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

Answer:

Equilibrium concentration of Br₂ = 0.02 M

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.374\ g}{2.016\ g/mol}$

$Moles\= 0.68\ mol$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{70.31\ g}{159.808\ g/mol}$

$Moles\= 0.4400\ mol$

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