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).

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explanation: