Thermodynamics Important Questions And Answers

Thermodynamics Important Questions And Answers

Thermodynamics is the most important and interesting chapter of Chemistry. So the practice set of Thermodynamics with Important Questions And Answers helps students of class 11 and also for students studying for various competitive exams. Students are advised to practice and understand all the questions accordingly.

1.Given that C(g) + 4H(g) → CH₄ (g); ΔH =  −166kJ .The bond energy C−H will be:

a) 208 \(\frac{kJ}{mole}\)

b) 41.6 \(\frac{kJ}{mole}\)

c) 832 \(\frac{kJ}{mole}\)

d) None of these.

ANS:

\(\frac{-166}{4}\)

2.The dependence of Gibbs free energy on pressure for an isothermal process of an ideal gas is given by:

a) ΔGT= nRT ln \(\frac{p_2}{p_1}\)

b) ΔGT= nRT ln = \(\frac{V_1}{V_2}\)

c) ΔGT= nRT ln \(\frac{p_1}{p_2}\)

d) none of these

3.The energy required to dissociate 4 gm of gaseous hydrogen into free gaseous atoms is 208 kcal at 25^oC. The   bond energy of H−H bond will be:

a) 104 kcal

b) 10.4 kcal

c) 10.2

d) 10.3

ANS:

4gH₂ = 2moles.

\(\frac{208}{2}\)

Bond energy for 1 mole of H₂ = \(\frac{208}{2}\) = 104kcal.

4.The free energy change for a reversible reaction at equilibrium is:

a) Large positive

b) Small negative

c)Small positive

d) 0

ANS:

ΔG  at equilibrium = 0.

5.For a spontaneous change, free energy change ΔG is:

a) Positive

b) Negative

c)Zero

d) Can be positive or negative

ANS:

For spontaneous change  ΔG =−ve

6.The relation between ΔG and ΔH is:

a) ΔH= ΔG −TΔS

b) ΔG=ΔH− TΔS

c) TΔS−ΔG= ΔH

d) ΔH=TΔG+ ΔS

7.The relation  ΔG = ΔH− TΔS was given by:

a) Boltzmann

b) Faraday

c) Gibbs Helmholtz

d) Thomson

8.For a reaction at 25^oC enthalpy change and entropy changes are −11.7×10³Jmol⁻¹ and −10⁵Jmol⁻¹K⁻¹ respectively. What is the Gibbs free energy?

ΔG = ΔH − TΔS (Gibb’s free energy equation)

12.A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ΔH is the enthalpy change and ΔE is the change in internal energy, then:

a) ΔH>ΔE

b) ΔH<ΔE

c) ΔH=ΔE

d) The relationship depends on the capacity of the vessel

ANS:

Formation of CO₂ from CO is an exothermic reaction; heat is evolved from the system, i.e., energy is lowered. Thus, exothermic reactions occur spontaneously on account of decrease in enthapy of system. Thus, ΔE>ΔH

13.Which is not the correct relation between enthalpy (ΔH)and intrinsic energy (ΔE):

a) ΔH=ΔE+PΔV

b) ΔH=ΔE+nRT

c) ΔH=ΔE−PΔV

d) ΔE=ΔH−PΔV

14.The resultant heat change in a reaction is the same whether it takes place in one or several stages.This statement is called:

a) Lavoisier and Laplace law

b) Hess’s law

c) Joule’s law

d) Le-chatelier’s principle

15.The enthalpies of the elements in their standard states are assumed to be:

a) Zero at 298 K

b) Unit at 298 K

c) Zero at all temperatures

d) Zero at 273 K

ANS:

The enthalpies of all elements in their standard state at 25^oC or 298K are zero.

16.The heat Q for a reaction at constant volume is equal to:

a) H_P−H_R

b) H_R−H_P

c) E_P−E_R

d) None of these

ANS:

ΔE_v =  E_P−E_R.

17.In a reversible isothermal process, the change in internal energy is:

a) Zero

b) Positive

c) Negative

d) None of these

ANS:

ΔE = 0 for reversible isothermal process.

18.The work done by a system is 8 joule, when 40 joule heat is supplied to it. What is the increase in internal energy of system.

a) 25 J

b) 30 J

c) 32 J

d) 28 J

ANS:

q = 40J    w = −8J (work done by the system)

ΔE= q+w = 40−8 = 32J.

19.In which of the following neutralization reactions, the heat of neutralisation will be highest:

a) NH₄OH and CH₃COOH

b) NH₄OH and HCl

c) NaOH and CH₃COOH

d) NaOH and HCl

ANS:

Heat of neutralization between strong acid and a strong base is about −13.7Kcal.

20.From Kirchhoff’s equation which factor affects the heat of reaction

a) Pressure

b) Temperature

c) Volume

d) Molecularity

ANS:

Effect of temperature in heat of reaction is given by Kirchoff”s equation.

21.ΔH^of (298K) of methanol is given by the chemical equation:

a) CH₄(g) + \(\frac{1}{2}\) O₂(g) → CH₃OH(g)

b) C(graphite) +\(\frac{1}{2}\) O₂(g) + 2H₂(g)→CH₃OH(l)

c) C (diamond) +\(\frac{1}{2}\) O₂(g) + 2H₂(g)→CH₃OH(l)

d) CO(g)+2H₂(g) → CH₃OH(l)

22.The heat of transition (ΔH_t) of graphite into diamond would be,where
C(graphite) + O₂(g) → CO₂(g); ΔH = xkJ , C(diamond) +C₂(g) → CO₂(g);ΔH = ykJ.

a) (x+y)kJ mol⁻¹

b) (x−y)kJ mol⁻¹

c) (y−x)kJ mol⁻¹

d) None of these

ANS:

Graphite → diamond ΔHt = (x−y)kJ mol⁻¹

23.The mutual heat of neutralisation of 40 gm of NaOH and 60 gm CH₃COOH will be.

a) 56.1 kcal

b) Less than 56.1 kcal

c) More than 56.1 kcal

d) 13.7 kcal

ANS:

Heat of neutralisation is less than 56.1 Kcal.when a strong base and a weak acid reacts.

24.The heat of formations of CO(g) and CO₂(g) are −26.4 kcal and −94.0 kcal respectively. The heat of combustion of carbon monoxide will be.

a) +26.4kcal

b) −67.6kcal

c) −120.6kcal

d) +52.8kcal

ANS:

Aim: CO + \(\frac{1}{2}\) O₂ → CO₂

ΔH = ΔH^of (CO₂)−[ΔH^of(CO) +\(\frac{1}{2}\)ΔH^of(O₂)]

= −94.0−(−26.4)

=−67.6kcal.

25.The heat of reaction does not depend upon:

a) Temperature of the reaction

b) Physical state of reactants and products

c) Whether the reaction is carried out at constant pressure or at constant volume

d) The method by which the final products are obtained from the reactants

26.Which is the best definition of heat of neutralization?

a) The heat set free when one gram molecule of a base is neutralized by one gram molecule of an acid in dilute solution at a stated temperature

b) The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature

c) The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature

d) The heat set free or absorbed when a normal solution containing one gram equivalent of an acid is neutralized by a normal solution containing one gram equivalent of a base at a stated temperature

e) The heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in  dilute solution at a stated temperature

ANS:

It is the definition of heat of neutralization.

27.H₂ + \(\frac{1}{2}\)O₂ → H₂O; ΔH=−68.39kcal K+H₂O+Water →KOH(aq)+ \(\frac{1}{2}\)H₂;ΔH = −48kcal KOH+Water→KOH(aq); ΔH=−14kcal.The heat of formation of KOH is (in kcal).

a) −68.39+48−14

b) −68.39−48+14

c) 68.39−48+14

d) 68.39 + 48 + 14

ANS:

Aim: K(S)+\(\frac{1}{2}\)O₂(g) +\(\frac{1}{2}\)H₂(g) → KOH(S) eq. (ii) + eq. (i)

eq. (iii) gives.

ΔH =−48 (−68.39)−(−14) = −68.39−48 +14

28.The difference between heats of reaction at constant pressure and at constant volume for the reaction  2C₆H₆ (l)+15O₂(g)→12CO₂(g)+6H₂O(l)at 25^oCin kJ is:

a) -7.43

b) + 3.72

c) -3.72

d) +7.43

ΔH−ΔE = ΔnRT = 3 × 8.314× 298

= −7432J = −7.43kJ

29.For the reaction A(g) + 2B(g) →2C(g) + 3D(g), the value of ΔE at 27^oC is 19.0 kcal. The value of ΔH for the reaction would be (R=2.0calK⁻¹mol⁻¹)

a) 20.8 kcal

b)19.8 kcal

c) 18.8 kcal

d) 20.2 kcal

ANS:

Use ΔH =ΔE+ΔnRT ΔH =19+2×2×10⁻³×300

= 20.2kcal;  Δn=2.

30.Equal volumes of methanoic acid and sodium hydroxide are mixed. If x is the heat of formation of water, then heat evolved on neutralisation is:

a) More than x

b) Equal to x

c) Twice of x

d) Less than x

ANS:

As methanoic acid is weak acid, heat of neutralization < x.

31.In the reaction: H₂ + Cl₂ → 2HCl,ΔH=194kJ. Heat of formation of HCl is:

a) + 97 kJ

b) + 194 kJ

c) -194 kJ

d) -97 kJ

ANS:

Heat of formation = \(\frac{194}{2}\) = 97kJ

32.A system is changed from state A to state B by one path and from B to A another path. If E₁ and E₂ are the corresponding changes in internal energy, then:

a) E₁+ E₂=−ve

b) E₁+ E₂=+ve

c) E₁+ E₂=0

d) None of these.

ANS:

ΔE =0 for a cyclic process.

33.Molar heat of vaporisation of a liquid is 6 kJmol⁻¹. If the entropy change is 16Jmol⁻¹K⁻¹, the boiling   point of the liquid is:

a) 375^oC

b) 75K

c) 273K

d) 102^oC

ANS:

ΔS=16Jmole⁻¹K⁻¹

T_b.p= \(\frac{ΔH_{vapour}}{ΔS_{vapour}}\)

= \(\frac{6\times1000}{16}\) =375K

34.If enthalpies of methane and ethane are respectively 320 and 360 calories then the bond energy of C−C bond is:

a) 80 calories

b) 40 calories

c) 60 calories

d) 120 calories

ANS:

CH₄ →C + 4H, ΔH =320  E_C−H =90cal  C₂H₆ → 2C+6H, ΔH=360

∴ 360 = E_C−C+ 6E_C−H

∴E_C−C = 360−320 = 40cal.

35.Which one of the following bonds has the highest average bond energy \(\frac{kcal}{mol}\):

a) S=O

b) C≡C

c) C≡N

d) N≡N

ANS:

By definition of Hess”s law.

36.The equilibrium concentration of the species in the reaction A +B →C +D are 3, 5, 10 and 15 moleL⁻¹ respectively at 300K the ΔG for the reaction is:

a) +13.81

b) -1381.8

c) +138.18

d) -1391.6

ANS:

ΔG =−2.303RT log K′, Here R= 2cal,T= 300K

K′= \(\frac{10\times15}{3\times5}\) = 10

ΔG= −2.303 × 2 × 300 × log₁₀10

= −2.303 × 2 × 300 × 1 = −1381.8cal

37.For spontaneity of a cell, which is correct.

a) ΔG = 0, ΔE=0

b) ΔG =−ve, ΔE=0

c) ΔG = +ve, ΔE=+ve

d) ΔG = −ve, ΔE=+ve

ANS:

ΔG =ΔH−TΔS;  ΔG=ΔE+PΔV−TΔS

For spontaneity (ΔG=−ve)

∵ ΔG =−ve, ΔE =+ve

38.The free energy for a reaction having ΔH = 31400cal;. ΔS = 32calK⁻¹mol⁻¹ at 1000^oC is:

a) -9336 cal

b) +7386 cal

c) -1936 cal

d) +9336 cal

ANS:

ΔG =ΔH −TΔS

=31400−1273×32

= 31400−40736 = −9336cal

39.We assume if water vapour is to be a perfect gas, the molar enthalpy change for vapourization of 1 mole of water at 1bar and 100°C is 41kJ mol^–1. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C.

ANS:

The change H₂O (l ) → H₂O(g)

∆U = ∆H – ∆ng RT,

substituting the values, we get

∆U = 41.00 KJmol-1\(\times8.3 Jmol^{-1}K^{-1}\times{373K}\)

= 41.00 KJmol^-1 –  3.096 KJmol^-1

  =37.904 KJmol^-1

39.Diffrence between Isothermal Process and Isochoric process:

ANS:

Isothermal Process	Isochoric process
It is a thermodynamic process in which temperature remains constant	It is a thermodynamic process in which volume remains constant
(dt= 0, ΔU = 0).	ΔV = 0

40.Poisson’s ratio :-

a) Cp =\(\frac{3}{2}\)R+R = \(\frac{5}{2}\)R

b) \(\gamma=\frac{C_p}{C_v}=\frac{5}{3}\) = 1.66 

c) Both

d) None of these.

41.What is  Lavoisier Laplace Law?

ANS:

The enthalpy change during a reaction is equal in magnitude to the enthalpy change in the reverse process but it is opposite in sign.