Concepts of Chemical Equilibrium Greatly Expected to be asked in JEE Main

 Concepts of chemical equilibrium greatly expected  to be asked in JEE Main

As the JEE Main grows older, it asks more and more tricky questions. To be able to solve chemistry questions asked in JEE Main, one must get all important concepts cleared and  analyse previous year question papers. The ability to grasp the data given in the question and find appropriate formula or path to find the answer is frankly achieved through problem solving. The level of questions asked in JEE Main is the next important thing to realise. Here are the summary of concepts and questions asked from chemical equilibrium in the previous years, so that it will be easy to take a conclusion. (Though you will find little easier questions asked in 2020 due to corona pandemic...)

Rate vs. Time graph w.r.t. Equilibrium:

1. Q: (04/09/20 Morning) 

For the equilibrium A ⇌ B, the variation of the rate of the forward reaction (a) and reverse reaction (b) with time is given by:

Solution: 

Characteristics of chemical equilibrium constant:

 1. Q: 09/01/19 (evening): 

A2 (g) + B2 (g) ⇌ 2 AB, eq constant = k1

6 AB ⇌ 3A2 + 3B2 Eq. constant = K2, find the relationship between K1 and K2

Solution:

For the reverse process of first equilibrium, 

2 AB ⇌ A2 (g) + B2 (g)   Eq. Constant = 1/K1

Now if we multiply 3 throughout the above equation, 6 AB ⇌ 3A2 + 3B2, then its equiulibrium constant = K2 = given = (1/k1)³ 

2. Q: 09/04/19 (CBT : evening): S(s) + O2 (g) ⇌ SO2 (g); Eq. Constant = K1 = 10⁵²

                                                               2S (s) + 3O2 (g) ⇌ 2SO3 (g); k2 = 10¹²⁹,

Then Eq. Constant for the reaction, 2 SO2 (g) + O2 (g) ⇌ 2SO3, k3 = ?

Solution: Here eq. 3 = Eq. 2 - 2 (Eq. 1)

Thus k3 = 10¹²⁹ / 10^{52 x 2}

= 10²⁵

3. Q: (04/09/20 Evn) If the Eq. Const for the reaction A ⇌ B + C is K1 and that for 

B + C ⇌ P is K2 then eq. constant K3 for the reaction A  ⇌ P  is:

Solution: K3 = [P] / [A] = K1 X K2

4. Q: (06/09/20 Evn) The value of Kc is 64 at 800K for the reaction 

N2 (g) + 3H2 (g) ⇌ 2NH3 (g). 

The value of Kc of the reaction NH3 (g) ⇌ 1/2 N2 (g) + 3/2 H2 (g) is:

Relationship between Kc and Kp

1. Q: (12/04/19 CBT Evn) In which one of the following equilibria Kp not equal to Kc?

a.     2HI (g) ⇌ H2 (g) + I 2 (g)

b.     2NO (g) ⇌ N2 (g) + O2 (g)

c.     NO2 (g) + SO2 (g) ⇌ NO (g) + SO3 (g)

d.     2C (s) + O2 (g) ⇌ 2CO (g)

Solution: In d. Δng not equal to 0 hence its the answer.

2. Q: (10/01/19 Morning):  The Kp/ Kc for the following reactions at 300 k are respectively: (RT = 24.62 ltr atm per mol)

1.    N2(g) + O2(g) ⇌ 2NO(g)

2.    N2O4(g) ⇌ 2NO2(g)

3.    N2(g) + 3H2 (g) ⇌  2NH3(g) 

Solution: We know Kp/Kc = (RT) ^Δng

In 1.  Δng = 0,  Thus Kp = Kc

In 2. Δng =1 Thus Kp/Kc = RT = 24.62 ltr atm per mol

In 3. Δng = -2, Now find the answer

Relatioship between Kp and total pressure:

1. Q: (11/09/19 Morning): 

The eq const for the reaction N2(g) + 3H2 (g) ⇌  2NH3(g) is Kp. If pure ammonis is left to dissociate, the partial pressure of ammonia at eq. is given by: (assume that P of NH3 << P total at eq.)

Solution: 

Kp, Kc and Degree of dissociation:

Q: (10/01/19 Evn): 

5.1 g NH4SH is introduced in 3 L of evacuated flask at 327 degree centigrade. 30% of the solid of NH4SH decomposed to NH3 and H2S gasses. The Kp of the reaction at 327 degree centigrade is:

Solution: 

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