Answer to Question #347076 in General Chemistry for amela

Question

The reaction 2 NO(g) = N2(g) + O2(g) has a Kc value of 2400 at 2000 K. If 0.850 M each of N2 and O2 are initially present in a 3.00-L vessel, calculate the equilibrium concentrations of No, N2 and O2.

Answer 1

K_c = 2400

C_M (N₂) = C_M (O₂) = 0,85 mol/l

2NO = N₂ + O₂

Concentration of equilibrium:

NO = 2X

N₂ = 0,85 - X

O₂ = 0,85 - X

$K_c = \frac {[N_2][O_2]}{[NO]²}$

$2400= \frac {[0,85-x][0,85-x]}{[2x]²}$

$9600x² = 0,7225 -1,7x +x²$

$9599x² + 1,7x - 0,7225 = 0$

$X_{1,2} = \cfrac{-b ±\sqrt{b² -4ac} }{2a}$

$X_{1,2} = \frac{-1,7 ±\sqrt{1,7² -4*9599*(-0,7225)} }{2*9599}$

X= 0,00859

C_M (N₂) = C_M (O₂) = 0,85 - 0,00859 = 0,84141 mol/l

C_M (NO) = 0,00859 * 2 = 0,01718 mol/l

ANSWER:

C_M (N₂) = 0,84141 mol/l

C_M (O₂) = 0,84141 mol/l

C_M (NO) = 0,01718 mol/l