Question

At high pressures and moderate temperatures nitric oxide gas disproportionates rapidly according to the reaction 3NO(g) ⇌ N2O(g) + NO2(g) A 1-liter vessel is pressurized with NO(g) and maintained at a constant temperature until equilibrium is reached where K = 1.9×1016 (as expressed in terms of the concentrations of the gases in mol/L). At equilibrium, the amount of NO(g) in the flask is found to be 7.5×10-6 mol. What is the equilibrium concentration of N2O(g)?

Answer 1

Answer:

2.831mol/L

Explanation:

$k =\frac{[N_{2}O][NO_{2} ]}{[NO]^{3}} \\\\1.9*10^{16}=\frac{[x][x]}{[7.5*10^{-6} ]^{3} }\\1.9*10^{16}=\frac{x^{2} }{[7.5*10^{-6} ]^{3} } \\x= 1.9*10^{16}*[7.5*10^{-6} ]^{3}\\x = \sqrt{ 1.9*10^{16}*[7.5*10^{-6} ]^{3}} = 2.831mol/L$

Answer 2

Answer:

3.8 x 10⁵

Explanation:

For the equilibrium :  3NO(g) ⇌ N2O(g) + NO2(g), the equilibrium constant in the terms of the concentrations of the gases in mol/L is

Kc = (NO) (N2O)/ (NO) ³   where (NO), (N2O) , (NO2) are the concentrations of the gases in mol/L . So

K= (x mol/ 1 L) (x mol/1L) / (7.5 x 10⁻⁶ mol /1 L) ³

x  = mol of NO and NO2 at equilibrium

we have that

K = x²/ 7.5 x 10⁻⁶ = 1.9 x 10¹⁶

x = √ (7.5 x 10⁻⁶ x  1.9 x 10¹⁶) = 3.8 x 10⁵

∴ (N2O) = 3.8 x 10⁵