Question

For the reaction;

H2(g) + Br2 ⇌ 2HBr (g)
If I start with 0.10 M Hydrogen and 0.20 M bromine what are the equilibrium concentrations of each if Kc = 62.5?

Answer 1

At equilibrium:

[H2] = 0.005 M

[Br2] = 0.105 M

[HBr] = 0.189 M

Step-by-step explanation:

H2(g) + Br2(g) ⇄ 2HBr

an "x" value will be used from reactant to produced "2x"

so at equilibrium:

[H2] = 0.1 - x

[Br2] = 0.2 - x

[HBr] = 2x

we know that Kc=[HBr]²/[H2][Br2]

Thus 62.5 = (2x)²/(0.1-x)(0.2-x)

this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0

the x₁ = 0.23 x₂ = 0.09457

we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration

Then we substitute the "x₂" value at equilibrium:

[H2] = 0.1-0.09457 = 0.005 M

[Br2] = 0.2-0.09457 = 0.105 M

[HBr] = 2*0.09457 = 0.189 M