Class 11 | Chapter 6 (Thermodynamics) NCERT Chemistry Solutions

Question 1
Choose the correct answer

A thermodynamic state junction 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

(ii) whose value is independent of path (Correct Option)
In thermodynamics, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state.

A state function describes the equilibrium state of a system. 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.


Question 2
For the process to occur under adiabatic conditions, the correct condition is

(i) ∆T= 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0

Adiabatic Condition means that there’s no transfer of heat between the system and surroundings. Therefore option (iii) ∆q = 0 is correct


Question 3
The enthalpies of all elements in their standard states are
(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

By definition, the enthalpy of formation of elements in their standard state is taken as zero.

Therefore, The enthalpies of all elements in their standard states is zero irrespective of the element.

Thus option (ii) zero is correct.


Question 4
ΔU of combustion of methane is – X kJ mol-1

The value of ΔH is

(i) = ΔU

(ii) > ΔU

(iii) < ΔU

(iv) 0

Option (iii) < ΔU is correct

We know that

ΔH = ΔU + ΔngRT

Chemical Equation for combustion of methane is

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

Δng = Number of gaseous moles in product side – number of gaseous moles in reactant side

Number of gaseous moles in product side = 1

Number of gaseous moles in reactant side = 3

Δng = 1 – 3 = – 2

Δng = – 2
ΔU = – X

Replacing these values in equation ΔH = ΔU + ΔngRT

ΔH = – X – 2RT

Therefore
ΔH < ΔU

Thus (iii) < ΔU is correct option


Question 5
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


Question 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

(iv) possible at any temperature

(iv) possible at any temperature is correct option

As per Gibbs Free Energy Equation

ΔG = ΔH – TΔS

Where
ΔG = Change in Gibbs Free Energy
ΔH = Change in Enthalpy
T = Temperature
ΔS = Change in Entropy

For reaction
A + B → C + D + q

ΔS > 0

ΔH < 0 (Because heat is being released see q on right hand side of chemical equation)

Therefore as per equation ΔG = ΔH – TΔS will be always ΔG < 0 no matter what is value of temperature T.


Question 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?

Change in internal energy ΔU = q + w

As per question its given that
q = 701 J
w = – 394 J (negative because work is done by system)

Plugging these values in equation ΔU = q + w

ΔU = 701 – 394 = 307 J

ΔU = 307 J

Therefore if in a process 701 J of heat is absorbed by a system and 394 J of work is being done by system then change in internal energy during this process will be 307 J


Question 8
The reaction of cyanamide NH2CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742.7 KJ mol-1 at 298 K.

Calculate the enthalpy change for the reaction at 298 K

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


We know that

ΔH = ΔU + ΔngRT

Chemical Equation for combustion of methane is

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

Δng = Number of gaseous moles in product side – number of gaseous moles in reactant side

Number of gaseous moles in product side = 2

Number of gaseous moles in reactant side = 3/2

Δng = 2 – 3/2 = (4 – 3)/2 = 1/2

Δng = 1/2

ΔU = – 742.7 kJ mol-1

R = 8.314 × 10-3 kJ K-1 mol-1

T = 298 K
Replacing these values in equation ΔH = ΔU + ΔngRT

ΔH = – 742.7 + (1/2) × 8.314 × 10-3 × 298

ΔH = – 742.7 + 4.157 × 10-3 × 298

ΔH = – 742.7 + 1238.786 × 10-3

ΔH = – 742.7 + 1.238 = −741.462

ΔH = −741.462 = – 741.5 kJ

ΔH = – 741.5 kJ


Question 9
Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C.

Molar heat capacity of Al is 24 J mol-1 K-1

Relationship between molar heat capacity and heat is q = mc(T2 – T1)
Where
q = heat
m = amount of substance (moles)
c = molar heat capacity
T1 = initial temperature
T2 = final temperature

c = 24 J mol-1 K-1

m = 60/27 moles

T1 = 35 + 273 = 308 K

T2 = 55 + 273 = 328 K

Plugging all these values in equation q = mc(T2 – T1)

q = (60/27) × 24 × (328 – 308)

q = 2.22 × 24 × 20 = 1065.6 J

q = 1.0656 kJ


Question 10
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 °C to ice at – 10.0°C

ΔfusH = 6.03 kJ mol-1 at 0°C

Cp [H2O (l)] = 75.3 J mol-1 K-1

Cp [H2O (l)] = 36.8 J mol-1 K-1


Question 11
Enthalpy of combustion of carbon to carbon dioxide is – 393.5 J mol-1

Calculate the heat released upon formation of 35.2 g of CO from carbon and oxygen gas

Heat of combustion for 1 mole of CO2 is – 393.5 kJ

Number of moles in 35.2 g of CO2 = 35.2/44 = 0.8 mole

Thus heat of combustion for 0.8 mole of CO2 is – 393.5 × 0.8 = – 315 kJ

Therefore heat released upon combustion of 35.2 g of CO2 from carbon and oxygen gas is – 315 kJ


Question 12
Calculate the enthalpy of the reaction

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Given that
fH CO(g) = – 110 kJ mol-1

fH CO2(g) = – 393 kJ mol-1

fH N2O(g) = 81 kJ mol-1

fH N2O4(g) = 9.7 kJ mol-1

Enthalpy of reaction = total heat of formation for products – total heat of formation for reactants

= [81 + 3(- 393)] – [9.7 + 3(- 110)]

= [81 – 1179] – [9.7 – 330]

= – 1098 – (- 320.3)

= – 1098 + 320.3

= – 777.7 kJ mol-1

Therefore enthalpy of reaction is – 777.7 kJ mol-1


Question 13
Given N2(g) + H2(g) → 2 NH3 (g)

rH = – 92.4 kJ mol-1

What is the standard enthalpy of formation of NH3 gas?

Standard enthalpy of formation of NH3 gas = rH/2 = – 92.4/2 = – 46.2 kJ mol-1

Therefore standard enthalpy of formation of NH3 gas is – 46.2 kJ mol-1


Question 14
Calculate the standard enthalpy of formation of CH3OH (l) form the following data.

CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) ∆rH = – 726 kJ mol-1

C (s) + O2 (g) → CO2 (g) ∆cH = – 393 kJ mol-1

H2 (g) + O2 (g) → H2O (l) ∆fH = – 286 kJ mol-1

Therefore standard enthalpy of formation of CH3OH (l) is – 239 kJ mol-1


Question 15

Calculate the enthalpy change for the process

CCl4 (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)

Given
vap H° (CCl4) = 30.5 kJ mol-1 
fH°(CCl4) = – 135.5 kJ mol-1
aH° (C) = 715.0 kJ mol-1 where ∆a H° is enthalpy of atomisation
aH° (Cl2) = 242 kJ mol-1

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

(i) CCl4(l)    →    CCL4(g)      ΔvapH° = 30.5 kJ mol-1

(ii) C(s)    →   C(g)               ΔaH° = 715.0 kJ mol-1

(iii) Cl2(g)  →   2Cl(g)          ΔaH° = 242 kJ mol-1

(iv) C(g)  + 4Cl(g)  →  CCl4(g)  ΔfH = -135.5 kJ mol-1

Enthalpy change for the given process  C(g)  + 4Cl(g)  →  CCl4(g)   can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

ΔH = ΔaH°(C)  +  2ΔaH° (Cl2) –  ΔvapH° – ΔfH

= (715.0 kJ mol-1) + 2(242 kJ mol-1) – (30.5 kJ mol-1) – (-135.5 kJ mol-1)

∴ΔH = 1304 kJ mol-1

Bond enthalpy of C-Cl bond in CCl4(g) = 326 kJ mol-1


Question 16
For an isolated system ΔU = 0
What will be ΔS?

A system that is isolated from its surroundings will have no change in its internal energy, denoted by the symbol U, since it will not be exchanging any energy with its surroundings.

However, entropy has a tendency to rise whenever there is a spontaneous process. Therefore, S must be greater than 0 or positive.


Question 17
For a reaction at 298 K

2A + B → C
∆H = 400 kJ mol-1 and ∆S = 0.2 kJ mol-1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Gibbs-Helmholtz Equation
∆G = ∆H – T∆S

For reaction to be spontaneous ∆G = 0
Therefore

0 = ∆H – T∆S

∆H = T∆S

T = ∆H/∆S

Replacing ∆H = 400 kJ mol-1 and ∆S = 0.2 kJ mol-1

T = 400/0.2 = 2000 K

T = 2000 K

Therefore this reaction will be in a state of equilibrium at temperature 2000 K and will be spontaneous above this temperature. So for reaction to be spontaneous temperature should be greater than 2000 K.


Question 18
For the reaction 2 Cl (g) → Cl2 (g) what will be the signs of ∆H and ∆S?

Reaction 2 Cl (g) → Cl2 (g) is showing formation of chlorine molecule from chlorine atoms, which means bonds are being created therefore energy is released. Thus ∆H for reaction 2 Cl (g) → Cl2 (g) is negative.

Also in the reaction 2 Cl (g) → Cl2 (g) two atoms are combining to form one molecule therefore randomness of atoms is decreasing therefore spontaneity is decreasing as chlorine atoms combing to form chlorine molecule. Therefore ∆S for the reaction 2 Cl (g) → Cl2 (g) is negative.

Thus for reaction 2 Cl (g) → Cl2 (g) signs of both ∆H and ∆S are negative.


Question 19
For the reaction

2 A(g) + B(g) → 2D(g)

ΔU° = – 10.5 kJ

ΔS° = – 44.1 J K–1 mol-1

Calculate ΔG0 for the reaction, and predict whether the reaction may occur spontaneously.

For reaction 2 A(g) + B(g) → 2D(g)
ΔU° = – 10.5 kJ
ΔS° = – 44.1 J K–1 mol-1
value of ΔG° is + 5.12 kJ therefore reaction will not occur spontaneously.


Question 20
Equilibrium constant for a reaction is 10

Calculate the value of ΔG given that R = 8 J K-1 mol-1 and T = 300 K

Equation for calculating value of ΔG is

ΔG = – RT ln K
Where
ΔG = Gibbs Free Energy
R = Gas Constant
T = Temperature
K = Equilibrium Constant

Replacing
R = 8 J K-1 mol-1

T = 300 K

K = 10

Plugging all these values in equation ΔG = – RT ln K

ΔG = – 8 × 300 × ln 10

Value of ln 10 = 2.302

ΔG = – 8 × 300 × 2.302 = – 5524.8 J mol-1

ΔG = – 5.5248 kJ mol-1

Therefore if equilibrium constant for a reaction is 10 at temperature 300 K then value of Gibbs Free Energy will be
– 5.5248 kJ mol-1


Question 21
Comment on the thermodynamic stability of NO(g) and NO2(g) given

1/2 NO2 (g) + 1/2 O2 (g) → NO (g) ΔrH = 90 kJ mol-1

NO (g) + 1/2 O2 (g) → NO2 (g) ΔrH = – 74 kJ mol-1

If ΔrH is positive then it means that products have more energy as compared to reactants, therefore products are unstable.

If ΔrH is negative then it means that products have less energy as compared to reactants, therefore products are more stable as compared to reactants.

Based on these facts
For reaction 1/2 NO2 (g) + 1/2 O2 (g) → NO (g) value of ΔrH is positive therefore NO(g) is unstable in nature.

For reaction NO (g) + 1/2 O2 (g) → NO2 (g) value of ΔrH is negative therefore NO2(g) is stable in nature.


Question 22
Calculate the entropy change in surroundings when 1.0 mole of H2O (l) is formed under standard conditions.

Given that ΔfH = – 286 kJ mol-1

It is given that 286 kJ mol-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.

qsurr = +286 kJ mol-1

Entropy change (ΔSsurr) for the surroundings =  qsurr / T

= 286/298

ΔSsurr = 0.959 J mol-1K-1


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