Answer:
a. 0.139 M → [H₂] ; 0.217 M → [Br₂] ; 0.01 M → [HBr]
b. Kc = 3.31x10⁻³
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
<em>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.</em>
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
