Question

A mixture of 1.374 g of H₂ and 70.31 g of Br₂ is heated in a 2.00 L vessel at 700 K . These substances react as follows: H₂(g)+Br₂(g)⇌2HBr(g) At equilibrium the vessel is found to contain 0.566 g of H₂. Calculate the equilibrium concentration of H₂.

Answer 1

Equilibrium concentration of Br₂ = 0.02 M

Step-by-step explanation:

Moles of hydrogen gas :

Given, Mass of H₂ = 1.374 g

Molar mass of H₂ = 2.016 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (1.374\ g)/(2.016\ g/mol)`

`Moles\= 0.68\ mol`

Moles of Bromine gas :

Given, Mass of Br₂ = 70.31 g

Molar mass of Br₂ = 159.808 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (70.31\ g)/(159.808\ g/mol)`

`Moles\= 0.4400\ mol`

Considering the ICE table for the equilibrium as:

H₂(g) + Br₂(g) ⇌ 2HBr(g)

t = o 0.68 0.44 0

t = eq -x -x +2x

--------------------------------------------- -----------------------------

Moles at eq: 0.68-x 0.44-x 2x

Given that: At equilibrium the vessel is found to contain 0.566 g of H₂

Moles = 0.566 g / 2.016 g/mol = 0.28 moles

Thus, 0.68 - x = 0.28

x = 0.40 moles

Volume = 2.00 L

Equilibrium moles of Br₂ = 0.44 - 0.40 moles = 0.04 moles

Equilibrium concentration of Br₂ = 0.04 moles/ 2 L = 0.02 M