# If the initial concentration of NO is 0.100M and O2 is 0.050M. What are the equilibrium concentrations for NO, O2 and NO2? For this reaction, Kc = 3 * 10(6). 2NO + O2 <-> 2NO2

Feb 23, 2015

I'll emphasize the strategy you should use for this type of problems this time.

So, you know that you have a dynamic equilibrium that needs to be established between the reactants, $N O$ and ${O}_{2}$, and the product, $N {O}_{2}$.

SInce you only start with reactants, the reaction will move to the right and $N {O}_{2}$ will be produced. Now, the equilibrium constant tells you how much of the reactants and how much of the product you'll have at equilibrium.

In this case, the ${K}_{c}$ is very, very large, which implies that, at equilibrium, the vessel will contain mostly product, $N {O}_{2}$, with the concentrations of the two reactants dropping significantly.

To actually solve for the equilibrium concentrations, you must use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart). You set it up like this

....$2 N {O}_{\left(g\right)} + {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 N {O}_{2 \left(g\right)}$
I...0.100........0.050...................0
C...(-2x)..........(-x)......................(+2x)
E..0.100-2x...(0.050-x)............2x

Remember that the concentrations increase or decrease proportional to the stoichimetric coefficients of the compounds, that is why the concentration of $N {O}_{2}$ dropped by 2x, while the concentration of ${O}_{2}$ dropped by just x.

Use the expression of the equilibrium constant to solve for $x$.

${K}_{c} = \frac{{\left(2 x\right)}^{2}}{{\left(0.100 - 2 x\right)}^{2} \cdot \left(0.050 - x\right)} = 3 \cdot {10}^{6}$

Don't be intimidated by such equations; this particular one is a little ugly, but it can be solved using a cubic equation calculator such as this one - http://www.1728.org/cubic.htm; all you have to do is get it to the $a {x}^{3} + b {x}^{2} + c x + d = 0$ form.

This equation will produce one real solution, $x = 0.0499$, and two imaginary ones; using the value for $x$ that we've calculated will get us

$\left[N {O}_{2}\right] = 2 \cdot x = 2 \cdot 0.0499 = \text{0.0998 M}$
$\left[N O\right] = 0.100 - 2 x = \text{0.000200 M}$
$\left[{O}_{2}\right] = 0.0500 - 0.0499 = \text{0.000100 M}$

Finally, take a second to ckeck the result. We've predicted that, at equilibrium, the concentrations of the two reactans will be very small and the concentration of the product will be, by comparison, very large.

Indeed, the very large ${K}_{c}$ determined that the vessel will mostly contain $N {O}_{2}$ at equilibrium and that the concentrations of the two reactants will be very, very small.