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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

Edited By Sumit Saini | Updated on Aug 18, 2022 04:14 PM IST

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Thermodynamics is the branch of science which deals with the transfer of energy or changes in energy in physical or chemical processes. This is an important chapter for class 11th students because some concepts of this chapter will also be applied in the next chapter which is equilibrium and also some in the 12th class NCERT book chapter like chemical kinetics etc.

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In NCERT solutions for class 11 chemistry chapter 6 Thermodynamics, there are 22 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 6 thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you in the preparation of class 11 final examination as well as in the 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.

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Exercise Questions


Question 6.1 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.

Answer :

A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of path.
Like- p,V,T depends on the state of the system, not on the path.

So, the correct option is (ii)

Question 6.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) ΔT=0
(ii) Δp=0
(iii) q = 0
(iv) w = 0

Answer :

In adiabatic condition, the heat exchange between the system and the surrounding through its boundary is not allowed.
So, the correct option is (iii) q = 0

Question 6.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Answer :

The enthalpies of all elements in their standard states are Zero.
So, the correct option is (ii)

Question 6.4 ΔU of combustion of methane is XKJmol1 . The value of ΔH is
(i) =ΔU
(ii) >ΔU
(iii) <ΔU
(iv) = 0

Answer :

Equation of combustion of methane-

CH4(g)+2O2(g)CO2(g)+2H2O(l)

Since, ΔHo=ΔUo+ΔngRT and ΔUo=XKJmol1

from the equation Δng=(npnR)=13=2

ΔHo=X2RT

ΔHo<ΔUo

so the correct option is (iii).

Question 6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, 890.KJmol1,393.5KJmol1 and 285.8KJmol1 respectively. Enthalpy of formation of CH4(g) will be

(i) –74.8 kJmol1

(ii) –52.27 kJmol1

(iii) +74.8 kJmol1

(iv) +52.26 kJmol1

Answer :

CH4+2O2CO2+2H2O ΔH ................. -890.3 kJ/mol
C+2O2CO2 ΔH ..................-393.5 kJ/mol
2H2+O22H2O ΔH ..................-258.8 kJ/mol

So, the required equation is to get the formation of CH4(g) by combining these three equations-

Thus, C+2H2CH4

ΔfHCH4=ΔcHc+2ΔcHH2
= [-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)

Question 6.6 A reaction, A + B C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature

Answer :

For the reaction to be feasible the ΔG should be negative
ΔG=ΔHTΔS

According to question,
ΔS = positive unit
ΔH = negative (as heat is evolved in the reaction)
Overall the ΔG is negative.

Therefore the reaction is possible at any temperature. So, the correct option is (iv)

Question 6.7 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?

Answer :

The first law of Thermodynamics state that,
ΔU=q+W

where, ΔU = 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 equation of first law we get,

ΔU=701+(394)=307J

So, the change in internal energy for the process is 307 J

Question 6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be 742.7KJmol1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH2CN(g)+32O2(g)N2+CO2(g)H2O(l)

Answer :

Given information,
ΔU=742.7kJmol1
T = 298 K
R = 8.314 ×103

Δng= ng (products) ng (reactants)
= (2 - 1.5)
= 0.5 moles

The enthalpy change for the reaction is expressed as;

ΔH=ΔU+ngRT
where, ΔU = change in internal energy and
Δng= change in no. of moles

By putting the values we get,
ΔH=(742.7)+0.5(298)(8.314×103)
=741.5kJmol1

Question 6.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35C to 55C . Molar heat capacity of Al is 24Jmol1K1

Answer :

We know that
q=mcΔT
m = mass of the substance
c = heat capacity
ΔT = change in temperature

By putting all these values we get,
q=(6027mol)(24J/mol/K)(20K)
= 1066.7 J
= 1.07 kJ

Question 6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0C to ice at 10.0C .

ΔfusH=6.03KJmol1 at 0C .

Cp[H2O(l)]=75.3Jmol1K1.Cp[H2O(s)]=36.8Jmol1K1

Answer :

Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from 10oC to 0oC of 1 mol of water (water water )
(ii) from 0oC to 0oC of 1 mol of ice (water ice)
(iii) from 0oC to 10oC of 1 mole of ice (ice ice)

So, the total enthalpy change
ΔH=Cp[H2O(l)](ΔT)+ΔH(freezing)+Cp[H2O(s)]ΔT

=(75.3)(10)+(6.03×103)+(36.8)(10)=7536030368=7151Jmol1=7.151kJ/mol
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol

Question 6.11 Enthalpy of combustion of carbon to CO2 is 393.5KJmol1 Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

Answer :

Formation of carbon dioxide from carbon and dioxygen reaction is-

C+O2CO2 ............. ΔHf = -393.5 kJ/mol

1 mole of CO2 = 44 g
So, the heat released in the formation of 44g of CO2 = -393.5 kJ/mol
therefore, In 35.5 g of CO2 the amount of heat released =
393.544×35.2=314.8kJ/mol

Question 6.12 Enthalpies of formation of CO(g),CO2(g),N2O(g)andN2O4(g) are –110, – 393, 81 and 9.7 KJmol1 respectively. Find the value of ΔrH for the reaction:

N2O4(g)+3CO(g)N2O(g)+3CO2(g)

Answer :

Given,
Enthalpies of formation of CO(g),CO2(g),N2O(g)andN2O4(g) are –110, – 393, 81 and 9.7 KJmol1
N2O4(g)+3CO(g)N2O(g)+3CO2(g)
We know that the ΔrH=ΔfH (product) ΔfH (reactants)

For the above reaction,

ΔrH=[ΔfH(N2O)+3ΔfH(CO2)][ΔfH(N2O4)+3ΔfH(CO)]
Substituting the given values we get,
=[81+3(393)][9.7+3(110)]=777.7kJ/mol
Thus the value of ΔrH of the reaction is -777.7 kJ/mol

Question 6.13 Given N2(g)+3H2(g)2NH3(g);ΔrH=92.4kJmol1 .What is the standard enthalpy of formation of NH3 gas?

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;
12N2+32H2NH3

Therefore, standard enthalpy of formation of ammonia is = 12ΔrHΘ
= 1/2 (-92.4)
= -46.2 kJ/mol

Question 6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH(l)+3/2O2(g)CO2(g)+2H2O(l)ΔrH=726kJmol1.C(graphite)+O2(g)CO2(g);ΔcH=393kJmol1.H2(g)+1/2O2(g)H2O(l);ΔfH=286kJmol1.

Answer :

for the formation of CH3OH the reaction is ,

C+2H2O+12O2CH3OH
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)

ΔfH[CH3OH]=ΔcHΘ+2ΔfH[H2O]ΔrHΘ
=(393)+2(286)(726)
=239kJmol1

Question 6.15 Calculate the enthalpy change for the process CCl4(g)C(g)+4Cl(g) and calculate bond enthalpy of CCl in CCl4(g)
ΔvapH(CCl4)=30.5kJmol1..ΔfH(CCl4)=135.5kJmol1..ΔaH(C)=715.0kJmol1,whereΔaHisenthalpyofatomisation.ΔaH(Cl2)=242kJmol1

Answer :

We have the following chemical reactions equations-

(1) CCl4(l)CCl4(g) ................ ΔvapHo=30.5kJ/mol
(2) C(s)C(g) ........................ ΔaHo=715.0kJ/mol
(3) Cl2(g)2Cl(g) .................. ΔaHo=242kJ/mol
(4) C(g)+4ClCCl4(g) ............ ΔfH=135.5kJ/mol
The enthalpy change for the process CCl4(g)C(g)+4Cl(g) by the above reaction is calculated as;

ΔH=ΔaHo(C)+2(ΔaHo(Cl2))ΔvapHoΔfH
=[(715)+2(242)(30.5)(135.5)]kJ/mol
=1304kJmol1

And the bond enthalpy of C-Cl bond in CCl4 (g) = 1304/4 = 326 kJ/mol

Question 6.16 For an isolated system, ΔU=0 , what will be ΔS ?

Answer :

Since ΔU=0
So, the ΔS>0 (positive) therefore the reaction will be feasible.

Question 6.17 For the reaction at 298 K,
2A + B C
ΔH=400kJmol1andΔS=0.2kJK1mol1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.

Answer :

From the equation,

ΔG=ΔHTΔS
Suppose the reaction is at equilibrium, So the change in temperature is given as;

T=ΔHΔGΔS
=4000.2=2000K ( ΔG at equilibrium is zero)
To reaction should be spontaneous, ΔG should be neagtive. So, that for the given reaction T should be greater than 2000 K

Question 6.18 For the reaction, 2Cl(g)Cl2(g) , what are the signs of ΔHandΔS ?

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, ΔH is negative.

Two moles of an atom have more randomness than the one mole of chlorine. So, spontaneity is decreased. Hence ΔS is negative.

Question 6.19 For the reaction
2A(g)+B(g)2D(g)
ΔU=10.5KJandΔS=44.1JK1
Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.

Answer :

Given reaction is 2A(g)+B(g)2D(g)

We know that, ΔGo=ΔHoTΔSo
and ΔHo=ΔUo+ΔngRT
here Δng=2(3)=1

and ΔU=10.5KJandΔS=44.1JK1

substituting the given values in equations-
ΔG=(10.5+8.314×103×298)298(44.1)
ΔG=(12.98+13.14)kJΔG=+0.16kJ

Hence the reaction will not occur spontaneously because the ΔG value is positive for this reaction.

Question 6.20 The equilibrium constant for a reaction is 10. What will be the value of

ΔG?R=8.314JK1mol1,T=300K

Answer :

Given values are,
R=8.314JK1mol1,T=300K
and equilibrium constant = 10

It is known that,
ΔGo=2.303 RTlogKeq
=2.303(8.314)(300)log10=5744.14Jmol1=5.744kJmol1

Hence the value of ΔGo is -5.744 kJ/mol

Question 6.21 Comment on the thermodynamic stability of NO(g), given

1/2N2(g)+1/2O2(g)NO(g);ΔrH=90kJmol1.NO(g)+1/2O2(g)NO2(g):ΔrH=74kJmol1

Answer :

The formation of NO is unstable because ΔrH is positive it means heat is absorbed during the reaction. So, NO (g) has higher energy than its reactants N2 and O2 .
On the other hand, NO2 is stable because ΔrH is negative means heat is released during the formation of NO2 . It is stablised with minimum energy. Hence unstable NO changes to stable NO2 .

Question 6.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfH=286KJmol1

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.

qsurr=+286 kJ/mol
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298

= 959.73 J/mol/K

More About Thermodynamics Class 11 NCERT Chemistry Chapter

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 surroundings. After completing NCERT solutions for class 11 chemistry chapter 6 Thermodynamics you will be able to explain the terms like system, surroundings, and boundary; explain three types of system open, close and isolated systems; explain work, heat, and internal energy, etc. In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically, internal energy, enthalpy, entropy, Gibbs energy. You will find NCERT solutions for class 11 chemistry chapter 6 Thermodynamics by scrolling down.

Important points of NCERT solutions for class 11 chemistry chapter 6 Thermodynamics-

1. System and Surroundings - in thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. The universe = The system+ The Surroundings .

2. The First Law of Thermodynamics- according to it under any circumstances, energy cannot be created and nor be destroyed, it can be converted from one form to another form with the interaction of work, heat and internal energy. Mathematically, this is represented as: ΔU=q+w , where w is the work done on or by the system, q is the heat exchanged between the system and its surroundings and ΔU is the total change in internal energy of the system .

3. The second law of thermodynamics- it states that entropy of the entire universe(total entropy of both the system and its surroundings), as an isolated system, will always increase over time.

NCERT Syllabus Class 11 Chemistry Chapter 6 Thermodynamics-Topics

6.1 Thermodynamic Terms

6.2 Applications

6.3 Measurement of ∆U and ∆H: Calorimetry

6.4 Enthalpy Change, ∆H of a Reaction – Reaction Enthalpy

6.5 Enthalpies for Different Types of Reactions

6.6 Spontaneity

6.7 Gibbs Energy Change and Equilibrium

NCERT Solutions for Class 11 Chemistry

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

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be a function of the molecular mass of the substance.

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Option 1)

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Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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twice that in 60 g carbon

Option 2)

6.023 × 1022

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Option 4)

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Option 1)

less than 3

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more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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