JEE Exam  >  JEE Tests  >  Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - JEE MCQ

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - JEE MCQ


Test Description

15 Questions MCQ Test - Test: Law of Mass Action, Equilibrium Constants (Kc and Kp)

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) for JEE 2024 is part of JEE preparation. The Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) questions and answers have been prepared according to the JEE exam syllabus.The Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) below.
Solutions of Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) questions in English are available as part of our course for JEE & Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) 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 Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) | 15 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 1

At T( K), the equilibrium constant of H2(g) + I2(g) ⇌ 2HI(g) is 49. If [H2],[I2] at equilibrium at the same temperature are 2.0 × 10−2M and 8.0 × 10−2M respectively, the [HI] at equilibrium in molL−1 is

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 1

Given,

For, the reaction,


On putting the values in the above equation, we have

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 2

For a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be.........

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 2

The value of equilibrium constant is independent of initial concentrations of reactants and products. So, options (a), (b) and (c) are not correct, i.e. so, for a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be remains same.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 3

Which among the following denotes the correct relationship between  and for the reaction

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 3

So we know the relation between the equilibrium constant and equilibrium pressure constant
So the formula is

Whereas we know that is the difference of the gaseous number of moles of reactant side and product side

Thus, the relation is

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 4

Kc for the reaction N2(g) + O2(g) ⇌ 2NO(g) at 300 K is 4.0 × 10−6. Kp for the above reaction will be (R = 2calmol−1 K−1)

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 4

K= Kc(RT)Δn since Δn = 0, Kp = Kc.

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 5

The equilibrium constant of the following are :

The equilibrium constant of the reaction :
, will be

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 5

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 6

The degree of dissociation of PCl5(α) obeying the equilibrium PCl⇌ PCl+ Cl2 is related to the equilibrium pressure by

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 6

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 7

moles of and moles of are mixed and allowed to attain equilibrium at . At equilibrium, the number of moles of is found to be mole. The equilibrium constant for the formation of is

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 7


                                           
                    
Equilibrium constant

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 8

For the reaction H2(g) + I2( g) ⇌ 2HI(g) at 721 K, the value of equilibrium constant is 50, when equilibrium concentration of both is 5M. Value of Kp under the same conditions will be

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 8



Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 9

For a reversible gaseous reaction at equilibrium, if some moles of are replaced by same number of moles of (T is tritium, isotope of and assume isotopes do not have different chemical properties) without affecting other parameter, then:

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 9

Since has similar chemical properties as so upon mixing, and are formed.

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 10

The ratio for the reaction  is:

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 10


For the reaction


Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 11

Which of the following is true at chemical equilibrium?

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 11

The condition for the equilibrium is
(ΔG)T, P = 0 and (ΔS)U,V = 0

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 12

Consider the expression and indicate the correct statement at equilibrium

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 12

Van't Hoff reaction isotherm is
When the reaction is in a state of equilibrium
Then

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 13

If the synthesis of ammonia from Haber's process is carried out with exactly the same starting conditions (of partial pressure and temperature) but using (deuterium) in place of . Then

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 13

The reaction mixtures starting either with or reach equilibrium with the same composition, except that and are present instead of and .

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 14

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? (K = equilibrium constant)

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 14

For a reaction,
Hence,
if , then [Product] > [Reactant]

Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 15

The increase of pressure on ice water system at constant temperature will lead to

Detailed Solution for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) - Question 15

Volume of ice is greater than that of water. The direction in which the reaction will proceed can be predicted by applying Le-Chatelier's principle Pressure So equilibrium, will shift forward.

Information about Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) Page
In this test you can find the Exam questions for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Law of Mass Action, Equilibrium Constants (Kc and Kp), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE