Question

For the reaction H2+I2----2HI Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?

Answer 1

`["HI"] = 0.778M`

Explanation:

We're asked to find the equilibrium concentration of `"HI"` with an initial concentration of `1` `"mol H"_2` and `1` `"mol I"_2`.

The equilibrium constant expression for this reaction is

`K_c = (["HI"]^2)/(["H"_2]["I"_2]) = 49`

The initial concentrations of both `"H"_2` and `"I"_2` are

`(1color(white)(l)"mol")/(2color(white)(l)"L") = 0.5M`

So our initial concentrations for each species are

Initial:

• `"H"_2`: `0.5M`

• `"I"_2`: `0.5M`

• `"HI"`: `0`

From the coefficients of the chemical equation, we can predict the changes in concentration with the quantity `x`:

Change:

• `"H"_2`: `-x`

• `"I"_2`: `-x`

• `"HI"`: `+2x`

and the final concentrations are the sum of the initial and change:

Final:

• `"H"_2`: `0.5M-x`

• `"I"_2`: `0.5M-x`

• `"HI"`: `2x`

It's algebra time! Let's plug these into our equilibrium constant expression to start solving for `x`:

`K_c = ((2x)^2)/((0.50-x)(0.50-x)) = 49`

`4x^2 = 49(0.25-x+x^2)`

`49x^2 - 49x + 12.25 = 4x^2`

`45x^2 - 49x + 12.25 = 0`

`x = (49 +-sqrt((-49)^2-4(45)(12.25)))/(2(45)) = 0.7` or `0.389`

We can neglect the solution that is greater than `0.5`, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is `color(blue)(0.389`.

Therefore, the final equilibrium concentration of `"HI"` is

`["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M`