Question

Equal molar amounts of H2 and I2 are placed into an evacuated container and heated to 445°C, where the following equilibrium is established. H2(g) + I2(g) ⇌ 2 HI(g) K = 50.2 Once equilibrium is reached, the concentration of HI is found to be 0.030 M. What were the initial concentrations of H2 and I2 in this experiment?

Answer 1

Answer: The initial concentration of
`H_2\text{ and }I_2` are 0.0192 M and 0.0192 M respectively.

Step-by-step explanation:

We are given:

Equilibrium concentration of HI = 0.030 M

Moles of hydrogen gas = Moles of iodine gas (concentration will also be the same)

For the given chemical equation:

`H_2+I_2\rightleftharpoons 2HI`

Initial: x x -

At eqllm: x-c x-c 2c

Calculating the value of 'c'

`2c=0.030\\\\c=(0.030)/(2)=0.015M`

The expression of
`K_(eq)` for above reaction follows:

`K_(eq)=([HI]^2)/([H_2]* [I_2])`

We are given:

`K_(eq)=50.2`

`[H_2]=(x-c)=(x-0.015)`

`[I_2]=(x-c)=(x-0.015)`

Putting values in above equation, we get:

`50.2=((0.030)^2)/((x-0.015)* (x-0.015))\\\\x=0.0108,0.0192`

Neglecting the value of x = 0.0108 M, because the initial concentration cannot be less than the equilibrium concentration.

x = 0.0192 M

Hence, the initial concentration of
`H_2\text{ and }I_2` are 0.0192 M and 0.0192 M respectively.