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)?

Question

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)?

Owen Ambrose · Accepted Answer

#4.08xx10^8#

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

In an equilibrium reaction, the equilibrium constant #"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"N"_2#, hence #["N"_2]^2# in the equilibrium constant expression.)

Divide the mole amounts by the volume to determine the concentrations (molarity).

#["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"#

#"K"=((1.00xx10^-2)^2)/((1.40xx10^-4)^2(1.25xx10^-5))=4.08xx10^8#

There is no unit in the equilibrium constant.

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

Luna Chen · Answer

undefined $$ K = \frac{{[N_2O]^2}}{{[N_2]^2 [O_2]}} $$

Given:
$$ [N_2] = 2.80 \times 10^{-4} \, \text{mol/L} $$
$$ [O_2] = 2.50 \times 10^{-5} \, \text{mol/L} $$
$$ [N_2O] = 2.00 \times 10^{-2} \, \text{mol/L} $$

$$ K = \frac{{(2.00 \times 10^{-2})^2}}{{(2.80 \times 10^{-4})^2 (2.50 \times 10^{-5})}} $$

$$ K = \frac{{4.00 \times 10^{-4}}}{{1.96 \times 10^{-7} \times 6.25 \times 10^{-10}}} $$

$$ K ≈ 1.03 \times 10^6 $$

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) --&gt; 2 N2O(g)?

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)?