# At a particular temperature a 2.00-L flask at equilibrium contains 2.80 10-4 mol N2, 2.50 10-5 mol O2, and 2.00 10-2 mol N2O. How would you calculate K at this temperature for the following reaction: 2 N2(g) + O2(g) --> 2 N2O(g)?

Dec 16, 2015

$4.08 \times {10}^{8}$

#### Explanation:

"K"=(["N"_2"O"]^2)/(["N"_2]^2["O"_2])

In an equilibrium reaction, the equilibrium constant $\text{K}$ is found through taking the concentrations of the products over the concentrations of the reactants. (If you don't know why, ask.)

The stoichiometric coefficients in the equation are used as exponents on the concentrations. (See how the equilibrium equation references $2 {\text{N}}_{2}$, hence ${\left[{\text{N}}_{2}\right]}^{2}$ in the equilibrium constant expression.)

To find the concentrations (molarity), divide the mole amounts by the volume.

["N"_2"O"]=(2.00xx10^-2"mol")/(2.00"L")=1.00xx10^-2"M"

["N"_2]=(2.80xx10^-4"mol")/(2.00"L")=1.40xx10^-4"M"

["O"_2]=(2.50xx10^-5"mol")/(2.00"L")=1.25xx10^-5"M"

$\text{K} = \frac{{\left(1.00 \times {10}^{-} 2\right)}^{2}}{{\left(1.40 \times {10}^{-} 4\right)}^{2} \left(1.25 \times {10}^{-} 5\right)} = 4.08 \times {10}^{8}$

The equilibrium constant is unitless.

An equilibrium constant ≫1 also tells us that the reaction is heavily product-favored.