Question

QUESTION FROM CHEMICAL EQUILIBRIUM...​

Answer 1

Answer:

b) 1.4l mol‐¹ is answer dear friend

Answer 2

Answer:

2.35 L/mol

Explanation:

Given: 1.5 moles of 〖SO〗_2 and 1 mole of O_2

Volume of container = 2L

At equilibrium, [〖SO〗_3] = 0.35 mol⁄L

Reaction is:

2 〖SO〗_2 (g) + O_2 (g) ⇌ 2 〖SO〗_3 (g)

To find: K_c of the above given reaction.

For the given above reaction,

2〖SO〗_2 (g) + O_2 (g) ⇌ 2 〖SO〗_3 (g)

Initial moles: 1.5 1 0

At equilibrium: 1.5 - 2a 1-a 2a

According to the question,

Concentration of 〖SO〗_3 at equilibrium = 0.35 mol⁄L

So, 0.35 = ( 2a)/2 (Volume of container is 2L)

a = 0.35

Now,

K_c = 〖[SO_3]〗^2/(〖[SO_2]〗^2 [O_2])

K_c = 〖(2a)〗^2/((1.5-2a)^2 (1-a)) = ((〖0.7/2)〗^2)/(〖((1.5-0.7)/2)〗^2*((1-0.35)/2)) = 0.1225/0.052 = 2.35 L⁄mol Hence, equilibrium constant of the above given reaction is 2.35 L⁄mol.