Answer to Question #112691 in General Chemistry for Adam Rosenberger

Question

Suppose the following system reaches equilibrium. N2(g) + O2(g) ←→ 2NO(g) Analysis of the mixture in a 1-L flask gives the following results: N2 = 0.50 mol, O2 = 0.50 mol, and NO = 0.020 mol. Calculate Keq for the reaction.

Answer 1

Amount of $N_2 = 0.50\ mol, O_2 = 0.50\ mol,$ and $NO=0.020\ mol$ in a $1$ litre flask.

And,Concentration is amount of solute in $1$ litre of solution.

Therefore,$[N_2]=0.5\ mol/l;[O_2]=0.5\ mol/l$ $;[NO]=0.020 \ mol/l$

For the given reaction,$N_2(g) + O_2(g) ←→ 2NO(g)$

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}=\frac{(0.020)^2}{0.5\times0.5}=0.080$