NEET Exam  >  NEET Tests  >  NCERT Based Tests for NEET  >  Test: Thermodynamics of Chemical Equilibrium - NEET MCQ

Test: Thermodynamics of Chemical Equilibrium - NEET MCQ


Test Description

15 Questions MCQ Test NCERT Based Tests for NEET - Test: Thermodynamics of Chemical Equilibrium

Test: Thermodynamics of Chemical Equilibrium for NEET 2025 is part of NCERT Based Tests for NEET preparation. The Test: Thermodynamics of Chemical Equilibrium questions and answers have been prepared according to the NEET exam syllabus.The Test: Thermodynamics of Chemical Equilibrium MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Thermodynamics of Chemical Equilibrium below.
Solutions of Test: Thermodynamics of Chemical Equilibrium questions in English are available as part of our NCERT Based Tests for NEET for NEET & Test: Thermodynamics of Chemical Equilibrium solutions in Hindi for NCERT Based Tests for NEET course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Thermodynamics of Chemical Equilibrium | 15 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study NCERT Based Tests for NEET for NEET Exam | Download free PDF with solutions
Test: Thermodynamics of Chemical Equilibrium - Question 1

Direction (Q. Nos. 1-9) This section contains 9 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. For the reaction in equilibrium,
I. CO(g) + 1/2O2(g)  

II. 2CO(g) + O2 (g)  2CO2 (g)
 


Then

Detailed Solution for Test: Thermodynamics of Chemical Equilibrium - Question 1

CO(g) + ½ O2(g)  <====> CO2(g)
For this reaction, k1 = [CO2]/[CO][O2
2CO(g) +  O2(g)  <====> 2CO2(g)
For this reaction, k2 = [CO2]2/[CO]2[O2] = ([CO2]/[CO][O2]½)2 = (k1)2
So, k2 = k12
∆G = -RTln keqm
∆G2/∆G1 = ln k1/ln k12 = 2
∆G2 = 2∆G1

Test: Thermodynamics of Chemical Equilibrium - Question 2

Equilibrium constant for the following equilibrium,

is 1.0 x 1028 at a temperature at which RT = 2500 J mol-1. Gibbs free energy change (ΔG°) is

Detailed Solution for Test: Thermodynamics of Chemical Equilibrium - Question 2

ΔG° = -RT lnk
ΔG° = -(2500) (ln1028)
ΔG° = -161.2 kJ
Hence C is Correct Answer.

Test: Thermodynamics of Chemical Equilibrium - Question 3

Kp for the equilibrium  N2O4 (g)  2NO2 (g)  is 0.141 at 25°C and 1 bar. This reaction is spontaneous at 25°C and partial pressures of

Test: Thermodynamics of Chemical Equilibrium - Question 4

For the following equilibrium at 373 K,
H2O(g)H2O(g)
ΔAG° is

Test: Thermodynamics of Chemical Equilibrium - Question 5

For the following electrochemical cell reaction at 298 K,
 
E°cell = 1.10 V

Detailed Solution for Test: Thermodynamics of Chemical Equilibrium - Question 5

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq),

E= +1.10V

∴ Eo = 0.0591/n log10Keq

because at equilibrium, 

Ecell = 0

(n = number of electrons exchanged = 2)

1.10 = 0.0591/2 log10Keq
2.20/0.0591 = log10Keq

Keq = antilog37.225

Test: Thermodynamics of Chemical Equilibrium - Question 6

Given 
COCl2(g)  CO(g) + Cl2 (g)
Kp = 8 x 10-9 atm
ΔS° = 30 cal K-1 at 373 K.
Temperature at which phosgene will be 0.1 % dissociated at 2 atm
(R = 2 cal mol-1K-1)is

Test: Thermodynamics of Chemical Equilibrium - Question 7

For the following equilibrium at 298 K,
CO(g) + H2O (g)  CO2(g) + H2(g);
Δ f G° (in kcal mol-1) of CO = - 32.81, CO2 = - 94.26, H2O = - 54.64, H2 = 0.0 then, degree of dissociation of CO(g) is

Test: Thermodynamics of Chemical Equilibrium - Question 8

At 1 atmospheric pressure, N2O4 is 50% dissociated at 300 K and 80% dissociated at 400 K as given
N2O4 (g)  2NO2 (g)
Thus, enthalpy of dissociation is

Test: Thermodynamics of Chemical Equilibrium - Question 9

At 800 K, ΔG° for the reaction

NiO (s) + H2(g)  Ni (s) + H2O (g)

is - 9.0 kcal. Thus, ratio of pressures of water vapours and hydrogen in equilibrium with NiO and Ni at 800 K in terms of logarithm is

Test: Thermodynamics of Chemical Equilibrium - Question 10

Direction (Q. Nos. 10) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.

Q. Statement I

For a reaction  2NO2(g)  N2O4 (g) variation of (log10 K) with (T-1) is represented as

Statement II

Association o f NO2 to N2O4 is an exotherm ic efiange.

*Multiple options can be correct
Test: Thermodynamics of Chemical Equilibrium - Question 11

Direction (Q. No. 11) This section contains 4 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equation

Graphically it is shown as given θ = tan (0 .5) and OP = 10

Thus,

Test: Thermodynamics of Chemical Equilibrium - Question 12

Direction (Q. Nos. 12-13) This section contains  a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)

1 mole H2(g) and 0.2 mole CO2(g) are introduced in a vacuum flask at 450°C and 0.5 atm.

H2(g) + CO2  H2O(G) + CO(g)

Analysis shows that mixture contains 10 moles per cent steam. Also equilibrium constant increases by one per cent per degree around 450°C.(log 1.1 = 0.0414)

Q. Equilibrium constant Kp is

Test: Thermodynamics of Chemical Equilibrium - Question 13

1 mole H2(g) and 0.2 mole CO2(g) are introduced in a vacuum flask at 450°C and 0.5 atm.

H2(g) + CO2 (g)  H2O (g) + CO (g)
Analysis shows that mixture contains 10 moles per cent steam. Also equilibrium constant increases by one per cent per degree around 450°C.(log 1.1 = 0.0414)

Q. ΔH (enthalpy change) by taking equilibrium constant at 450° C and 460° C is

Test: Thermodynamics of Chemical Equilibrium - Question 14

Direction (Q. No. 14) Choice the correct combination of elements and column I and coloumn II  are given as option (a), (b), (c) and (d), out of which ONE option is correct.

The hydrogenation of pyridine (C5H5N) to piperidine (C5H11N)

C5H5N(g) + 3H2(g)  C5H11N(g)

is an equilibrium process whose equilibrium constant (Kp)is given by

Match the thermodynamics parameters in column I with their respective values in column II.

*Answer can only contain numeric values
Test: Thermodynamics of Chemical Equilibrium - Question 15

Direction (Q. Nos. 15) This section contains 1 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

For a reversible reaction 
Rate constant k1 (forward)  = 1015e -2000/T and k2 (backward ) = 1012e -2000/T

What is the value of 


748 tests
Information about Test: Thermodynamics of Chemical Equilibrium Page
In this test you can find the Exam questions for Test: Thermodynamics of Chemical Equilibrium solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Thermodynamics of Chemical Equilibrium, EduRev gives you an ample number of Online tests for practice
Download as PDF