NCERT Solutions For Class 11 Chemistry Chapter 6

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics

NCERT solutions class 11 chemistry chapter 6 thermodynamics function is one of the key tools to prepare chemistry for 11th standard examination. The NCERT solutions for class 11 chemistry chapter 6 thermodynamics function is provided here to help students prepare for their examination more effectively. The NCERT solutions for class 11 chemistry chapter 6 is prepared by expert teachers according to the latest syllabus of the Central Board of Secondary Education. The NCERT solutions for class 11 chemistry chapter 6 will also help students to get familiar with the exam pattern and marking scheme of the examination. Check the NCERT solutions for class 11 chemistry chapter 6 pdf given below.

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. Δp=0

4. ΔT=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 ΔUΘ is – Y kJmol1. Then value of ΔHΘ is ____.

(a) > ΔUΘ

(b) = ΔUΘ

(c) = 0

(d) < ΔUΘ

Ans:

(d) < ΔUΘ

Reason:

ΔHΘ=ΔUΘ+ΔngRT ; ΔUΘ = – Y kJmol1,

ΔHΘ=(Y)+ΔngRTΔHΘ<ΔUΘ

Q-5: For, Methane, di-hydrogen and and graphite the enthalpy of combustion at 298K are given -890.3kJ mol1, -285.8kJmol1 and -393.5kJmol1 respectively. Find the enthalpy of formation of Methane gas?

(a) -52.27kJmol1

(b) 52kJmol1

(c) +74.8kJmol1

(d) -74.8kJmol1

Ans:

(d) -74.8kJmol1

à CH4(g)  + 2O2(g) CO2(g) + 2H2O(g)

ΔH=890.3kJmol1

à C(s) + O2(g)   CO2(g)

ΔH=393.5kJmol1

à 2H2(g) + O2(g) 2H2O(g)

ΔH=285.8kJmol1

à C(s) + 2H2(g) CH4(g)

ΔfHCH4  = ΔcHc + 2ΔfHH2ΔfHCO2

= [ -393.5 +2(-285.8) – (-890.3)] kJmol1

= -74.8kJmol1

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

ΔG should be –ve, for spontaneous reaction to occur

ΔG = ΔH – TΔS

As per given in question,

ΔH is –ve ( as heat is evolved)

ΔS is +ve

Therefore, ΔG 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 ΔU for the given process.

 

Ans:

As per Thermodynamics 1st law,

ΔU = q + W(i);

ΔU internal energy = heat

W = work done

W = -594 J (work done by system)

q = +801 J (+ve as heat is absorbed)

Now,

ΔU = 801 + (-594)

ΔU = 207 J

Q-8: The reaction given below was done in bomb calorimeter, and at 298K we get, ΔU = -753.7 kJ mol1. Find ΔH at 298K.

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

Ans:

ΔH is given by,

ΔH=ΔU+ΔngRT………………(1)

Δng = change in number of moles

ΔU = change in internal energy

Here,

Δng=ng(product)ng(reactant)

  = (2 – 2.5) moles

 Δng = -0.5 moles

Here,

T =298K

ΔU = -753.7 kJmol1

R  = 8.314×103kJmol1K1

Now, from (1)

ΔH=(753.7kJmol1)+(0.5mol)(298K)(8.314×103kJmol1K1)

= -753.7 – 1.2

ΔH = -754.9 kJmol1

Q-9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from 45Cto65C. For aluminium molar haet capacity is 24 Jmol1K1

Ans:

Expression of heat(q),

q=mCPΔT;………………….(a)

ΔT = Change in temperature

c = molar heat capacity

m = mass of substance

From (a)

q=(5027mol)(24mol1K1)(20K)

q = 888.88 J q

Q-10: Calculate ΔH for transformation of 1 mole of water into ice from10Cto(10)C.. ΔfusH=6.03kJmol1at10C.

Cp[H2O(l)]=75.3Jmol1K1

Cp[H2O(s)]=36.8Jmol1K1

 Ans:

ΔHtotal = sum of the changes given below:

(a) Energy change that occurs during transformation of 1 mole of water from 10Cto0C.

(b) Energy change that occurs during transformation of 1 mole of water at 0C  to  1 mole of ice at 0C.

(c) Energy change that occurs during transformation of 1 mole of ice from 0Cto(10)C.

ΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Os]ΔT

= (75.3 Jmol1K1)(0 – 10)K + (-6.03*1000 Jmol1(-10-0)K

= -753 Jmol1 – 6030Jmol1 – 368Jmol1

= -7151 Jmol1

= -7.151kJmol1

Thus, the required change in enthalpy for given transformation is -7.151kJmol1.

Q-11: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393kJmol1,-110kJmol1, 9.7kJmol1 and 81kJmol1 respectively. Then, ΔrH = _____.

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(g)

 Ans:

ΔrH for any reaction is defined as the fifference between ΔfH value of products and ΔfH value of reactants.”

ΔrH=ΔfH(products)ΔfH(reactants)

Now, for

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(g)

ΔrH=[(ΔfH(N2O)+(3ΔfH(CO2))(ΔfH(N2O4)+3ΔfH(CO))]

Now, substituting the given values in the above equation, we get:

ΔrH = [{81kJmol1 + 3(-393) kJmol1} – {9.7kJmol1 + 3(-110) kJmol1}]

ΔrH = -777.7 kJmol1

 

Q-12 Enthalpy of combustion of C to CO2 is -393.5 kJmol1. 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); ΔfH = -393.5 kJmol1

1 mole CO2 = 44g

Heat released during formation of 44g CO2 = -393.5 kJmol1

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

= 393.5kJmol144g×37.2g

= -332.69 kJmol1

 

Q-13: N2(g) + 3H2(g) à 2NH3(g) ; ΔrHΘ = -92.4 kJmol1

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) ΔrHΘ

= (0.5)(-92.4kJmol1)

= -46.2kJmol1

 

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); ΔrHΘ = -726 kJmol1

C(g) + O2(g) à CO2(g); ΔcHΘ = -393 kJmol1

H2(g) + (1/2)O2(g) à H2O(l); ΔfHΘ = -286 kJmol1

 

Ans:

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

CH3OH(l) can be obtained as follows,

ΔfHΘ [CH3OH(l)] = ΔcHΘ

2ΔfHΘΔrHΘ

= (-393 kJmol1) +2(-286kJmol1) – (-726kJmol1)

= (-393 – 572 + 726) kJmol1

= -239kJmol1

Thus, ΔfHΘ [CH3OH(l)] = -239kJmol1

 

Q-15: Calculate ΔH for the following process

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

ΔvapHΘ (CCl4) = 30.5 kJmol1.

ΔfHΘ (CCl4) = -135.5 kJmol1.

ΔaHΘ (C) = 715 kJmol1,

ΔaHΘ  is a enthalpy of atomisation

ΔaHΘ (Cl2) = 242 kJmol1.

 

Ans:

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

(1) CCl4(l) à CCl4(g) ; ΔvapHΘ = 30.5 kJmol1

(2) C(s) à C(g) ΔaHΘ = 715 kJmol1

(3) Cl2(g) à 2Cl(g) ; ΔaHΘ = 242 kJmol1

(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ = -135.5 kJmol1

ΔH for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:

ΔH=ΔaHΘ(C)+2ΔaHΘ(Cl2)ΔvapHΘΔfH

= (715kJmol1) + 2(kJmol1) – (30.5kJmol1) – (-135.5kJmol1)

Therefore, H=1304kJmol1

The value of bond enthalpy for C-Cl in CCl4(g)

= 13044kJmol1

= 326 kJmol1

 

Q-16: ΔU = 0 for isolated system, then what will be ΔU?

Ans:

ΔU is positive ; ΔU > 0.

As, ΔU = 0 thenΔS will be +ve, as a result reaction will be spontaneous.

Q-17:

Following reaction takes place at 298K,

2X + Y à Z

ΔH = 400 kJmol1

ΔH = 0.2kJmol1K1

Find the temperature at which the reaction become spontaneous considering ΔS and ΔH to be constant over the entire temperature range?

Ans:

Now,

ΔG=ΔHTΔS

Let, the given reaction is at equilibrium, then ΔT will be:

T = (ΔHΔG)1ΔS

ΔHΔS; (ΔG = 0 at equilibrium)

= 400kJmol1/0.2kJmol1K1

Therefore, T = 2000K

Thus, for the spontaneous, ΔG must be –ve and T > 2000K.

 

Q-18: 2Cl(g) à Cl2(g)

In above reaction what can be the sign for ΔS and ΔH?

Ans:

ΔS and ΔH 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,  ΔH is negative.

Also, 2 moles of Chlorine atoms is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, ΔS is negative.

 

Q-19: 2X(g) + Y(g) à 2D(g)

ΔUΘ = -10.5 kJ and ΔSΘ = -44.1JK1

Determine ΔGΘ for the given reaction, and predict that whether given reaction can occur spontaneously or not.

Ans:

2X(g) + Y(g) à 2D(g)

Δng = 2 – 3

= -1 mole

Putting value of ΔUΘ in expression of ΔH:

ΔHΘ=ΔUΘ+ΔngRT

= (-10.5KJ) – (-1)( 8.314×103kJK1mol1)(298K)

= -10.5kJ -2.48kJ

ΔHΘ = -12.98kJ

Putting value of ΔSΘ and ΔHΘ in expression of ΔGΘ:

ΔGΘ=ΔHΘTΔSΘ

= -12.98kJ –(298K)(-44.1JK1)

= -12.98kJ +13.14kJ

ΔGΘ = 0.16kJ

As, ΔGΘ is positive, the reaction won’t occur spontaneously.

 

Q-20: Find the value of ΔGΘ for the reaction, if equilibrium is given 10.given that T = 300K and R =  8.314×103kJK1mol1.

Ans:

Now,

ΔGΘ = 2.303RTlogeq

= (2.303)( 8.314×103kJK1mol1)(300K) log10

= -5744.14Jmol1

-5.744kJmol1

 

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

(1/2)N2(g) + (1/2)O2(g); ΔrHΘ=90kJmol1

NO(g) + (1/2)O2(g) à NO2(g) ; ΔrHΘ=74kJmol1

Ans:

The +ve value of ΔrH 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 ΔrH 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 ΔS in surrounding given that  mole of H2O(l) is formed I standard condition. ΔrHΘ=286kJmol1.

 Ans:

ΔrHΘ=286kJmol1 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 = +286kJmol1.

Now, ΔSsurr = Qsurr/7

= 286kJmol1298K

Therefore, ΔSsurr=959.73Jmol1K1

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