Question

A mixture of 1.374g of H2 and 70.31g 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 0.566g of H2.
a. Calculate the equilibrium concentrations of H2, Br2 and HBr.
b. Calculate Kc.

Answer 1

a. 0.139 M → [H₂] ; 0.217 M → [Br₂] ; 0.01 M → [HBr]

b. Kc = 3.31x10⁻³

Step-by-step explanation:

H2(g) + Br2(g) ⇄ 2HBr(g)

Initial 1.374 g 70.31 g -

reacts X X 2x

eq. (1.374 - x) (70.31-x) 2x

In equilibrium I see, the grams I initially had minus some mass which has reacted. In products I have the double of that mass, because the stoichiometry.

So I have the mass in equilibrium, of H2 and of course I can know the mass which has reacted.

1.374g - x = 0.556 g

1.374g - 0.556 g = x = 0.808 g (This is the mass which has reacted)

70.31g - 0.808 g = 69.502 g (Mass in equilibrium of Br2)

2 . 0.808 g = 1.616 g (Mass in equibrium of HBr)

By molar mass, we can kwow the moles.

Molar mass H2: 2 g/m

Moles = mass / molar mass → 0.556 g / 2 g/m = 0.278 moles

Molar mass Br2: 159.80 g/m

Moles = mass / molar mass → 69.502 g / 159.80 g/m = 0.434 moles

Molar mass HBr: 80.9 g/m

Moles = mass / molar mass → 1.616 g / 80.9 g/m = 0.02 moles

The moles are not molarity. In equilibrium, to calculate Kc we need molarity (moles/L). The moles we have calculated are in 2 L of mixture so:

moles / L = molarity

0.278 moles / 2L = 0.139 M → [H₂]

0.434 moles / 2L = 0.217 M → [Br₂]

0.02 moles / 2L = 0.01 M → [HBr]

Kc = [HBr]² / ([H₂] . [Br₂])

Kc = 0.01² / (0.139 . 0.217) = 3.31x10⁻³