## CBSE NCERT Class XI (11th) | Chemistry

Chapter 6 - Thermodynamics

Q1 :

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

Answer :

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.

Q2 :

Answer :

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

(i) Δ

*T*= 0
(ii) Δ

*p*= 0
(iii)

*q*= 0
(iv)

*w*= 0Answer :

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.

Q3 :

Answer :

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

Answer :

The enthalpy of all elements in their standard state is zero.

Therefore, alternative (ii) is correct.

Q4 :

Answer :

Δ

*U*ÃŽÂ¸of combustion of methane is -*X*kJ mol-1. The value of Δ*H*ÃŽÂ¸is
(i) = Δ

*U*ÃŽÂ¸
(ii) > Δ

*U*ÃŽÂ¸
(iii) < Δ

*U*ÃŽÂ¸
(iv) = 0

Answer :

SinceΔ

*H*ÃŽÂ¸= Δ*U*ÃŽÂ¸+ Δ*n**g*R*T*and Δ*U*ÃŽÂ¸= -*X*kJ mol-1,
Δ

*H*ÃŽÂ¸= (-*X*) + Δ*n**g*R*T.*
⇒ Δ

*H*ÃŽÂ¸< Δ*U*ÃŽÂ¸
Therefore, alternative (iii) is correct.

Q5 :

Answer :

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 CH4(

*g*) will be
(i) -74.8 kJ mol-1 (ii) -52.27 kJ mol-1

(iii) +74.8 kJ mol-1 (iv) +52.26 kJ mol-1.

Answer :

According to the question,

Thus, the desired equation is the one that represents the formation of CH4 (

*g*)i.e.,
Enthalpy of formation of CH4(

*g*) = â€“74.8 kJ molâ€“1
Hence, alternative (i) is correct.

Q6 :

Answer :

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

(iv) possible at any temperature

Answer :

For a reaction to be spontaneous, Δ

*G*should be negative.
Δ

*G*= Δ*H*-*T*Δ*S*
According to the question, for the given reaction,

Δ

*S*= positive
Δ

*H*= negative (since heat is evolved)
⇒ Δ

*G*= negative
Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Q7 : In a process, 701 J of heat is absorbed by a system and 394 J of

Answer :

Answer :

According to the first law of thermodynamics,

Δ

*U*=*q*+*W*(i)
Where,

Δ

*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

Δ

*U*= 701 J + (-394 J)
Δ

*U*= 307 J
Hence, the change in internal energy for the given process is 307 J.

Q8 :

Answer :

The reaction of cyanamide, NH2CN(

*s*),with dioxygen was carried out in a bomb calorimeter, and Δ*U*was found to be -742.7 kJ mol-1at 298 K. Calculate enthalpy change for the reaction at 298 K.Answer :

Enthalpy change for a reaction (Δ

*H*) is given by the expression,
Δ

*H*= Δ*U*+ Δ*n**g*R*T*
Where,

Δ

*U*= change in internal energy
Δ

*n**g*= change in number of moles
For the given reaction,

Δ

*n**g*= ∠‘*n**g*(products) - ∠‘*n**g*(reactants)
= (2 - 1.5) moles

Δ

*n**g*= 0.5 moles
And,

Δ

*U*= -742.7 kJ mol-1*T*= 298 K

R = 8.314 x 10-3 kJ mol-1 K-1

Substituting the values in the expression of Δ

*H*:
Δ

*H*= (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1 K-1)
= -742.7 + 1.2

Δ

*H*= -741.5 kJ mol-1
Q9 : Calculate the number of kJ of heat necessary to raise the temperature

Answer :

Answer :

From the expression of heat (

*q*),*q*=

*m*. c. Δ

*T*

Where,

c = molar heat capacity

*m*= mass of substance

Δ

*T*= change in temperature
Substituting the values in the expression of

*q*:*q*= 1066.7 J

*q*= 1.07 kJ

Q10 :

Answer :

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ

*fus**H*= 6.03 kJ mol-1 at 0°C.*C*

*p*[H2O(l)] = 75.3 J mol-1 K-1

*C*

*p*[H2O(s)] = 36.8 J mol-1 K-1

Answer :

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°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at â€“10°C.

= (75.3 J molâ€“1 Kâ€“1) (0 â€“ 10)K + (â€“6.03 × 103 J molâ€“1) + (36.8 J molâ€“1 Kâ€“1) (â€“10 â€“ 0)K

= â€“753 J molâ€“1â€“ 6030 J molâ€“1â€“ 368 J molâ€“1

= â€“7151 J molâ€“1

= â€“7.151 kJ molâ€“1

Hence, the enthalpy change involved in the transformation is â€“7.151 kJ molâ€“1.

Q11 : Enthalpy of combustion of carbon to CO

Answer :

Answer :

Formation of CO2 from carbon and dioxygen gas can be represented as:

(1 mole = 44 g)

Heat released on formation of 44 g CO2= â€“393.5 kJ molâ€“1

Heat released on formation of 35.2 g CO2

= â€“314.8 kJ molâ€“1

Q12 :

Answer :

Enthalpies of formation of CO(

*g*), CO2(*g*), N2O(*g*) and N2O4(*g*) are â€“110 kJ molâ€“1, â€“ 393 kJ molâ€“1, 81 kJ molâ€“1and 9.7 kJ molâ€“1respectively. Find the value of Δ*r**H*for the reaction:
N2O4(

*g*)+ 3CO(*g*) N2O(*g*)+ 3CO2(*g*)Answer :

Δ

*r**H*for a reaction is defined as the difference between Δ*f**H*value of products and Δ*f**H*value of reactants.
For the given reaction,

N2O4(

*g*)+ 3CO(*g*) N2O(*g*)+ 3CO2(*g*)
Substituting the values of Δ

*f**H*for N2O, CO2, N2O4,and CO from the question, we get:
Hence, the value of Δ

*r**H*for the reaction is.
Q13 :

Answer :

Given

; Δ

*r**H*θ= â€“92.4 kJ molâ€“1
What is the standard enthalpy of formation of NH3gas?

Answer :

Standard enthalpy of formation of a compound is the change in 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.

Re-writing the given equation for 1 mole of NH3(

*g*),
Standard enthalpy of formation of NH3(

*g*)
= ½ Δ

*r**H*θ
= ½ (â€“92.4 kJ molâ€“1)

= â€“46.2 kJ molâ€“1

Q14 :

Answer :

Calculate the standard enthalpy of formation of CH3OH(

*l*) from the following data:
CH3OH(

*l*) + O2(*g*) CO2(*g*) + 2H2O(*l*) ; Δ*r**H*θ = â€“726 kJ molâ€“1
C(

*g*) + O2(*g*) CO2(*g*) ; Δ*c**H*θ = â€“393 kJ molâ€“1
H2(

*g*) +Answer :

The reaction that takes place during the formation of CH3OH(

*l*)can be written as:
C(

*s*)+ 2H2O(*g*)+ O2(*g*) CH3OH(*l*) (1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) â€“ equation (i)

Δ

*f**H*θ[CH3OH(*l*)] = Δc*H*θ + 2Δ*f**H*θ[H2O(*l*)] â€“ Δ*r**H*θ
= (â€“393 kJ molâ€“1) + 2(â€“286 kJ molâ€“1) â€“ (â€“726 kJ molâ€“1)

= (â€“393 â€“ 572 + 726) kJ molâ€“1

Δ

*f**H*θ[CH3OH(*l*)] = â€“239 kJ molâ€“1
Q15 :

Answer :

Calculate the enthalpy change for the process

CCl4(

*g*) → C(*g*) + 4Cl(*g*)
and calculate bond enthalpy of C-Cl in CCl4(

*g*).
Δ

*vap**H*ÃŽÂ¸ (CCl4) = 30.5 kJ mol-1.
Δ

*f**H*ÃŽÂ¸ (CCl4) = -135.5 kJ mol-1.
Δ

*a**H*ÃŽÂ¸ (C) = 715.0 kJ mol-1, where Δ*a**H*ÃŽÂ¸ is enthalpy of atomisation
Δ

*a**H*ÃŽÂ¸ (Cl2) = 242 kJ mol-1Answer :

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

Δ

*vap**H*θ = 30.5 kJ molâ€“1
Δ

*a**H*θ = 715.0 kJ molâ€“1
Δ

*a**H*θ = 242 kJ molâ€“1
Δ

*f**H*= â€“135.5 kJ molâ€“1
Enthalpy change for the given process can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) â€“ Equation (i) â€“ Equation (iv)

Δ

*H*= Δ*a**H*θ(C) + 2Δ*a**H*θ (Cl2) â€“ Δ*vap**H*θ â€“ Δ*f**H*
= (715.0 kJ molâ€“1) + 2(242 kJ molâ€“1) â€“ (30.5 kJ molâ€“1) â€“ (â€“135.5 kJ molâ€“1)

Δ

*H*= 1304 kJ molâ€“1
Bond enthalpy of Câ€“Cl bond in CCl4 (

*g*)
= 326 kJ molâ€“1

Q16 : For an isolated system, Δ

Answer :

Answer :

ΔS will be positive i.e., greater than zero

Since Δ

*U*= 0, Δ*S*will be positive and the reaction will be spontaneous.
Q17 :

Answer :

For the reaction at 298 K,

2A + B → C

Δ

*H*= 400 kJ mol-1and Δ*S*= 0.2 kJ K-1mol-1
At what temperature will the reaction become spontaneous considering Δ

*H*and Δ*S*to be constant over the temperature range?Answer :

From the expression,

Δ

*G*= Δ*H*â€“*T*Δ*S*
Assuming the reaction at equilibrium, Δ

*T*for the reaction would be:
(Δ

*G*= 0 at equilibrium)*T*= 2000 K

For the reaction to be spontaneous, Δ

*G*must be negative. Hence, for the given reaction to be spontaneous,*T*should be greater than 2000 K.
Q18 :

Answer :

For the reaction,

2Cl(

*g)*→ Cl2(*g*),what are the signs of Δ*H*and Δ*S*?Answer :

Δ

*H*and Δ*S*are negative
The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, Δ

*H*is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, Δ

*S*is negative for the given reaction.
Q19 :

Answer :

For the reaction

2A(

*g*) + B(*g*) → 2D(*g*)
Δ

*U*ÃŽÂ¸ = -10.5 kJ and Δ*S*ÃŽÂ¸= -44.1 JK-1.
Calculate Δ

*G*ÃŽÂ¸ for the reaction, and predict whether the reaction may occur spontaneously.Answer :

For the given reaction,

2 A(

*g*) + B(*g*) → 2D(*g*)
Δ

*n**g*= 2 - (3)
= -1 mole

Substituting the value of Δ

*U*ÃŽÂ¸ in the expression of Δ*H*:
Δ

*H*ÃŽÂ¸ = Δ*U*ÃŽÂ¸ + Δ*n**g*R*T*
= (-10.5 kJ) - (-1) (8.314 x 10-3 kJ K-1 mol-1) (298 K)

= -10.5 kJ - 2.48 kJ

Δ

*H*ÃŽÂ¸ = -12.98 kJ
Substituting the values of Δ

*H*ÃŽÂ¸ and Δ*S*ÃŽÂ¸ in the expression of Δ*G*ÃŽÂ¸:
Δ

*G*ÃŽÂ¸ = Δ*H*ÃŽÂ¸ -*T*Δ*S*ÃŽÂ¸
= -12.98 kJ - (298 K) (-44.1 J K-1)

= -12.98 kJ + 13.14 kJ

Δ

*G*ÃŽÂ¸ = + 0.16 kJ
Since Δ

*G*ÃŽÂ¸ for the reaction is positive, the reaction will not occur spontaneously.
Q20 : The equilibrium constant for a reaction is 10. What will be the value

Answer :

Answer :

From the expression,

Δ

*G*ÃŽÂ¸= -2.303 R*T*log*K**eq*
Δ

*G*ÃŽÂ¸for the reaction,
= (2.303) (8.314 JK-1mol-1) (300 K) log10

= -5744.14 Jmol-1

= -5.744 kJ mol-1

Q21 :

Answer :

Comment on the thermodynamic stability of NO(

*g*),given
N2(

*g*) + O2(*g*) Ã¢â€ ’ NO(*g*); Δ*r**H*θ= 90 kJ molâ€“1
NO(

*g*)+O2(*g*) Ã¢â€ ’ NO2(*g*) : Δ*r**H*θ= â€“74 kJ molâ€“1Answer :

The positive value of Δ

*r**H*indicates that heat is absorbed during the formation of NO(*g*). This means that NO(*g*) has higher energy than the reactants (N2 and O2). Hence, NO(*g*) is unstable.
The negative value of Δ

*r**H*indicates that heat is evolved during the formation of NO2(*g*) from NO(*g*)and O2(*g*). The product, NO2(*g*) is stabilized with minimum energy.
Hence, unstable NO(

*g*) changes to stable NO2(*g*).
Q22 : Calculate the entropy change in surroundings when 1.00 mol of H

Answer :

Answer :

It is given that 286 kJ molâ€“1of heat is evolved on the formation of 1 mol of H2O(

*l*). Thus, an equal amount of heat will be absorbed by the surroundings.*q*

*surr*= +286 kJ molâ€“1

Entropy change (Δ

*S**surr*) for the surroundings =
Δ

*S**surr*= 959.73 J molâ€“1Kâ€“1**<< Back to**

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