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)?

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

it takes on an electronical energy