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Thermodynamics - Class 12 MCQ


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30 Questions MCQ Test - Thermodynamics

Thermodynamics for Class 12 2024 is part of Class 12 preparation. The Thermodynamics questions and answers have been prepared according to the Class 12 exam syllabus.The Thermodynamics MCQs are made for Class 12 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics below.
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Thermodynamics - Question 1

The formation of water from H2(g) and O2(g) is an exothermic reaction because:    

Detailed Solution for Thermodynamics - Question 1

ΔH = Hproduct - H reactant;
HR>HP. Thus -ve

Thermodynamics - Question 2

The enthalpy changes of formation of the gaseous oxides of nitrogen (N2O nd NO) are positive because of:    

Detailed Solution for Thermodynamics - Question 2

Due to high bond energy of , more heat is absorbed to breakup N2 molecule.

Thermodynamics - Question 3

Which is not correct?    

Detailed Solution for Thermodynamics - Question 3

ΔH = ΔU + ΔnRT
Δn = +1/2
Thus,ΔH>ΔU

Thermodynamics - Question 4

A positive change in enthalpy occurs in:

Detailed Solution for Thermodynamics - Question 4

Decomposition of MgCO3 occurs only on heating.

Thermodynamics - Question 5

The temperature of a 5 mL of strong acid increases by 50C when 5 mL of a strong base is added to it, If 10 mL of each are mixed, temperature should increase by:    

Detailed Solution for Thermodynamics - Question 5

No doubt heat evolved during neutralisation of 10 mL of each acid and base is twice the heat evolved during  neutralisation of 5 mL of each acid and base but the quantity of solution taking heat is also doubled thus, same temperature rise is noticed.

Thermodynamics - Question 6

The value of ΔH0 for the reaction Cu+(g) + l-(g) → Cul(g) is -446 kJ mol-1. If the ionisation energy of Cu(g) is 745 kJ mol-1 and the electron affinity of I(g) is -295 kJ mol-1, then the value of ΔH0 for the formation of one mole of CuI(g) from Cu(g) and l(g) is:    

Detailed Solution for Thermodynamics - Question 6

Thermodynamics - Question 7

The heat of combustion for C,H2 and CH4 are -349.0,-241.8 and -906.7 kJ respectively. The heat of formation of CH4 is:    

Detailed Solution for Thermodynamics - Question 7

By. Eqs. [(i)+2x(ii)]-(iii),
C + 2H2 → CH4;ΔH = 74.1

Thermodynamics - Question 8

Equal volume of  M HCI and 1 M H2SO4 are neutralised by dilute NaOH solution and x and y kcal of heat are liberated respectively. Which of the following is true?                    

Detailed Solution for Thermodynamics - Question 8

1 M H2SO4 = 2 eq. H2SO4
1 M HCI = 1 eq. HCI
Thus, for equal volume of two acis to be neutralized separately with NaOH, heat evolved will be twice in case of H2SO4 to that of HCI.

Thermodynamics - Question 9

When a certain amount of ethyene was burnt, 6226 kJ was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of O2 (at NTP) that entered into the reaction is:    

Detailed Solution for Thermodynamics - Question 9


Thus, 
= 296.5 litre

Thermodynamics - Question 10

For the reaction,

which of the following statements is correct?

Detailed Solution for Thermodynamics - Question 10

ΔH for a reaction is equal but opposite to backward reaction.

Thermodynamics - Question 11

Heat of combusion ΔH for C(s),H2(g) and CH4(g) are -94, -68 and -213 kcal/mole,than

  is:    

Detailed Solution for Thermodynamics - Question 11

Solve using Hess’s law.

Thermodynamics - Question 12

Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔU?

Detailed Solution for Thermodynamics - Question 12

Δn = 0 and thus from 

Thermodynamics - Question 13

ΔHf of graphite is 0.23 kJ/mol and ΔHf of diamond is 1.896 kJ/mol. ΔHTrsnsition from graphite to diamond is:    

Detailed Solution for Thermodynamics - Question 13


= 1.896-0.23=1.666 kJ/mol

Thermodynamics - Question 14

For the reaction
X2O4(l) → 2XO2(g)
ΔU = 2.1 kcal, ΔS = 20 cal K-1 at 300K
Hence,ΔG is:

Detailed Solution for Thermodynamics - Question 14

Δn = 2-0 = 2
ΔH = ΔU + Δn(g)RT
= 2100 + 2*2*300 =3300 cal
ΔG = ΔH - TΔS
= 3300 - 300*20 = -2700 cal
= -2.700kcal

Thermodynamics - Question 15

The temperature coefficient of e.m.f of a cell can be given by:
 

Detailed Solution for Thermodynamics - Question 15

   (Gibbs - Helmholtz equation)
Also, ΔG = ΔH - TΔS; and -ΔG = 
∴ ΔG - ΔH = -TΔS
or 
or  Similarly drive for other values

Thermodynamics - Question 16

ΔS0 will be highest for the reaction:

Detailed Solution for Thermodynamics - Question 16

Gaseous product is formed:
more positive is Δn, more is entropy

Thermodynamics - Question 17

Change in entropy is negative for:

Detailed Solution for Thermodynamics - Question 17

The gaseous phase have more entropy and thus, ΔS is +ve in (a) and (b). Also decrease in pressure increases disorder and thus ΔS is +ve in (c). In (d) the disorder decreases in liquid state due to decrease in temperature.Thus ,ΔS = -ve

Thermodynamics - Question 18

For the precipitation of AgCI by Ag+ ions and HCI

Detailed Solution for Thermodynamics - Question 18

Ag + Cl- → AgCl is a spontaneous reaction

Thermodynamics - Question 19

Which is an extensive property of the system ?    

Detailed Solution for Thermodynamics - Question 19

Properties which are mass independent are intensive properties and others which are mass dependent are extensive properties.

Thermodynamics - Question 20

A container has hydrogen and oxygen mixture in ratio of 4:1 by mass, then on mixing:    

Detailed Solution for Thermodynamics - Question 20

On mixing gases entropy increases due to increase in disorderness.

Thermodynamics - Question 21

Which of the following is correct option for free expansion of an ideal gas ?

Detailed Solution for Thermodynamics - Question 21

Free expansion is expansion against vacuum,so,q= ΔT & w all are zero

Thermodynamics - Question 22

Enthalpy of vaporisation for water is 186.5 kJ mol-1. The entropy change during vaporisation is ...kJ K-1mol-1    

Detailed Solution for Thermodynamics - Question 22


ΔH = 186.5KJ
T = 373 K
∴ ΔS = 186.5/373 = 0.5 kJK-1mol-1

Thermodynamics - Question 23

The work done by a system is 8 J, when 40 J heat is supplied to it. The change in internal energy of the system during the process is:    

Detailed Solution for Thermodynamics - Question 23

q = ΔU - W; W is work done by the system
ΔU = 40-8 = 32J  (∵ -W = 8)

Thermodynamics - Question 24

The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of 5 dm3 isothermally until the volume become 25dm3 is:    

Detailed Solution for Thermodynamics - Question 24

- W = +2.30nRT log V2/V1
- W = +2.30 * 16/32 * 300 * 8.314 log 25/5
- W = 2.01 * 103J

Thermodynamics - Question 25

The entropy change for the reaction given below,
2H2(g) + O2 (g) → 2H2O(l)
is....at 300 K. Standard entropies of H2(g),O2(g) and H2O(l) are 126.6, 201.20 and 68.0 JK-1 mol-1 respectively.

Detailed Solution for Thermodynamics - Question 25



= 2x68-[2x126.6+201.20]
= 318.4 JK-1mol-1

Thermodynamics - Question 26

One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is:    

Detailed Solution for Thermodynamics - Question 26


or ΔH = 273 * (60.1 - 38.20) = 5954.13 J mol-1

Thermodynamics - Question 27

Boiling point of a liquid is 50 K at 1 atm and ΔH vap= 460.6 cal mol-1 . What will be its b.p. at 10 atm?    

Detailed Solution for Thermodynamics - Question 27


∴ T2 = 100K

Thermodynamics - Question 28

The ratio of slopes of log P vs log V for reversible adiabatic process and reversible isothermal process of an ideal gas is equal to :

Detailed Solution for Thermodynamics - Question 28

PVγ= constant for adiabatic expansion
and PV = constant for isothermal expansion
∴ log P =-γlogγ   slope = -γ
 log P =-logV       slope = -1

Thermodynamics - Question 29

Internal energy and pressure of a gas of unit volume are related as:

Detailed Solution for Thermodynamics - Question 29

Px1=RT
Also internal energy,U=3/2 RT
∴ U=3/2 P
or   P = 2/3 U

Thermodynamics - Question 30

The molar heat capacity of water at constant pressure P, is 75 JK-1mol-1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand,the increase in temperature of water is :    

Detailed Solution for Thermodynamics - Question 30

q = mCvΔT
1000 = 100/18 * 75ΔT
∴  ΔT = 2.4 

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