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Thermodynamics

Class 11th Chemistry Part I Bihar Board Solution
Exercise
  1. Choose the correct answer. A thermodynamic state function is a quantity (i) used to…
  2. For the process to occur under adiabatic conditions, the correct condition is: (i) (ii)…
  3. The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) 0…
  4. of combustion of methane is X kJ mol1. The value of His (i) = (ii) (iii) (iv) = 0…
  5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, 890.3 kJ mol1…
  6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction…
  7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the…
  8. The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter,…
  9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of…
  10. Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at -10.0°C.…
  11. Enthalpy of combustion of carbon to CO2 is -393.5 kJ mol-1. Calculate the heat released…
  12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110, - 393, 81 and 9.7…
  13. Given N2(g) + 3H2(g) → 2NH3(g); = -92.4 kJ mol-1 What is the standard enthalpy of…
  14. Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH…
  15. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond…
  16. For an isolated system, ΔU = 0, what will be ΔS?
  17. For the reaction at 298 K, 2A + B → C = 400 kJ mol-1 and = 0.2 kJ K-1 mol-1 At what…
  18. For the reaction, 2 Cl(g) → Cl2(g), what are the signs of and ?
  19. For the reaction 2 A(g) + B(g) → 2D(g) = -10.5 kJ and = -44.1 JK-1. Calculate for the…
  20. The equilibrium constant for a reaction is 10. What will be the value of ? R = 8.314 JK-1…
  21. Comment on the thermodynamic stability of NO(g), given N2(g) +O2(g) → NO(g); = 90 kJ mol-1…
  22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under…

Exercise
Question 1.

Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes

(ii) whose value is independent of the path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only.


Answer:

Thermodynamic state functions are functions that depend only on the initial and final state of the system. It is independent of the path taken to change from one state to another.

Thus, answer is (ii) whose value is independent of path



Question 2.

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

(i)

(ii)

(iii) q = 0

(iv) w = 0


Answer:

Adiabatic process is processes that take place without any transfer of heat between the system and surrounding. i.e, in adiabatic process q = 0

Thus answer is (iii) q = 0


[Option (i) is an isothermal process. In the isothermal process, the temperature of the system remains constant.


Option (ii) is for isobaric process; where pressure of system is kept constant.]



Question 3.

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element


Answer:

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, right answer is (ii) zero



Question 4.

of combustion of methane is – X kJ mol–1. The value of ΔH is
(i) =

(ii) >

(iii) <

(iv) = 0


Answer: iii) < ΔU0

According to the first law of thermodynamics,
∆H°=∆U°+∆ngRT

we need to find ∆ng by writing the chemical equation of combustion of methane.
where, ∆ng=total mole of gaseous products - total mole of gaseous reactants.


Combustion of methane:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)

∆ng = 1 - (1 + 2) = -2

Now,
∆H°=∆U°+∆ngRT
=-X°-2RT
=-(X°+RT) < -X°

hence, ∆H°< ∆U°


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
(1) –74.8 kJ mol–1
(2) –52.27 kJ mol–1
(3) +74.8 kJ mol–1
(4) +52.26 kJ mol–1


Answer:

Given,


We need to find enthalpy change for reaction


ΔH5 = ΔH(CH4)


ΔH1 = ΔH(CO2) + 2ΔH(H2O) – ΔH(CH4)


ΔH2 = ΔH(CO2)


ΔH4 = 2ΔH(H2O)


Thus,


ΔH(CH4) = ΔH2 + ΔH4 – ΔH1
So, ΔH= (-393.5-2×285.8+ 890.3) KJ/mol
= -74.8 KJ/mol


Thus, answer is (i) –74.8 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

(v) possible at any temperature


Answer:

A reaction is said to be feasible of possible when change in Gibbs free energy for that reaction at particular temperature is negative.

We know, change in Gibbs free energy,


ΔG = ΔH – TΔS


Where


ΔH is change in enthalpy


ΔS is change in entropy


T is temperature at which reaction occurs.


Given Energy is released when a forward reaction occurs (since q is on product side), therefore ΔH is negative.


It is also given that entropy change for the reaction is positive.


Thus –TΔS would be negative for all temperatures.


Therefore,


ΔG = ΔH – TΔS would always be negative.


Thus (iv) possible at any temperature is the right answer.



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?


Answer:

We know that,

ΔU = ΔH + W


Where ΔU is change in internal energy


Δ H is change in enthalpy


And W is work done on the system.


Given, 701 J of heat is taken by the system


Thus ΔH = 701 J


And 394 J of work is done by the system


Thus W = - 394


∴ ΔU = 701 – 394


= 307 J



Question 8.

The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and DU was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.



Answer:

We know that,

ΔU = ΔH - ΔngRT


The balanced chemical equation of reaction is,


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


Number of moles of gaseous particles in product side is 2 (one N2 and one CO2)


Number of moles of gaseous particles on reactant side is 3/2 (O2)


Thus, change in number of moles of gaseous particle,


Δng = 2 – 3/2 = 1/2


∴ ΔU = ΔH - ΔngRT

Given,

ΔU = -742.7 kJ

R =8.314 J K-1 mol-1

= 8.414 × 10-3 kJ K-1 mol-1

Δng =(2-2.5) =-0.5 mole

T = 298 K

Thus,

ΔH = (-742.7 kJ mol-1) – (1/2mol)×(8.314 × 10-3 kJ K-1mol-1)×298

= -742.7 - 1.239

= -743.9 kJ


Question 9.

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.


Answer:

We Know that,

Q = ncΔT


Where n is number of mole of substance


c is molar heat capacity of substance


ΔT is rise in temperature


Given,


c = 24 J mol-1 K-1


ΔT = 55 – 35 = 20°C = 20 K


m = 60 g


number of mole of Aluminium in m gram is given as


n = m/M


Where M is molar mass of aluminium.


∴ n = 60/27 = 2.22


Thus Q = ncΔT


= (2.22 mol)×(24 J mol–1 K–1)×(20K)


= 1065.6 J


= 1.065 kJ



Question 10.

Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. .

Cp[H2O(l)] = 75.3 J mol–1 K–1

Cp[H2O(s)] = 36.8 J mol–1 K–1


Answer:

Given: ΔfusH = 6.03 kJ mol-1 at 0° C

Cp [H2O(l)] = 75.3 J mol–1 K–1


Cp [H2O(s)] = 36.8 J mol–1 K–1


To calculate the enthalpy change, we use Hess’s Law


Total ΔH = ΔH1 + ΔH2 + ΔH3


Where ΔH = CpΔT


ΔCp = Heat capacity at constant pressure


ΔT = T2-T1


∴ Total ΔH = (1 mol water at 10° C →1 mol water at 0° C) +


(1 mol water at 0° C → 1 mol water at 0° C) +


(1 mol water at 0° C → 1 mol water at -10° C)


∴ Total ΔH = Cp[H2O (l)] × ΔT + ΔHfreezing + Cp [H2O (s)] × ΔT


By substituting the given values, we get


Total ΔH = 75.3 J mol–1 K–1 × (0-10) K +(-6.03 kJ mol-1) +36.8 J mol–1 K–1× (-10) K


⇒Total ΔH = -753 Jmol–1 -6.03 kJ mol-1 -368 Jmol-1


⇒Total ΔH = -0.753 kJmol–1 - 6.03 kJ mol-1 - 0.368 kJmol-1 (1J=)


⇒Total ΔH = -7.151 kJmol-1


Thus, the enthalpy change involved in the process is -7.151 kJ/mol


Note: Hess’s Law states that “ if a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e., the total enthalpy change is independent of intermediate steps involved in the change. The enthalpy depends on the initial and final stages only.



Question 11.

Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.


Answer:

Given: Enthalpy of combustion of carbon (ΔH) = –393.5 kJ mol–1.

Mass of CO2 = 35.2g


Molar mass of CO2 = 12 + 2× 16


= 44 g/mol


∴ 1 mol of carbon = 44g


Heat released when 44g CO2 is formed = 393.5 kJ


Heat released when 35.2g CO2 is formed = × 35.2g


⇒314.8 kJ


Thus, heat released when 35.2g CO2 formed is 314.8 kJ



Question 12.

Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)


Answer:

Given: For the given reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)


Enthalpy of formation (ΔHf (CO)) = -110 kJ mol-1


ΔHf (CO2) = -393 kJ mol-1


ΔHf (N2O) = 81kJ mol-1


ΔHf (N2O4) = 9.7 kJ mol–1


Now, to calculate ΔrH for the reaction, we apply the formula given below:


ΔrH = (Sum of enthalpy of formation of products) – (sum of enthalpy of reactants)


or, ΔrH = ∑ ΔfH (Products) - ∑ ΔfH (Reactants)


In the given reaction,


N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)


Products are N2O and CO2


Reactants are N2O4 and CO


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


By substituting the values given, we get


⇒ΔrH = [81 kJ/mol + 3 × (-393 kJ/mol)] – [9.7 kJ/mol + 3 × (-110 kJ/mol)]


⇒ΔrH = -777.7 kJ/mol


Thus, the value of ΔrH for the reaction is -777.7 kJ/mol.


Note: Enthalpy of reaction (ΔrH) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted.



Question 13.

Given

N2(g) + 3H2(g) → 2NH3(g); = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?


Answer:

Given: For the given reaction,

N2(g) + 3H2(g) → 2NH3(g)


Enthalpy of reaction (ΔrH°) = –92.4 kJ mol–1


To calculate standard enthalpy of formation of NH3 gas, first we will divide the whole reaction by 2, so that we get 1 mol of NH3


N2 (g) + H2(g) → NH3(g)


Hence, standard enthalpy of NH3 is equal to the 1/2 ΔrH°


As, ΔrH° = –92.4 kJ mol–1


∴ Standard enthalpy of NH3 =


⇒Standard enthalpy of NH3 = -46.2 kJ mol-1


Thus, standard enthalpy of formation of NH3 gas = -46.2 kJ mol-1


Note: Standard enthalpy of formation is the amount of heat absorbed or evolved when 1 mole of the substance is directly obtained from its constituent elements.



Question 14.

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

CH3OH (I) + O2(g) → CO2(g) + 2H2O(l); = –726 kJ mol–1

C(graphite) + O2(g) → CO2(g); = –393 kJ mol–1

H2(g) +O2(g) → H2O(l); = –286 kJ mol–1.


Answer:

Given:

CH3OH (I) + O2(g) → CO2(g) + 2H2O(l); ΔrH° = –726 kJ mol–1(1)


C(graphite) + O2(g) → CO2(g); ΔcH° = –393 kJ mol–1 (2)


H2(g) +O2(g) → H2O(l); ΔfH°= –286 kJ mol–1 (3)


To calculate the standard enthalpy of formation of CH3OH(l), first we will make the formation reaction of CH3OH by using carbon, hydrogen, oxygen.


C (s) + 2H2 (g) + 1/2 O2 (g)→ CH3OH (l)


This is the required reaction


Now, we make the required reaction from the given data.


Step 1: Add the reaction (2) and reaction (3)


C(graphite) + O2(g) → CO2(g); ΔcH° = –393 kJ mol–1


2H2(g) +O2(g) → 2H2O(l); ΔfH°= –572 kJ mol–1. [2× reaction (2)]


C + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l); ΔfH° = -965 kJ mol-1


Step 2: Subtract reaction (1) from the above reaction, we get


C + 2H2(g) +2O2(g) → CO2(g) + 2H2O(l); ΔfH° = -965 kJ mol-1


CO2(g) + 2H2O(l) → CH3OH (l) + O2(g); ΔrH°= –726 kJ mol–1


C (s) + 2H2 (g) + 1/2 O2 (g)→ CH3OH (l); ΔrH°= –239 kJ mol–1


(formation of required reaction)


Thus, the 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 CCl4(g).

.

.

, where is enthalpy of atomisation

(Cl2) = 242 kJ mol–1


Answer:

Given:

ΔvapH° (CCl4) = 30.5 kJ mol–1


ΔfH° (CCl4) = -135.5 kJ mol–1


ΔaH° (C) = 715.0 kJ mol–1


ΔaH° (Cl2) =242 kJ mol–1


We can write,


CCl4(l) → CCl4(g); ΔH = 30.5 kJ mol–1 (1)


C(s) + 2Cl2(g) → CCl4(l); ΔH = -135.5 kJ mol–1 (2)


C(s) → C(g); ΔH = 715.0 kJ mol–1 (3)


Cl2(g) → 2Cl(g); ΔH =242 kJ mol–1 (4)


To get the required reaction, CCl4(g) → C(g) + 4 Cl(g)


We follow the steps given below:


Step 1: Add t the both reaction (1) and (2), we get


C(s) + 2Cl2(g) → CCl4(l); ΔH = -135.5 kJ mol–1


CCl4(l) → CCl4(g); ΔH = 30.5 kJ mol–1


C(s) + 2Cl2(g) → CCl4(g); ΔH = -105 kJ mol–1 (5)


Step 2: Add the both reaction (3) and (4), we get


C(s) → C(g); ΔH = 715.0 kJ mol–1


2Cl2 (g) → 4Cl (g); ΔH = 484 kJ mol–1 [ 2×reaction 4]


C(s) + 2Cl2(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol–1 (6)


Step 3: Subtract the both reaction (5) and (6), we get


C(s) + 2Cl2(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol–1


CCl4(g) → C(s) + 2Cl2(g); ΔH = -105 kJ mol–1


CCl4(g) → C(g) + 4 Cl(g); ΔH = 1304 kJ mol–1


(formation of required reaction)


Thus, the enthalpy change for the process is 1304 kJ mol–1


⇒Calculation the bond enthalpy of C-Cl in CCl4(g)


Bond enthalpy of C-Cl bond in CCl4 is equal to one fourth of the energy of dissociation of CCl4


∴ Bond enthalpy of C – Cl in CCl4(g) =


As ΔH =1304 kJ mol–1



= 326 kJ mol-1


Note: Bond enthalpy is defined as the amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 am pressure and the specified temperature is called bond enthalpy.



Question 16.

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


Answer:

An isolated system is the one in which neither energy nor matter can enter/leave the system.

∆S is the change in entropy. Entropy is a measure of randomness or disorder in a system. More the randomness, more is the entropy of a system.


According to the second law of thermodynamics the total entropy can never decrease over time for an isolated system.


The total entropy is constant in ideal cases where the system is in equilibrium, or undergoing a reversible process and approaches 0 at absolute zero only.


In all spontaneous processes, the total entropy always increases and the process is irreversible.


Therefore, ∆S is always greater than 0, ∆S>0.



Question 17.

For the reaction at 298 K, 2A + B → C

= 400 kJ mol–1 and = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering and to be constant over the temperature range.


Answer:

According to Gibbs Helmholtz equation

∆G = ∆H - T∆S, where


∆G = change in Gibbs energy


∆H = change in enthalpy, given as 400 kJ mol-1


∆S = change in entropy, given as 0.2 kJ K-1 mol-1


T = temperature at which reaction occurs


For a reaction to be spontaneous, ∆G < 0, means it must be negative.


If ∆G = 0, it is in equilibrium.


So, for reaction to be spontaneous, ∆G < 0; ∆H - T∆S < 0; ∆H < T∆S;


(∆H / ∆S) < T; 400 / 0.2 = 2000 K < T


This implies that the temperature must be more than 2000K for the reaction to be spontaneous.



Question 18.

For the reaction,

2 Cl(g) → Cl2(g), what are the signs of and ?


Answer:

In the reaction, chlorine molecule is being formed from chlorine atoms. Hence, bond formation takes place, leading to release of energy. As it is exothermic reaction, ∆H is negative (-ve).


In the reaction, 2 moles of atoms having higher randomness combine to form 1 mole of molecules, having lesser randomness.


Therefore, ∆S is negative (-ve ), which is represented as Sproducts – Sreactants.



Question 19.

For the reaction

2 A(g) + B(g) → 2D(g)

= –10.5 kJ and = –44.1 JK–1.

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


Answer:

∆U = -10.5 kJ

Also, ∆U = ∆H – ∆nRT


Where,


∆U = change in energy


∆H = change in enthalpy


∆n = moles of products – moles of reactants = 2-3 = -1


R = 8.314 × 10-3 kJ mol-1 K-1 - constant


T = temperature; here room temperature is taken as 298 K


Substituting all the above values, we get


∆H = -10.5kJ + (-1 mol × (8.314× 10-3kJ mol-1 K-1) × 298 K


= -12.977kJ


we know that


∆G = ∆H - T∆S, where


∆G = change in Gibbs energy


∆H = change in enthalpy, got as -12.977 kJ


∆S = change in entropy, given as -44.1 JK-1 = -44.1 × 10-3 kJK-1


T = temperature at which reaction occurs = 298K


Substituting the above values


∆G = -12.977 – (298 × -44.1 × 10-3)


= -12.977 – ( -13.14)


= 0.163kJ


As ∆G > 0, reaction doesn’t occur spontaneously. Reactions are spontaneous only when ∆G < 0.



Question 20.

The equilibrium constant for a reaction is 10. What will be the value of ? R = 8.314 JK–1 mol–1, T = 300 K.


Answer:

Using,

∆G = -RTlnK


⇒ ∆G = -2.303RTlogK


∆G = change in Gibbs energy


R = 8.314 × 10-3 kJ mol-1 K-1 - constant


T = temperature; given as 300K


K = equilibrium constant


Substituting the above values


∆G = -2.303 × 8.314 × 10-3 × 300 × log 10


∆G = -5.744 kJ mol-1



Question 21.

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

N2(g) +O2(g) → NO(g); = 90 kJ mol–1

NO(g) +O2(g) → NO2(g): = –74 kJ mol–1


Answer:

The first reaction is endothermic. It consumes or absorbs energy to form products. So, NO has more energy than its reactants. Hence, NO is unstable.

In the second reaction, it reacts with oxygen to form NO2, and gives off energy. It’s an exothermic reaction. So, NO2 has less energy than its reactants, and is stable. It is stabilised with minimum energy.


Hence unstable NO gets converted into stable NO2.



Question 22.

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. = –286 kJ mol–1.


Answer:

It is given that an energy of 286 kJ is released per mole, when water is formed under standard conditions. This means 286 kJ of energy is absorbed by the surroundings in this process. So,

q = - ∆H = 286 kJ mol-1


∆S = q/T


⇒ ∆S = 286 kJ/ 298 K


⇒ ∆S = 959.73 J mol-1K-1


T = Temperature = 298K