Question

At a certain temperature, the equilibrium constant, kc, for this reaction is 53.3. h2(g)+i2(g)â½âââ2hi(g)kc=53.3 at this temperature, 0.700 mol of h2 and 0.700 mol of i2 were placed in a 1.00-l container to react. what concentration of hi is present at equilibrium?

Answer 1

Step-by-step explanation:

5.11= 9.3001x

x=0.5494

2 mol HI produced. Thus,

(2)(0.5494)=1.0989 = 1.09 = [HI] present at equilibrium.

Answer 2

5.11 mol/L.

Step-by-step explanation:

• For the reaction:

H₂(g) + I₂(g) ⇄ 2HI(g), Kc = 53.3.

∵ Kc = [HI]²/[H₂][I₂]

[H₂] = 0.70 mol/1.0 L = 0.70 mol/L.

[I₂] = 0.70 mol/1.0 L = 0.70 mol/L.

∵ Kc = [HI]²/[H₂][I₂]

∴ [HI]² = Kc[H₂][I₂] = (53.3)(0.70)(0.70) = 26.117.

∴ [HI] = √26.117 = 5.11 mol/L.