Equilibrium Concentration problem

Question

For the reaction: H2(g) Br2(l) → 2HBr(g)

Kc = 4.8x108

An evacuated container is filled with 0.300M of H2(g) and excess of Br2(l). What is the equilibrium concentration of H2(g)?

A. 2.45x10-9M

B. 4.9x10-10M

C. 5.3x10-9M

D. 1.2x10-10M

E. 7.5x10-10M

Answer 1

E  7.5×10⁻¹⁰M

Explanation:

You've got a heteregenous euqilibrium. Since bromine is present as pure liquid it makes not contribution to the equilbrium equation. So the eueilibrium concentrations in gas phase are given by:

Kc = [HBr]² / [H₂]

According to reaction equation the amount hydrogen bromide formed is twice the amount of hydrogen consumed. So:

[HBr] = 2 · ( [H₂]₀ - [H₂] )

([H₂]₀ is the inital concetration of hydrogen )

Because the equlibrium is rather large, the equilibrium postion is far right and the hydrogen concentration is rather small compared to initial concentration. Hence you can approximate:

[HBr] ≈ 2·[H₂]₀

Insert this to equilibrium equation and solve for hydrogen concentration:

Kc = (2·[H₂]₀)² / [H₂]

<=>

[H₂] = (2·[H₂]₀)² / Kc

= (2 · 0.3 )² / 4.8×10⁻⁸

= 7.5×10⁻¹⁰M