Question

The reaction 20F2(9) —2F2(9) + O2(9) has an equilibrium constant K, in terms of pressures) at 25°C of 6.19x10-4. Calculate the concentration of F2 that will be present at 25°C in equilibrium with OF (at a concentration of 2.49x10-2 mol L-1) and O2 (at a concentration of 1.58 10-2 mol L

Answer 1

The concentration of F₂in equilibrium is 4,93x10⁻₃M

Step-by-step explanation:

For the reaction:

2 OF₂ ⇄ 2 F₂ + O₂

The equilibrium constant, k, is defined as:

k = [F₂]²[O₂] / [OF₂]²

If you have in equilibrium [OF₂] = 2,49x10⁻² and [O₂] = 1,58x10⁻²

The concentration of F₂ is:

6,19x10⁻⁴ = [F₂]²[1,58x10⁻²] / [2,49x10⁻²]²

2,43x10⁻⁵ = [F₂]²

4,93x10⁻³ = [F₂]

I hope it helps!