Question

A mixture of 2 moles of N, and 8 moles of H2
are heated in a 2 lit vessel, till equilibrium is
established. At equilibrium, 0.4 moles of N2
was present. The equilibrium concentration
of H, will be
1) 2 mole/lit
2) 4 mole/lit
3) 1.6 mole/lit 4) 1 mole/lit​

Answer 1

Answer:

H

2

+3H

2

⇌2NH

3

t=0 2 8 −

t eg. 2.x 8−3x 2x

2−x=0.4

⇒x=1.6 ∴(H

2

)

eq

=8−4.8=3.2

[H

2

]

eq

=

2

3.2

=1.6mol/lit

Answer 2

Question :

A mixture of 2 moles of $\sf N_2$ and 8 moles of $\sf H_2$ are heated in a 2 lit vessel, till equilibrium is established. At equilibrium 0.4 moles of $\N_2$ was present. What will be the equilibrium concentration of $\sf H_2$.

Answer :

Given :

A mixture of 2 moles of $\sf N_2$ and 8 moles of $H_2$ are heated in a 2 lit vessel.

To find :

The equilibrium constant of $\sf H_2$ will be ?

Solution :

• N₂  + 3H₂ → 2NH₃.

Here, 1 mole of N₂ reacts with 3 moles of H₂  to give 2 moles of NH₃

No of moles present initially,

• N₂  = 2 moles.
• H₂  = 8 moles.
• NH₃ = 0.

At equilibrium,

• N₂ = 0.4 moles
• Moles of N₂ used = 2-0.4 = 1.6 moles.

1.6 moles of N₂ will react with 1.6 x 3 = 4.8 moles of H₂ to give 2 x 1.6 = 3.2 moles of NH₃

Moles of H₂  left = 8 - 4.8 = 3.2 moles

Volume of solution = 2 litre

Moles / litre = 3.2 / 2 = 1.6 mol / litre

The equilibrium concentration of the solution is 1.6 mol/lit.