Question

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?

Answer 1

Answer: 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).

Answer 2

Answer:

$[HI]_{eq}=0.47M$

Explanation:

Hello,

In this case, for the given reaction, the law of mass action in terms of the change $x$ due to equilibrium is:

$53.3=\frac{(2x)^2}{(0.300M-x)(0.300M-x)}$

For which $x$ turns out:

$x=0.235M$

Therefore, the concentration of HI is:

$[HI]_{eq}=2*0.235M=0.47M$

Best regards.