Question

Given `K_c = 0.36` at `2000^@ "C"` for `"N"_2"O"_4(g) rightleftharpoons "NO"_2(g)`, if the initial concentration of `"NO"_2` is `"1 M"`, what are the equilibrium concentrations of `"N"_2"O"_4(g)` and `"NO"_2(g)`?

Answer 1

`"N"_2"O"_4: 0.33` `M`

`"NO"_2: 0.34` `M`

Explanation:

We're asked to find the equilibrium concentrations of `"N"_2"O"_4` and `"NO"_2`, given the initial `"NO"_2` concentration.

The equilibrium constant expression is given by

`K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = ul(0.36)" "` `(2000color(white)(l)""^"o""C")`

We'll do our I.C.E. chart in the form of bullet points, for fun. Then, our initial concentrations are

INITIAL

• `"N"_2"O"_4:` `0`

• `"NO"_2:` `1` `M`

According to the coefficients of the reaction, the amount by which `"NO"_2` decreases is two times as much as the amount by which `"N"_2"O"_4` increases:

CHANGE

• `"N"_2"O"_4:` `+x`

• `"NO"_2:` `-2x`

And so the final concentrations are

FINAL

• `"N"_2"O"_4:` `x`

• `"NO"_2:` `1` `M` `- 2x`

Plugging these into the equilibrium constant expression gives us

`K_c = ((1-2x)^2)/(x) = ul(0.36)" "` `(2000color(white)(l)""^"o""C"`, excluding units`)`

Now we solve for `x`:

`(4x^2 - 4x + 1)/x = 0.36`

`4x^2 - 4x + 1 = 0.36x`

`4x^2 - 4.36x + 1 = 0`

Use the quadratic equation:

`x = (4.36+-sqrt((4.36)^2 - 4(4)(1)))/(8) = 0.328color(white)(l)"or"color(white)(l)0.762`

If we plug the larger solution in for `x` in the final `"NO"_2` concentration `(1-2x)`, we would obtain a worthless negative value. Therefore, we use the smaller solution to calculate the final concentrations:

`color(red)("final N"_2"O"_4) = x = color(red)(ulbar(|stackrel(" ")(" "0.33color(white)(l)M" ")|)`

`color(blue)("final NO"_2) = 1-2x = 1-2(0.328) = color(blue)(ulbar(|stackrel(" ")(" "0.34color(white)(l)M" ")|)`

each rounded to `2` significant figures.