through the solution

This is a Matching Type Questions as classified in NCERT Exemplar

(i) (b)  

(ii) (c) 

(iii) (a)

This is a Matching Type Questions as classified in NCERT Exemplar

(i)-  (c)

(ii)- (a)

(iii)- (b)

This is a Matching Type Questions as classified in NCERT Exemplar

(i)-  (b), (d)     

(ii)- (b)

(iii)- (c)

(iv)- (a)

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option (ii)

Explanation: The enthalpy of the reactants is always greater than the enthalpy of the product in a combustion reaction.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans:  (ii)

Explanation: The energy factor for a spontaneous process should be favourable (i.e., -ve) and the randomness should be positive.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option  (i)

Explanation: The entropy of a liquid reduces as it crystallises. Because the molecules are more organised in crystalline form.