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?

At equilibrium:

[H2] = 0.005 M

[Br2] = 0.105 M

[HBr] = 0.189 M

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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.