﻿ NEET Important Questions of Thermodynamics | Zigya

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Thermodynamics Multiple Choice Questions

1.

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is

• 38.3 J mol-1 K-1

• 35.8 J  mol-1 K-1

• 32.3 J  mol-1 K-1

• 42.3 J mol-1 K-1

A.

38.3 J mol-1 K-1 789 Views

2.

The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and F2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ → X 2+ + Y will be spontaneous when

• X = Ni, Y = Fe

• X = Ni, Y = Zn

• X =Fe, Y= Zn

• X = Zn, Y = Ni

D.

X = Zn, Y = Ni

X = Zn, Y = Ni
Zn + Ni2+  →Zn2+ + Ni 206 Views

3.

The standard enthalpy of formation of NH3 is– 46.0 kJmol–1. If the enthalpy of formation of H2 from its atoms is – 436 kJ mol–1 and that of N2 is – 712 kJ mol–1,the average bond enthalpy of N – H bond is NH3 is

• -964 kJ mol-1

• +352 kJ mol-1

• +1056 kJ mol-1

• -1102 kJ mol-1

B.

+352 kJ mol-1 670 Views

4.

The energy required to break one mole of Cl— Cl bonds in Cl2 is 242 kJ mol. The longest wavelength of light capable of breaking a single Cl — Cl bond is
(c= 3 x 108 ms–1and NA = 6.02 x 1023 mol–1)

• 594 nm

• 640 nm

• 700 nm

• 494

D.

494

Energy, E = NA 232 Views

5.

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:(R = 8.314 J/mol K) ( ln 7.5 = 2.01)

• q =+208J, W = - 208 J

• q =-208 J, W =-208 J

• q=-208J, W = +208 J

• q =+208 J, W = +208 J

A.

q =+208J, W = - 208 J

From first law of thermodynamics, ΔE = q+W for an isothermal expansion.
Hence, q =-W
q= +208 J
W =-208 J [expansion work]

541 Views

6.

The following reaction is performed at 298 K
2NO(g) + O2 (g) ⇌ 2NO2 (g)
The standard free energy of formation of NO (g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2 (g) at 298 K? (KP = 1.6 x 1012)

• R (298) In (1.6 x 1012)-86600

• 86600 + R (298) In (1.6 x 1012)

• 86600 - In(1.6 x 1012)/R(298)

• 0.5[2 x 86600-R(298)In (1.6 x 1012)]

D.

0.5[2 x 86600-R(298)In (1.6 x 1012)]

For the given reaction,
2NO(g) + O2 (g) ⇌ 2NO2 (g)
Given , Now, we have, 564 Views

7.

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

• 676.5

• -676.5

• -110.5

• 110 s

C.

-110.5

C(s) + O2 (g) → CO2 (g); ΔH = -393.5 kJ mol-1 ... (i)

CO + O2/2 → CO2 (g); ΔH = - 283.5 kJ mol-1 ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get

C (s) + O2/2 (g) → CO (g)

ΔH = (-393.5 + 283.5) kJ mol-1 = - -110 kJ mol-1

593 Views

8.

For the complete combustion of ethanol, C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(l), the amount of heat produced as measured in a bomb calorimeter, is 1364.47 kJ mol-1 at 25oC. Assuming ideality the enthalpy of combustion, CH, for the reaction will be (R = 8.314 JK-1 mol-1)

• -1366.95 kJ mol-1

• -1361.95 kJ mol-1

• -1460.50 kJ mol-1

• -1350.50 kJ mol-1

A.

-1366.95 kJ mol-1

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2(l),
∆U = - 1364.47 kJ/mol
∆H = ∆U +∆ngRT
∆ng = -1
∆H = - 1364.47 + [Here, value of R in unit of J must be converted into kJ]
= - 1364.47-2.4776
= -1366.94 kJ/mol
980 Views

9.

The incorrect expression among the following is

• • In isothermal process • • C. Option C has incorrect expression. The correct expression is, 334 Views

10.

The standard Gibbs energy change at 300 K for the reaction, 2A  ⇌ B +C is 2494.2J at a given time, the composition of the reaction mixture is and , The reaction proceeds in the [R= 8.314 JK/mol, e = 2.718]

• forward direction because Q>Kc

• reverse direction because Q>Kc

• forward direction because Q < Kc

• reverse direction because Q < Kc

B.

reverse direction because Q>Kc

We know,
ΔG = ΔGo + RTlnQ .. (i)
Given,
ΔGo  = 2494.2J thus,
putting the value in equation (i)
= 2494.2 +8.314 + 300 In 4
= 28747.27 J
= positive value
Also, we have If ΔG is positive, Q >K
therefore, reaction shifts in the reverse direction

600 Views