Thermodynamics

≠ ΔrG0 = -nFE0cell → 

Q1. Which of the following are not state functions?

(I) q + w

(II) q

(III) w

(IV) H – TS

(a) (I), (II) and (III)

(b) (II) and (III)

(c)  (I) and (IV)

(d) (II), (III) and(IV)

Ans. b.

Q2. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true?

(a) E = W≠0, q = 0 (b) E=W=q≠ 0 (c) E = 0, W = q≠0 (d) W = 0, E = q≠0

Ans. a.

Q3. Which of the following is the correct equation? 

(a)  U = W + Q (b) U = Q – W

(c)  W = U + Q (d) None of these 

Ans. b

Q4. The correct option for free expansion of an ideal gas under adiabatic condition is

(a) q = 0, ΔT = 0 and w = 0

(b) q = 0, ΔT < 0 and w > 0

(c) q < 0, ΔT = 0 and w = 0

(d) q > 0, ΔT > 0 and w > 0

Ans. a.

Q5. Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1 L bar = 100 J]

(a) 30 J (b) –30 J

(c) 5 kJ (d) 25 J 

Ans. b.

Explanation: W = - Pext x V = -30J

Q6. Reversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure.

AB  → Isothermal expansion

AC → Adiabatic expansion

Which of the following options is not correct ?

(a) ΔSisothermal > ΔSadiabatic

(b) TA = TB

(c) Wisothermal > Wadiabatic

(d) Tc > TA

Ans. d.

Q7. An ideal gas expands isothermally from 10–3  m3  to 10–2 m3 at 300 K against a constant pressure of 105 N m–2. The work done on the gas is

(a) +270 kJ (b) –900 J

(c) +900 kJ (d) –900 kJ

Ans. b.

Q8. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy U of the gas in joules will be

(a) –500 J (b) –505 J

(c) +505 J (d) 1136.25 J

Ans.b.

Q9. Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (CP/CV) of the mixture will be

(a) 0.83 (b) 1.50

(c) 3.3 (d) 1.67 (2012)

Ans. d.

Q10. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition? 

(a) q = 0, ΔT ≠  0, w = 0 

(b)  q ≠  0, ΔT = 0, w = 0 

(c) q = 0, ΔT = 0, w = 0 

(d) q = 0, ΔT < 0, w≠ 0

Ans. c

Q11. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be

(a) infinite (b) 3 Joules

(c) 9 Joules (d) zero.

Ans. d.

Q12. Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE? 

(a) 2CO(g) + O2(g) →   2CO2(g)

(b) H2(g) + Br2(g)  →  2HBr(g)

(c) C(s) + 2H2O(g)  2H2(g) + CO2(g)

(d)  PCl5(g)  PCl3(g) + Cl2(g)

Ans. b.

Q13. The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)

(a)  – 6 J (b) – 608 J

(c)  + 304 J (d)  – 304 J

Ans. b.

Q14. For the reaction,

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)

at constant temperature, H – E is

(a) + RT (b) –3RT

(c)  +3RT (d) –RT (2003)

Ans. b.

Q15. The molar heat capacity of water at constant pressure, C , is 75 J K–1 mol–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

(a) 1.2 K (b) 2.4 K

(c) 4.8 K (d) 6.6 K

Ans. b.

Q16 When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308K. Heat supplied to the gas is 500 J. Then which statement is correct?

(a) q = w = 500 J, Δ E = 0 (b) q = Δ E = 500 J, w = 0 (c) q = w = 500 J, ΔE = 0

(d) ΔE = 0, q = w = –500 J

Ans. b.

Q17. For the reaction,

C2H5OH(l) + 3O2(g) →  2CO2(g) + 3H2O(l)

which one is true?

(a) Δ H = Δ E – RT (b) Δ H = Δ E + RT

(c) Δ H = Δ E + 2RT (d) Δ H = Δ E – 2RT (2000)

Ans. a.

Q18. In an endothermic reaction, the value of H is

(a) negative (b) positive

(c) zero (d) constant.

Ans. b.

Q19. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre  to  10 litres. The E for this process is

(R = 2 cal mol–1 K–1)

(a) 1381.1 cal (b) zero

(c) 163.7 cal (d)  9 L atm

Ans. b.

Q20. During isothermal expansion of an ideal gas, its

(a) internal energy increases

(b) enthalpy decreases

(c) enthalpy remains unaffected

(d) enthalpy reduces to zero.

Ans.c.

Q21. For the reactionm, N2 + 3H2 ⇌ 2NH3 , ΔH = ?

(a) ΔE + 2RT

(b) ΔE – 2RT

(c)  ΔH = RT

(d)  ΔE – RT

Ans. b. 

Q22. If ΔH is the change in enthalpy and ΔE, the change in internal energy accompanying a gaseous reaction, then

(a) ΔH is always greater than ΔE

(b) Δ H < Δ E only if the number of moles of the products is greater than the number of moles of the reactants

(c) Δ H is always less than Δ E

(d) Δ H < Δ E only if the number of moles of products is less than the number of moles of the reactants.

Ans. d.

Q24. Standard enthalpy of vaporisation ΔvapH° for water at 100°C is 40.66 kJ mol–1. The internal energy of vaporisation of water at 100°C (in kJ mol–1) is

(a) +37.56 (b)  – 43.76

(c) + 43.76 (d)  + 40.66

(Assume water vapour to behave like an ideal gas)

Ans.a.

Q25. Consider the following processes : 

..............................H (kJ/mol)

1/2A → B............+150

3B → 2C + D ........125

E + A → 2D.........+350

For B + D →  E + 2C, ΔH will be

(a) 525 kJ/mol (b) –175 kJ/mol

(c) –325 kJ/mol (d) 325 kJ/mol

Ans. b.

Q26. The following two reactions are known

Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g); H = –26.8 kJ 

FeO(s) + CO(g) → Fe(s) + CO2(g); H = – 16.5 kJ

The value of H for the following reaction

Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) is 

Equilibrium 

Equilibrium acts/ works / comes into force in reversible reactions.

A + B ⇌ C + D 

If concentration of reactants 

-increases then the reaction proceeds in the forward direction

-decrease then the reaction proceeds in the backward direction

If concentration of products

- increases, then the reaction proceeds in the backward direction

- decreases, then the reaction proceeds in the forward direction

Dynamic Equilibrium -  the rates of forward and reverse reactions are equal and there is no net change in composition

A + B ⇌ C + D 

Kc = [C] [D] / [A] [B]

Kc is equilibrium constant.

4NH3 (g) + 5O2 (g) ⇌ 4NO(g) + 6H2O (g)

Kc = [NO]4[H2O]6/[NH3 ]4[O2 ]5 

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

4NH3 (g) + 5O2 (g) ⇌ 4NO(g) + 6H2O (g)

Kc = [NO]4[H2O]6/[NH3 ]4[O2 ]5 

but for the reverse reaction/ backward reaction the K'c=1/Kc 

pv = nRT

p = n/v RT and we can write n/V = C    and C is concentration

p = CRT     and R = 0.0831 bar litre/mol K

Kp=Kc(RT)Δn       ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants) 

1pascal, Pa=1Nm-2, and 1bar = 105 Pa

If Kc > 103 , products predominate over reactants, i.e., if Kc is very large, the reaction proceeds nearly to completion. 

If Kc < 10-3, reactants predominate over products, i.e., if Kc is very small, the reaction proceeds rarely. 

If Kc is in the range of 10-3 to 103 , appreciable concentrations of both reactants and products are present. 

If Qc < Kc , net reaction goes from left to right 

If Qc > Kc , net reaction goes from right to left. 

If Qc = Kc , no net reaction occurs 

-∆G is negative, then the reaction is spontaneous and proceeds in the forward direction. 

-∆G is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants. 

-∆G is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. 

∆G0 = – RT ln Kc 

Le Chatilier's Principle: - change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. 

Conjugate is difference in H+ ion

HF is an acid, then its conjugate base is F-

NH3 is a base, then its conjugate acid is NH4+

HCl (acid) + H2O(base) ⇌ H3O+ (conjugate acid) + Cl- (conjugate base) 

*strong acids have very weak conjugate bases and vice versa.

H2O(base)  + H2O(acid) ⇌ H3O+ (conjugate acid) + OH- (conjugate base) 

Kw = [H3O+] [OH-] /[H2O] 

Kw is ionic product of water and its value is equal to 10-14

Acidic: [H3O+] > [OH-] 

Neutral: [H3O+] = [OH-] 

Basic : [H3O+] < [OH-] 

pH = - log [H+]

ionisation constant for weak acid is Ka and for weak base is Kb

Q1. In liquid-gas equilibrium, the pressure of vapours above the liquid is constant at

(a) constant temperature

(b) low temperature

(c) high temperature

(d) none of these.

Ans. a. constant temperature

Q20 20. If the value of equilibrium constant for a particular reaction is 1.6 × 1012, then at equilibrium the system will contain

(a) mostly products

(b)    similar amounts of reactants and products

(c)    all reactants

(d)    mostly reactants.

Ans. (a) mostly products.

Q21. In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

(a) 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen

(b) 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

(c) 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen

(d) 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

Ans. N2 + 3H2 ⇌ 2NH3

b. 10 L ammonia, 25 L nitrogen, 15 L hydrogen

Q26. Which one of the following conditions will favour maximum formation of the product in the reaction 

A2(g) + B2(g) ⇌ X2(g), rH = –X kJ ?

(a) Low temperature and high pressure

(b) Low temperature and low pressure

(c) High temperature and high pressure

(d) High temperature and low pressure

Ans. a. low temp and high pressure.

PDPs

Q1 The equilibrium constant for the reaction N2 + 3H2 = 2NH3 is K. The equilibrium constant for 

1/2 N2 + 3/2 H2 ⇌  NH3 will be

a. K/2 

b. 2K

c. √ K

d. K2 

Ans. c. √ K

Q2 If an inert gas is added to the equilibrium mix of the dissociation of PCl5 in a closed vessel,

a. Conc of Cl2 increases

b. Conc of PCl3 increases

c. Conc of PCl5 increases

d. the equilibrium conc will remain unaffected.

Ans. d. the equilibrium conc will remain unaffected.

PCl5 ⇌  PCl3 + Cl2

Q3 If pressure is increased on the equilibrium N2 + O2 ⇌ 2NO, the equilibrium will 

a. shift in the forward direction

b. shift in the backward direction

c. remain undisturbed 

d. may shift in the forward or backward direction

Ans. c. remain undisturbed.

Q4 For the reaction: PCl3 (g) + Cl2 (g) ⇌ PCl5(g) and the value of Kc at 250 oC is 26. The value of Kp at this temp will be?

a. 0.61

b. 0.57

c. 0.83

d. 0.46

Ans. a. 0.61

Solution: Kp=Kc(RT)Δn

Δn= no of moles of gaseous product - number of moles of gaseous reactant =1−2=−1

R = 0.0821, T = 250 + 273.15 = 523.15 and Kc= 26

Kp=0.61

Q5 According to Le Chatelier's Principle - adding heat to a solid and liquid in equilibrium will cause the:

a. amount of solid to decrease

b. amount of liquid to decrease

c. temperature to rise

d. temperature to fall.

Ans. a. amount of solid to decrease

Q6 In a reaction A2(g) + 4B2(g) ⇌ 2AB4(g), ΔH<0. The formation of AB4(g) will be favoured by

a. low temp and high pressure

b. high temp and low pressure

c. low temp and low pressure

d. high temp and high pressure

Ans. a. low temp and high pressure

Solution: as the reaction is exothermic, when temp is lowered; according to Le Chatelier's principle the reactants will react to release heat so decreasing the temp will favour the forward reaction. Also when we put pressure in reactant side we have 5 molecules in total however 2 molecules in product side. High pressure will make reactant molecules which are larger in no. to collide and form product.

Q7 The reaction which proceeds in the forward direction is: 

a. Fe3O4 + 6HCl ⇌ 2FeCl3 + 3H2O

b. NH3 + H2O + NaCl ⇌ NH4Cl + NaOH

c. SnCl4 + Hg2Cl ⇌ SnCl2 + 2HgCl2

d. 2CuI + I2 + 4K+ ⇌ 2Cu2+ + 4KI

Ans. a. Fe3O4 + 6HCl ⇌ 2FeCl3 + 3H2O

Due to absence of hydrolysis of FeCl3, backward reaction will not take place.

Q8 For which of the following reaction Kp = Kc

a. 2NOCl(g) ⇌ 2NO(g) + Cl2(g)

b. N2(g) + 3H2(g) ⇌ 2NH3(g)

c. H2(g) + Cl2(g) ⇌ 2HCl(g)

d. 2N2O4(g) ⇌ 2NO2(g)

Ans.c. H2(g) + Cl2(g) ⇌ 2HCl(g)

Solution: if we get change in no. of moles as zero then Kp = Kc

Kp=Kc(RT)Δn

Kp=Kc(RT)0

Q9 In a vessel containing SO3, SO2 and O2 at equilibrium some helium gas is introduced so that the total pressure increases while temp and vol remain constant. According to Le Chatelier's Principle, the dissociation of SO3;

a. increases

b. decreases

c. remains unaltered

d. changes unpredictably.

Ans. c. remains unaltered

Solution: SO3 ⇌ SO2 + O2

Q10 An equilibrium mixture for the reaction 

2H2S (g) ⇌ 2H2(g) + S2(g)   has 1 mole of H2S, 0.2 mole of H2 and 0.8  mole of S2 in a 2L flask. The value of Kc in mol / L is 

a. 0.004

b. 0.08

c. 0.016

d. 0.160

Ans. d.0.16

Q1 Which of the following statements is most applicable to hydrogen? It can act

a. as a reducing agent

b. as an oxidising agent

c. both as oxidising agent and reducing agent

d. neither as an oxidising nor as a reducing agent

Ans. c. both as oxidising agent and reducing agent. 

Explanation: hydrogen when gets bonded with metals, it oxidises metals example Na+H-

Q2 Hydrogen combines with other elements by 

a. losing an electron

b. gaining an electron

c. sharing an electron

d. losing, gaining and sharing of an electron 

Ans.  d. losing, gaining and sharing of an electron.

Q3. The first ionisation enthalpy (kJ/mol) for H, Li, F, Na has one of the following values: 1681, 520, 1312, 495. Which of these values corresponds to that of hydrogen.

a. 1681

b. 1312

c. 520

d. 495

Ans. b. 1312 kJ/mol

Q4 Which of the following metals cannot be used for liberating hydrogen from dil HCl?

a. Zn

b. Copper

c. Iron

d. Magnesium

Ans. b. Copper

Explanation: Electrode potential of Copper is more positive than other metals. Hence, it does not substitute H in HCl.

Q5. Which of the following is used as a moderator in nuclear reactors?

a. Hard water

b. Heavy Water

c. Deionised water

d. Mineral water. 

Ans. b. Heavy water

Q6 The temporary hardness of water due to Calcium bicarbonate can be removed by adding.

a. CaCO3

b. Ca(OH)2

c. CaCl2

d. HCl

Ans. b. Ca(OH)2

Explanation: Permanent hardness is due to the presence of Chlorides & Sulfates of Ca and Mg. 

Temporary hardness is due to the presence of bicarbonates of Ca and Mg.

Q7 In the calgon process of softening of water, which of the following is used?

a. Sodium polymetaphosphate

b. Hydrated sodium aluminium silicate

c. Cation exchange resins

d. Anion exchange resins

Ans. a. Sodium polymetaphosphate

Q8 

Q8 Hydrogen will not reduce

a. heated cupric oxide

b. heated ferric oxide

c. heated stannic oxide

d. heated aluminium oxide

Ans. d. heated aluminium oxide will never be reduced by hydrogen as Reduction potential of aluminium is more negative than hydrogen.

Q9 The oxidation states exhibited by hydrogen in its various compounds are 

a. -1 only

b. 0 only

c. +1, -1, 0

d. +1 only

Ans. c. +1, -1, 0

Q10 The oxidation states of the most electronegative element in the products of the reaction of BaO2 with dil. H2SO4 are

a. 0 and -1

b. -1 and -2

c. -2 and 0

d. -2 and +1

Ans. b. -1 and -2. 

Explanation: BaO2 + H2SO4 = BaSO4 + H2O2

Here in this reaction oxidation state of the most electronegative element i.e. oxygen is -2 everywhere besides in H2O2 where the oxidation state of oxygen is -1. 

Q11. 30 volumes H2O2 means. 

a. 30% H2O2

b. 30cm3 of the solution contains 1g of H2O2

c. 1cm3 of the solution liberates 30 cm3 of O2 at STP

d. 30 cm3 of the solution contains one mole of H2O2

Ans. c. 1cm3 of the solution liberates 30cm3 of O2 at STP.

Explanation.