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Test: Entropy & Gibbs Energy - NEET MCQ


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10 Questions MCQ Test - Test: Entropy & Gibbs Energy

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

In the evaporation of water, the entropy

Detailed Solution for Test: Entropy & Gibbs Energy - Question 1

As the randomness increases on evaporation, entropy also increases. 
The greater the randomness in a system, greater is its entropy.

Test: Entropy & Gibbs Energy - Question 2

The value of ΔS for spontaneous process is

Detailed Solution for Test: Entropy & Gibbs Energy - Question 2

The second law of thermodynamics states that for any spontaneous process, the overall ΔS must be greater than or equal to zero.

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Test: Entropy & Gibbs Energy - Question 3

What is the change in the entropy of water, When ice melts into water?

Detailed Solution for Test: Entropy & Gibbs Energy - Question 3

The greater the randomness in a system, greater is its entropy. The randomness is greater in liquid state as compared to solid state so the entropy increases when ice melts into water.

Test: Entropy & Gibbs Energy - Question 4

 For the reaction Hg (l) ———> Hg (g) , the entropy change will have the sign:

Detailed Solution for Test: Entropy & Gibbs Energy - Question 4

+ve (because gases are highly random than liquid) ΔS = S(g) - S(l) = +ve

Test: Entropy & Gibbs Energy - Question 5

Which of the following is extensive property?

Detailed Solution for Test: Entropy & Gibbs Energy - Question 5

Intensive properties: Properties which are independent of the amount of substance (or substances) present in the system are called intensive properties, e.g. pressure, density, temperature, viscosity, surface tension, refractive index, emf, chemical potential, sp. heat etc, These are intensive properties.
So, we can say that entropy is an extensive property.

Test: Entropy & Gibbs Energy - Question 6

 Which of the following is true about a spontaneous process?

Detailed Solution for Test: Entropy & Gibbs Energy - Question 6

In cases where ΔG is: negative, the process is spontaneous and proceed in the forward direction.

Test: Entropy & Gibbs Energy - Question 7

For a spontaneous chemical process, the free energy change is

Detailed Solution for Test: Entropy & Gibbs Energy - Question 7

The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative, the process is spontaneous and may proceed in the forward direction as written. positive, the process is non-spontaneous as written, but it may proceed spontaneously in the reverse direction.

Test: Entropy & Gibbs Energy - Question 8

Which of the following statements is correct about the process of evaporation of water from an open beaker?

Detailed Solution for Test: Entropy & Gibbs Energy - Question 8

Evaporation of water from an open pot is a spontaneous process. As it starts on its own without any external agent. However it takes heat from surrounding means that it is endothermic.

Test: Entropy & Gibbs Energy - Question 9

Which of the following is NOT a state function?

Detailed Solution for Test: Entropy & Gibbs Energy - Question 9

A state function is the property of the system whose value depends only on the initial and final state of the system and is independent of the path. It is a state function because it is independent of the path. Heat (q) and work (W) are not state functions being path dependent.

Test: Entropy & Gibbs Energy - Question 10

The equilibrium constant for a reaction is 10. Calculate the value of ΔG° at 300K , R = R =8.314 JK−1mol−1

Detailed Solution for Test: Entropy & Gibbs Energy - Question 10

The value of equilibrium constant(k) = 10
∆Gθ = -2.303RTlogk
Δ Gθ for the reaction,
= (2.303)(8.314JK−1mol−1)(300K)log10 = –5744.14 Jmol−1 = –5.744 kJ mol−1

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