Answer to Question #112691 in General Chemistry for Adam Rosenberger

Question #112691
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.
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Expert's answer
2020-04-28T12:19:47-0400

Amount of N2=0.50 mol,O2=0.50 mol,N_2 = 0.50\ mol, O_2 = 0.50\ mol, and NO=0.020 molNO=0.020\ mol in a 11 litre flask.

And,Concentration is amount of solute in 11 litre of solution.

Therefore,[N2]=0.5 mol/l;[O2]=0.5 mol/l[N_2]=0.5\ mol/l;[O_2]=0.5\ mol/l ;[NO]=0.020 mol/l;[NO]=0.020 \ mol/l

For the given reaction,N2(g)+O2(g)2NO(g)N_2(g) + O_2(g) ←→ 2NO(g)

Keq=[NO]2[N2][O2]=(0.020)20.5×0.5=0.080K_{eq}=\frac{[NO]^2}{[N_2][O_2]}=\frac{(0.020)^2}{0.5\times0.5}=0.080




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