JEE Exam  >  JEE Tests  >  Thermodynamics MCQ - 4 (Advanced) - JEE MCQ

Thermodynamics MCQ - 4 (Advanced) - JEE MCQ


Test Description

10 Questions MCQ Test - Thermodynamics MCQ - 4 (Advanced)

Thermodynamics MCQ - 4 (Advanced) for JEE 2024 is part of JEE preparation. The Thermodynamics MCQ - 4 (Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Thermodynamics MCQ - 4 (Advanced) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics MCQ - 4 (Advanced) below.
Solutions of Thermodynamics MCQ - 4 (Advanced) questions in English are available as part of our course for JEE & Thermodynamics MCQ - 4 (Advanced) solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Thermodynamics MCQ - 4 (Advanced) | 10 questions in 40 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
Thermodynamics MCQ - 4 (Advanced) - Question 1

 

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbis energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to Zero.

Q.

The value of log10 keg is equal to Kg is the equilibrium constant]

 

Thermodynamics MCQ - 4 (Advanced) - Question 2

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbis energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to Zero.

Q.

Which diagram represents the large value of equilibrium constant for the reversible reaction

1 Crore+ students have signed up on EduRev. Have you? Download the App
Thermodynamics MCQ - 4 (Advanced) - Question 3

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbis energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to Zero.

For a reaction M2 O(s)→ 2M(S) →2M(s) + 1/2O2(g) ΔH = 30 KJ/mol and ΔS = 0.07 KJ/mol /K at latm. The reaction would not be spontaneous at  temperatures                       

Thermodynamics MCQ - 4 (Advanced) - Question 4

The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (ΔG).

ΔG = ΔH – TΔS

where,        ΔH = Enthalpy change ; ΔS = Entropy change;

                  T = Temperature in kelvin.

If ΔG < 0, Process is spontaneous ; ΔG = 0, Process is at equilibrium, AG > 0, Process is non-spontaneous.

Q.

For the change

H20 (s, 273 K, 2 atm) → H20 (l, 273K, 2 atom),

choose the correct option.               

Detailed Solution for Thermodynamics MCQ - 4 (Advanced) - Question 4

H2O (s) →H2O ( l ),

As P ­ , more H2O ( l ) is formed.
At P = 1 atm, Δ G = 0.

Thermodynamics MCQ - 4 (Advanced) - Question 5

The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (ΔG).

ΔG = ΔH – TΔS

where,        ΔH = Enthalpy change ; ΔS = Entropy change;

                  T = Temperature in kelvin.

If ΔG < 0, Process is spontaneous ; ΔG = 0, Process is at equilibrium, AG > 0, Process is non-spontaneous.

5 mol of liquid water is compressed from 1 bar to 10 bar at constant temperature. Change in Gibb's energy (ΔG) in Joule is t [Density of water = 1000 kg/m3].                       

Detailed Solution for Thermodynamics MCQ - 4 (Advanced) - Question 5

Δ G = V (P2 - P1) = 5 × 18 × 10-6 (101 - 1) × 105 Joule = 5 × 18 × 10-1 × 102 = 900 Joule.

Thermodynamics MCQ - 4 (Advanced) - Question 6

The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (ΔG).

ΔG = ΔH – TΔS

where,        ΔH = Enthalpy change ; ΔS = Entropy change;

                  T = Temperature in kelvin.

If ΔG < 0, Process is spontaneous ; ΔG = 0, Process is at equilibrium, AG > 0, Process is non-spontaneous.

Q.

Quick lime, Ca0 is produced by heating limestone, CaC03 to drive off C02 gas.

CaC03 (s) → CaO (s) + C02 (g), (ΔH°) = 180 KJ, ΔS° = 150 J/K.

Assuming that variation of enthalpy change and entropy change with temperature to be negligible, choose the correct option :

Detailed Solution for Thermodynamics MCQ - 4 (Advanced) - Question 6

Thermodynamics MCQ - 4 (Advanced) - Question 7

Thermodynamics MCQ - 4 (Advanced) - Question 8

Thermodynamics MCQ - 4 (Advanced) - Question 9

The feasibility of a chemical reaction can be explained based on H, S and G, so answer the following :

Thermodynamics MCQ - 4 (Advanced) - Question 10

Match the following :

Detailed Solution for Thermodynamics MCQ - 4 (Advanced) - Question 10

All the matching is obvious from the fact see the formula.

Information about Thermodynamics MCQ - 4 (Advanced) Page
In this test you can find the Exam questions for Thermodynamics MCQ - 4 (Advanced) solved & explained in the simplest way possible. Besides giving Questions and answers for Thermodynamics MCQ - 4 (Advanced), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE