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

`[HI]_(eq)=0.47M`

Step-by-step 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=((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.

Answer 2

Answer: 0.471 (3s.f)

Step-by-step 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).