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# Choose the best answer: Thermodynamics

Chemistry : Thermodynamics: Choose the best answer: one mark questions

Thermodynamics | Chemistry

1. The amount of heat exchanged with the surrounding at constant temperature and pressure is given by the quantity

a) ∆E

b) ∆H

c) ∆S

d) ∆G

2. All the naturally occurring processes proceed spontaneously in a direction which leads to

a) decrease in entropy

b) increase in enthalpy

c) increase in free energy

d) decrease in free energy

Answer: d) decrease in free energy

3. In an adiabatic process, which of the following is true ?

a) q = w

b) q = 0

c) ∆E = q

d) P ∆ V= 0

4) In a reversible process, the change in entropy of the universe is

a) > 0

b) > 0

c) < 0

d) = 0

5) In an adiabatic expansion of an ideal gas

a) w = – ∆u

b) w = ∆u + ∆H

c) ∆u = 0

d) w = 0

Answer: a) w = – ∆u

6) The intensive property among the quantities below is

a) mass

b) volume

c) enthalpy

d) mass/volume

7) An ideal gas expands from the volume of 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure at 1 × 105 Nm–2. The work done is

a) – 900 J

b) 900 kJ

c) 270 kJ

d) – 900 kJ

Solution:

w = – P  V

w = – (1 × 105 Nm–2)

(1× 10–2 m3 – 1 × 10–3 m3)

w= –105 (10–2 – 10–3) Nm

w = – 105 (10 – 1) 10–3) J

w = – 105 (9 × 10–3) J

w = –9 × 102 J w = –900 J

8. Heat of combustion is always

a) positive

b) negative

c) zero

d) either positive or negative

9. The heat of formation of CO and CO2 are – 26.4 kCal and – 94 kCal, respectively. Heat of combus-tion of carbon monoxide will be

a) + 26.4 kcal

b) – 67.6 kcal

c) – 120.6 kcal

d) + 52.8 kcal

Solution:

CO (g )+1/2O2 (g )→CO 2 (g)

ΔHC0 (CO) = ΔHf (CO2) – ΔHf(CO) + ΔHf (O2)]

ΔHC0 (CO) = ΔHf (CO2) – ΔHf(CO) + ΔHf (O2)]

ΔHC0 (CO) = – 94 KCal – [– 26.4 KCal + 0]

ΔHC0 (CO) = – 94 KCal + 26.4

KCal ΔHC0 (CO) = – 67.4 Kcal

10. C(diamond) → C(graphite), ∆H = –ve, this indicates that

a) graphite is more stable than diamond

b) graphite has more energy than diamond

c) both are equally stable

d) stability cannot be predicted

Answer: a) graphite is more stable than diamond

11. The enthalpies of formation of Al2O3 and Cr2O3 are – 1596 kJ and – 1134 kJ, respectively. ∆H for the reaction 2Al + Cr2O3 → 2Cr + Al2O3 is

a) – 1365 kJ

b) 2730 kJ

c) – 2730 kJ

d) – 462 kJ

Solution:

2Al + Cr2O3  2Cr + Al2O3

ΔHr0 = [2ΔHf (Cr) + ΔHf (Al2O3)]– [2ΔHf (Al) + ΔHf

(Cr2O3)] ΔHr0 = [0 + (– 1596 kJ)]– [0 + (– 1134)]

ΔHr0 = – 1596 kJ + 1134 kJ

ΔHr0 = – 462 kJ

12. Which of the following is not a thermodynamic function ?

a) internal energy

b) enthalpy

c) entropy

d) frictional energy

13. If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then

a) ∆H > ∆U

b) ∆H - ∆U = 0

c) ∆H + ∆U= 0

d) ∆H < ∆U

14. Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is

a) +1 kJ

b) – 5 kJ

c) +3 kJ

d) – 3 kJ

Solution:

ΔU = q + w

ΔU = – 1 kJ + 4 kJ

ΔU = + 3kJ

15. The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol–1) reacts with hydrochloric acid in an open beaker at 25° C

a) – 2.48 kJ

b) – 2.22 kJ

c) + 2.22 kJ

d) + 2.48 kJ

Solution:

Fe + 2HCl → FeCl2 + H2

1 mole of Iron liberates 1 mole of

Hydrogen gas

55.85 g Iron = 1 mole Iron

n = 1

T = 25° C = 298 K

w = – P Δ V

w =-P (nRT/P)

w = – nRT

w = –1 × 8.314 × 298 J

w = – 2477.57 J

w = – 2.48 kJ

16. The value of ∆H for cooling 2 moles of an ideal monatomic gas from 125° C to 25° C at constant pressure will be given [CP=5/2 R]

a) – 250 R

b) – 500 R

c) 500 R

d) + 250 R

Solution:

Ti = 125° C = 398 K

Tf = 25° C = 298 K

ΔH = nCp (Tf – Ti)

ΔH = 2x(5/2)R(298-398)

ΔH = – 500 R

17. Given that C(g) + O2 (g) → CO2 (g) ∆H0 = – a kJ; 2 CO(g) + O2(g) → 2CO2(g) ∆H0 = –b kJ; Calculate the ∆H0 for the reaction C(g) + ½O2(g) → CO(g)

a) (b+2a) / 2

b) 2a - b

c) (2a-b)/2

d) (b-2a)/2

Solution: 18. When 15.68 litres of a gas mixture of methane and propane are fully combusted at 00 C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat of released from this combustion in kJ is (∆HC (CH4) = – 890 kJ mol–1 and ∆HC (C3H8) = – 2220 kJ mol–1)

a) – 889 kJ

b) – 1390 kJ

c) – 3180 kJ

d) – 653.66 kJ

Solution:

Given :

ΔHC (CH4)= – 890 kJ mol–1

ΔHC (C3H8)= – 2220 kJ mol–1 19. The bond dissociation energy of methane and ethane are 360 kJ mol–1 and 620 kJ mol–1 respec-tively. Then, the bond dissociation energy of C-C bond is

a) 170 kJ mol–1

b) 50 kJ mol–1

c) 80 kJ mol–1

d) 220 kJ mol–1

Solution:

4EC–H= 360 kJ mol–1

EC–H= 90 kJ mol–1

EC–C + 6 EC–H = 620 kJ mol–1

EC–C + 6 × 90 = 620 kJ mol–1

EC–C + 540= 620 kJ mol–1

EC–C= 80 kJ mol–1

20. The correct thermodynamic conditions for the spontaneous reaction at all temperature is

a) ∆H < 0 and ∆S > 0

b) ∆H < 0 and ∆S < 0

c) ∆H > 0 and ∆S = 0

d) ∆H > 0 and ∆S > 0

Answer: a) ∆H < 0 and ∆S > 0

21. The temperature of the system, decreases in an ___________________

a) Isothermal expansion

b) Isothermal Compression

22. In an isothermal reversible compression of an ideal gas the sign of q, ∆S and w are respectively

a) +, –, –

b) –, +, –

c) +, –, +

d) –, –, +

Solution:

During compression, energy of the system increases, in isothermal condition, to main-tain temperature constant, heat is liberated from the system. Hence q is negative.

During compression entropy decreases.

During compression work is done on the system, hence w is positive

23. Molar heat of vapourisation of a liquid is 4.8 kJ mol–1. If the entropy change is 16 J mol–1 K–1, the boiling point of the liquid is

a) 323 K

b) 270 C

c) 164 K

d) 0.3 K

Solution: 24. ∆S is expected to be maximum for the reaction

a) Ca(S) + ½ O2(g) → CaO(S)

b) C(S) + O2(g) → CO2 (g)

c) N2(g) + O2(g) → 2NO(g)

d) CaCO3(S) → CaO(S) + CO2(g)

Answer: d) CaCO3(S) → CaO(S) + CO2(g)

Solution:

In CaCO3(S)  CaO(S) + CO2(g), entropy change is positive. In (a) and (b) entropy change is negative ; in (c) entropy change is zero.

25. The values of ∆H and ∆S for a reaction are respectively 30 kJ mol–1 and 100 JK–1 mol–1. Then the temperature above which the reaction will become spontaneous is

a) 300 K

b) 30 K

c) 100 K

d) 200 C

DG = DH – T D S

At 300K

DG =30000 J mol–1 – 300 K x 100 J K–1 mol–1

DG =  0

above 300 K ; ∆G will be negative and reaction becomes spontaneous.

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11th Chemistry : UNIT 7 : Thermodynamics : Choose the best answer: Thermodynamics | with Answers and Solution