Solve for Concentration

Question

For the reaction: H2(g) + Br2(g) ⇌ 2 HBr(g), Kc = 7.5 × 10^2 at a certain temperature. If 2 mole each of H2 and Br2 are placed in 2-L flask, what is the concentration of H2 at equilibrium?

Answer 1

[H₂] = 0.068 M at equilibrium

Step 1 — initial concentrations

K_c is written in concentrations, so convert the moles first. 2 mol in a 2 L flask:

[H₂]₀ = [Br₂]₀ = 1.0 M,  [HBr]₀ = 0

Skipping the division by volume is the most common way this problem goes wrong — the moles are not the concentrations here.

Step 2 — ICE table

H₂(g) + Br₂(g) ⇌ 2 HBr(g)

          H₂      Br₂     HBr
I:    1.0     1.0      0
C:  −x     −x    +2x
E: 1−x   1−x    2x

Step 3 — and here is the shortcut worth knowing

K_c = [HBr]² / ([H₂][Br₂]) = (2x)² / (1−x)(1−x) = 4x² / (1−x)² = 750

Both sides are perfect squares, so take the square root of the whole equation instead of expanding into a quadratic:

2x / (1−x) = √750 = 27.39

2x = 27.39 − 27.39x  →  29.39x = 27.39  →  x = 0.932

This works whenever the two reactants start at equal concentrations and have equal coefficients — it turns a quadratic into one line of algebra. Recognising it saves real time on an exam.

Step 4 — answer the question asked

[H₂] = 1.0 − x = 1.0 − 0.932 = 0.068 M

(Also: [Br₂] = 0.068 M and [HBr] = 1.86 M.)

Do not use the small-x approximation here. K_c = 750 is large, meaning the reaction runs strongly toward products — x turned out to be 93% of the starting concentration, nowhere near negligible. The 5% rule only applies when K is small. Assuming x << 1 here would have given a badly wrong answer.

Check the result: K_c = (1.86)² / (0.068)(0.068) = 3.47 / 0.00463 = 750 ✓

And it makes sense qualitatively: a large K should leave only a small amount of reactant behind, which is exactly what 0.068 M out of an initial 1.0 M represents.