At a certain temperature, the equilibrium constant, K c , for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 53.3 At this temperature, 0.300 mol H 2 and 0.300 mol I 2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

0.471 (3s.f)

Explanation:

Let x moles of HI being formed.

[H2]=0.3-2x

[I2]=0.3-x

Kc=[HI]^2/[I2][H2]

53.3=x^2/(0.3-x/2)(0.3-x/2)

53.3(0.3-0.5x)^2=x^2

When solved for we get,

x=0.47098

Concentration of HI present is therefore 0.47098.

Or 0.471 (3 significant figures).

alchemy hadn't scientific methods of study, appeal to mysticism and fantasy, the knowledge of chemistry and physics were modest at that time.

alchemy was - but as new rules were discovered it either changed into chemistry or it became "unscientific".