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NEET Exam  >  NEET Tests  >  Test: Thermodynamics (August 18) - NEET MCQ

Test: Thermodynamics (August 18) - NEET MCQ


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15 Questions MCQ Test - Test: Thermodynamics (August 18)

Test: Thermodynamics (August 18) for NEET 2024 is part of NEET preparation. The Test: Thermodynamics (August 18) questions and answers have been prepared according to the NEET exam syllabus.The Test: Thermodynamics (August 18) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Thermodynamics (August 18) below.
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Test: Thermodynamics (August 18) - Question 1
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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 Test: Thermodynamics (August 18) - Question 1

ΔS (per mole ) = 21.98 JK -1mol-1

Test: Thermodynamics (August 18) - Question 2
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For which one of the following equations is ΔHºreact equal to ΔHfº for the product? [2003]

Detailed Solution for Test: Thermodynamics (August 18) - Question 2

► 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.
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Test: Thermodynamics (August 18) - Question 3
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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 Test: Thermodynamics (August 18) - Question 3

H2(g) + Br2(g) → 2HBr(g)

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

Test: Thermodynamics (August 18) - Question 4
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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 Test: Thermodynamics (August 18) - Question 4

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

Test: Thermodynamics (August 18) - Question 5
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The absolute enthalpy of neutralisation of the reaction: MgO (s) + 2HCl (aq) —→ MgCl2(aq) + H2O (l) will be:[2005]
Detailed Solution for Test: Thermodynamics (August 18) - Question 5

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.

Test: Thermodynamics (August 18) - Question 6
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Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? [2005]
Detailed Solution for Test: Thermodynamics (August 18) - Question 6

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).

Test: Thermodynamics (August 18) - Question 7
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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 Test: Thermodynamics (August 18) - Question 7

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

Test: Thermodynamics (August 18) - Question 8
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Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE? [2006]
Detailed Solution for Test: Thermodynamics (August 18) - Question 8

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.

Test: Thermodynamics (August 18) - Question 9
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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 Test: Thermodynamics (August 18) - Question 9

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

Test: Thermodynamics (August 18) - Question 10
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Which of the following are not state functions? [2008]
(I) q + w
(II) q
(III) w
(IV) H - TS
Detailed Solution for Test: Thermodynamics (August 18) - Question 10

► 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).

Test: Thermodynamics (August 18) - Question 11
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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 Test: Thermodynamics (August 18) - Question 11

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

Test: Thermodynamics (August 18) - Question 12
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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 Test: Thermodynamics (August 18) - Question 12

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

Test: Thermodynamics (August 18) - Question 13
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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 Test: Thermodynamics (August 18) - Question 13

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

∴ Correct choice : (d)

Test: Thermodynamics (August 18) - Question 14
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Match List -I (Equations) with List-II (Type of processes) and select the correct option. [2010]

Options:
Detailed Solution for Test: Thermodynamics (August 18) - Question 14

∴ Correct choice : (d)

Test: Thermodynamics (August 18) - Question 15
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Consider the following processes :

 [2011M]
Detailed Solution for Test: Thermodynamics (August 18) - Question 15

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

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