Question

At a certain temperature K =1.8 litre^2 mole^-2 for
N2+3H2=2NH3. How many moles of NH3
must be placed in one litre vessel in order to get
6 mole litre-1 H2 at equilibrium?
(1) 27.88 mole
(2) 40 mole
(3) 31.88 mole
(4) 80 mole

Answer 1

Given:

• The equilibrium constant for the formation of ammonia (k) = 1.8 M⁻²
• Equilibrium concentration of hydrogen = 6 M
• Volume of the vessel = 1 L

To find:

The initial concentration of NH₃.

Solution:

• Let the initial concentration of NH₃ be C and the concentration of NH₃ converted be 2x.
• At equilibrium, [NH₃] = C-2x , [H₂] = 3x , [N₂] = x
• [H₂] = 3x = 6 M ⇒ x = 2M
• K = [NH₃]²/[N₂].[H₂]³ = (C-4)²/(2*6³) = 1.8 ⇒ C = 4 + √(1.8*2*216) = 31.88M

Answer:

The initial concentration of NH₃ = 31.88 M

Answer 2

Explanation:

Check the attachment , Don't forget that we need to put coefficient with X and subtract it from moles initially , For Eg:- if 2 mole is given and coefficient has 3 then equation will be 2-3x , where 3x is moles dissociated not degree of dissociation , degree of dissociation is x