Question

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 617°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 47 at 617°C If the initial concentrations of hydrogen gas and iodine vapor are both 0.041 mol/L and the concentration of hydrogen iodine is 0.115 mol/L what is the equilibrium concentration of hydrogen gas? Enter a number to 4 decimal places.

Answer 1

Answer: The equilibrium concentration of hydrogen gas is 0.0222 M

Step-by-step explanation:

We are given:

Initial concentration of hydrogen gas = 0.041 M

Initial concentration of iodine vapor = 0.041 M

Initial concentration of hydrogen iodide = 0.115 M

For the given chemical equation:

`H_2(g)+I_2(g)\rightleftharpoons 2HI(g)`

Initial: 0.041 0.041 0.115

At eqllm: 0.041-x 0.041-x 0.115+2x

The expression of
`K_c` for above equation follows:

`K_c=([HI]^2)/([H_2][I_2])`

We are given:

`K_c=47`

Putting values in above equation, we get:

`47=((0.115+2x)^2)/((0.041-x)* (0.041-x))\\\\x=0.0816,0.0188`

Neglecting the value of x = 0.0816 because, equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of hydrogen gas =
`(0.041-x)=(0.041-0.0188)=0.0222M`

Hence, the equilibrium concentration of hydrogen gas is 0.0222 M