1
$\begingroup$

I was recently given the question:

The equilibrium constant, K, for the following reaction:
N₂(g) + O₂(g) ↔ 2NO(g)
is 4.00 x 10⁻² at a very high temperature. The reaction is at equilibrium at this temperature with [N2] = [O2] = 0.100 M and [NO] = 0.0200 M in a 2.00 liter flask. If 0.120 mol of NO is suddenly added to the reaction mixture what will be the concentrations of all species when equilibrium is re-established?

I think filling out the ICE table would look like this:

N₂ O₂ 2NO
I .100 .100 .080
C +x +x -2x
E .100+x .100+x .080-2x

Which would make the the equilibrium formula: $$ 4.00\cdot 10^{-2}=\frac{(.080-2x)^{2}}{(.100+x)(.100+x)} $$

However, when solving for $x$ I get $x=\frac{8.2\pm\sqrt{7.84}}{198}$ or $.0\overline{5}$ and $.0\overline{27}$.
nvm I get it now but as this is my first time using stackexchange and this syntax took me too long I feel like posting it because the TeX looks nice to look at :)
noo why does the table only work in preview mode :(

Improve this question
$\endgroup$
3
  • $\begingroup$ Something wrong, I guess. $\endgroup$
    – ACR
    Commented Dec 22, 2022 at 7:06
  • $\begingroup$ The equation as written can be simplified and so not a quadratic $\endgroup$
    – porphyrin
    Commented Dec 22, 2022 at 9:32
  • $\begingroup$ @porphyrin It can be done in this case, as Sam202 explicitly shows in his answer. But it does not address a general case when it cannot be done. Then more or less obvious constraints of valid values must be considered. $\endgroup$
    – Poutnik
    Commented Dec 22, 2022 at 15:28

2 Answers 2

Reset to default
7
$\begingroup$

Scenarios leading to a quadratic equation have two mathematical solutions, real or complex. But, one of them often does not have interpretation in scientific context, as it's value is out of the acceptable value range.

Typically, it provides, directly or indirectly, a negative value of a quantity that cannot be negative. A trivial example:

What is the length of the shorter side of a rectangular with area $\pu{30 cm2}$, if the other side is longer by $\pu{1 cm}$?

$$x(x+1)=30$$ $$x^2 + x - 30 = 0$$ $$x=\frac{-1 \pm \sqrt{1+120}}{2}=\frac{-1 \pm 11}{2}$$

A side of the length $\pu{-6 cm}$ does not obviously make physical sense. But $\pu{5 cm}$ does.

In your particular case, the greater result value would lead to negative $\ce{NO}$ concentration, so it is not a valid result. There is the constraint $x \le \pu{0.04 mol L-1}$ for $[\ce{NO}] \ge \pu{0 mol L-1}$.

Improve this answer
$\endgroup$
3
$\begingroup$

$$4.00\cdot 10^{-2}=\frac{(0.080-2x)^{2}}{(0.100+x)(0.100+x)}=\frac{(0.080-2x)^2}{(0.100+x)^2}=\left(\frac{0.080-2x}{0.100+x}\right)^2=0.040$$

Taking the square root on both sides:

$$\frac{0.080-2x}{0.100+x}=0.2$$

Solving for x:

$$x=0.0273$$

Notice we did not consider the negative root above, since an equilibrium constant (or its square root) having a negative value is nonsense.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.