31 Year NEET Previous Year Questions: Thermodynamics - 1 - NEET MCQ

# 31 Year NEET Previous Year Questions: Thermodynamics - 1 - NEET MCQ

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## 36 Questions MCQ Test Chemistry Class 11 - 31 Year NEET Previous Year Questions: Thermodynamics - 1

31 Year NEET Previous Year Questions: Thermodynamics - 1 for NEET 2024 is part of Chemistry Class 11 preparation. The 31 Year NEET Previous Year Questions: Thermodynamics - 1 questions and answers have been prepared according to the NEET exam syllabus.The 31 Year NEET Previous Year Questions: Thermodynamics - 1 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for 31 Year NEET Previous Year Questions: Thermodynamics - 1 below.
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31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 1

### What is the entropy change (in JK–1 mol–1) when one mole of ice is converted into water at 0º C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol–1 at 0ºC) [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 1

ΔS (per mole ) = 21.98 JK -1mol-1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 2

### The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is: [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 2

ΔG = – PΔV = Work done
ΔV is the change in molar volume in the conversion of graphite to diamond.

Work done = – (–1.91 × 10–3) × P × 101.3 J

∵ 1 atm = 105 × 1.013 Pa
⇒ P = 9.92 x 108 Pa

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31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 3

### For which one of the following equations is ΔHºreact equal to ΔHfº for the product? [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 3

► The heat of formation is defined as the heat generated or observed when the compound is formed from its component elements in their standard state.

• By the definition (C) and (A) are incorrect as the starting material are compounds.
• (B) is incorrect as ozone is not the standard state for oxygen.
• (D) is correct as it satisfies the definition of the heat of formation.
31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 4

For the reaction C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) at constant temperature, ΔH – ΔE is: [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 4

ΔH = ΔE + nRT
Δng = 3 - (1 + 5) = -3
ΔH - ΔE = (-3 RT)

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 5

If the bond energies of H-H, Br-Br, and HBr are 433, 192 and 364 kJ mol–1 respectively, the ΔH° for the reaction H2(g) + Br2(g) → 2HBr(g) is: [2004]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 5

H2(g) + Br2(g) → 2HBr(g)

ΔHº = (B.E.)react - (B.E.)prod
⇒ ΔHº = (433 + 192) - (2 x 364) = 625 – 728 = – 103 kJ

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 6

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are -382.64 kJ mol–1 and -145.6 JK–1 mol–1, respectively.
Standard Gibb's energy change for the same reaction at 298 K is: [2004]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 6

ΔG = ΔH - TΔS
ΔG = -382.64 - (298 x 145.6 x 10-3) = -339.3 kJ mol-1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 7

Considering entropy(S) as a thermodynamic parameter, the criterion for the spontaneity of any process is: [2004]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 7

For a spontaneous process, ΔStotal is always positive.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 8

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: (1L atm  = 101.32 J) [2004]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 8

W = – pΔV
W = -3(6 - 4) = - 6 L atm = - 6 x 101.32 J = (approx) - 608 J

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 9

The absolute enthalpy of neutralisation of the reaction: MgO (s) + 2HCl (aq) —→ MgCl2(aq) + H2O (l) will be:[2005]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 9

As MgO is a oxide of  weak base hence  some energy is lost to break MgO (s). Hence enthalpy is less than –57.33 kJ mol–1.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 10

A reaction occurs spontaneously if: [2005]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 10

For a spontaneous reaction, ΔG  is always –ve.
Since, ΔG = ΔH – TΔS
Therefore, ΔS = +ve, ΔH = +ve and  TΔS > ΔH.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 11

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? [2005]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 11

The measure of the disorder of a system is nothing but Entropy.
For a spontaneous reaction, ΔG < 0. As per Gibbs Helmnoltz equation, ΔG = ΔH – TΔS.
Thus, ΔG is –ve, if ΔH = –ve (exothermic) and ΔS = +ve (increasing disorder).

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 12

The enthalpy of hydrogenation of cyclohexene is –119.5 kJ mol–1. If the resonance energy of benzene is –150.4 kJ mol–1, its enthalpy of hydrogenation would be: [2006]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 12

= – 348.5 kJ

The resonance energy provides extra stability to the benzene molecule so it has to be overcome for hydrogenation to take place.
So, ΔH = – 358.5 – (–150.4) = –208.1 kJ

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 13

The enthalpy and entropy change for the reaction Br2(l) + Cl2(g) → 2BrCl(g) are 30kJ mol–1 and 105 JK–1 mol–1 respectively.The temperature at which the reaction will be in equilibrium is: [2006]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 13

We know that, ΔG = ΔH – TΔS
When the reaction is in equilibrium, ΔG = 0

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 14

Identify the correct statement for change of Gibbs energy for a system (ΔGsystem) at constant temperature and pressure: [2006]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 14
• If ΔGsystem = 0 the system has attained equilibrium is correct.
• (d) is confusing as to when ΔG > 0, the process may be spontaneous when it is coupled with a reaction which has ΔG < 0 and total ΔG is negative.
31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 15

Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE? [2006]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 15

We know that: ΔH = ΔE + PΔV
In the reactions, H2 + Br2 → 2HBr there is no change in volume or ΔV = 0.
So, ΔH = ΔE for this reaction.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 16

Consider the following reactions: [2007]

(i)

(ii)

(iii)

(iv)

Enthalpy of formation of H2O (l) is

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 16

Reaction (ii) shows the formation of H2O, and the X2 represents the enthalpy of formation of H2O because as the definition suggests that the enthalpy of formation is the heat evolved or absorbed when one mole of a substance is formed from its constituent atoms.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 17

Given that bond energy of H–H and Cl–Cl is 430 kJ mol–1 and 240 kJ mol–1 respectively and ΔHf for HCl is - 90 kJ mol–1, bond enthalpy of HCl is: [2007]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 17

-90 = [1/2 x 430 + 1/2 x 240] - B.E. of HCl
∴  B.E. of HCl = 215 + 120 + 90 = 425 kJ mol–1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 18

Which of the following are not state functions? [2008]
(I) q + w
(II) q
(III) w
(IV) H - TS

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 18

► A property whose value doesn’t depend on the path taken to reach that specific value is known as state functions or point functions.
► In contrast, those functions which do depend on the path from two points are known as path functions.

Heat (q) and Work (w) are not state functions but (q + w) is a state functions.
H – TS (i.e. G) is also a state function.

Thus, II and III are not state functions so the correct answer is option (d).

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 19

For the gas phase reaction, which of the following conditions is correct? [2008]
PCl5(g) ⇌ PCl3​(g) + Cl2​(g)

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 19

The reaction given is an example of a decomposition reaction
Since, decomposition reactions are endothermic in nature, therefore, ΔH > 0.
Δn = (1+1) – 1= +1
Hence more molecules are present in products which shows more randomness i.e. ΔS > 0 (ΔS is positive).

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 20

Bond dissociation enthalpy of H2, Cl2, and HCl are 434, 242, and 431 kJ mol–1 respectively. Enthalpy of formation of HCl is: [2008]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 20

The reaction of formation of HCl: H2 + Cl2 → 2HCI
⇒ Enthalpy of formation of 2HCl = - (862 - 676) = -186 kJ.
∴ Enthalpy of formation of HCl = -(186/2) kJ = -93 kJ

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 21

The values of ΔH and ΔS for the reaction, C(graphite) + CO2 (g) → 2CO(g) are 170 kJ and 170 JK–1, respectively. This reaction will be spontaneous at: [2009]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 21

ΔG = ΔH – T Δ S
At equilibrium, ΔG = 0
⇒ 0 = (170 × 103 J) – T (170 JK– 1)
⇒ T = 1000 K
For spontaneity, ΔG is – ve, which is possible only if T > 1000 K.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 22

From the following bond energies: [2009]
H – H bond energy: 431.37 kJ mol–1
C = C bond energy: 606.10 kJ mol–1
C – C bond energy: 336.49 kJ mol–1
C – H bond energy: 410.50 kJ mol–1

Enthalpy for the reaction, will be:

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 22

Enthalpy of reaction = B.E(Reactant)– B.E(Product)
ΔHreaction = [606.1 + (4 × 410.5) + 431.37)] –  [336.49 + (6 × 410.5)]  = –120.0 kJ mol–1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 23

Standard entropies of X2 , Y2 and XY3 are 60, 40 and 50 JK–1mol–1 respectively. For the reaction
to be at equilibrium, the temperature should be: [2010]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 23

ΔS for the reaction

ΔS = 50 – (30 + 60) = – 40 J
For equilibrium ΔG = 0 = ΔH – TΔS

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 24

For vapor ization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJmol–1 and 108.8 JK–1 mol–1, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is: [2010]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 24

ΔH = 40630J mol –1
ΔS = 108.8JK–1 mol –1

∴ Correct choice : (d)

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 25

Match List -I (Equations) with List-II (Type of processes) and select the correct option. [2010]

Options:

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 25

∴ Correct choice : (d)

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 26

The following two reactions are known : [2010]

The value of ΔH for the following reaction

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 26

ΔH = –26.8 + 33.0 = + 6.2 kJ

∴   Correct choice : (a)

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 27

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol–1 at 27ºC, the entropy change for the process would be: [2011]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 27

Given ΔH = 30 kJ mol–1 T = 273 + 27 = 300 K
= 100J mol–1 K–1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 28

Enthalpy change for the reaction, 4H(g) → 2H2 (g)  is - 869.6 kJ. The dissociation energy of H–H bond is: [2011]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 28

Given:
4H(g) → 2H2 (g); ΔH = - 869.6 kJ
i.e. 2H2 (g) → 4H(g); ΔH = 869.6 kJ

⇒ H2 (g) → 2H(g); ΔH = (869.6) * 1/2 = 434.8 kJ

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 29

Consider the following processes :

[2011M]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 29

To calculate ΔH operate 2 × eq. (1) + eq. (2) – eq. (3)
ΔH = 300 – 125 – 350 = – 175KJ/mol

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 30

In which of the following reactions, standard entropy change (ΔS°) is positive and standard Gibb’s energy change (ΔG°) decreases sharply with increasing temperature? [2012]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 30

Since in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. ΔSº = +ve.

Since, ΔG° = ΔH° – TΔS
Hence the value of ΔG decrease on increasing temperature.

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 31

The enthalpy of fusion of water is 1.435 kCal/mol.The molar entropy change for the melting of ice at 0°C is: [2012]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 31

= 5.260 cal / mol-K

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 32

Standard enthalpy of vapourisation Δvap H°  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 : [2012]

(Assume water vapour to behave like an ideal gas).

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 32

vapH° = 40.66 kJ mol-1
H2O(l) ⇋ H2O(g)
∆ng = 1-0 = 1
∆H = ∆U + ∆ngRT
∆U = ∆H-∆ngRT
= 40,660 - 8.314×373
= 37558.878 J mol-1 or 37.56kJ mol-1

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 33

Equal volumes of two monoatomic gases, A and B, at the same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be: [2012 M]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 33

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 34

The Gibbs’ energy for the decomposition of Al2O3 at 500°C is as follows :
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500°C is at least : [2012 M]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 34

ΔG = -nFE°

for the reaction n

960 × 103 = – 4 × 96500 × E°
E° = – 2.5 volt
So, it needed 2.5 volt for reduction

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 35

When 5 liters of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 liters of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ is: [NEET Kar. 2013]
ΔHcomb (CH4) = 890 kJ mol–1
ΔHcomb (C3H8) = 2220 kJ mol–1

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 35

2x + 5(5– x) = 16
⇒  x = 3L
∴  Heat released

31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 36

Three thermochemical equations are given below:

(i)
(ii)
(iii)

Based on the above equations, find out which of the relationship given below is correct? [NEET Kar. 2013]

Detailed Solution for 31 Year NEET Previous Year Questions: Thermodynamics - 1 - Question 36

Applying Hess’s law, equation (i) can be obtained by adding equations (ii) and (iii).
∴ x = y + z

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