Question

10. At 25 °C, 0.200 mol of dinitrogen monoxide and 0.560 mol of oxygen gas were placed in a 10.0-L reaction vessel and allowed to react: 2 N2O(g) + 3 02(B) + 4NO2(e) When the system reached equilibrium at 25°C, the concentration of nitrogen dioxide was found to be 0.0200 M. What is the value of Kc for the oxidation of dinitrogen monoxide to nitrogen dioxide at 25°C? Setup ICE chart and show your work step by step:

Answer 1

Reaction Involved is

2N2O + 3 O2 --------> 4 NO2

	N2O	O2	NO2
Initial	0.2 moles	0.560 moles	0
Change	0.01 moles	0.015 moles	0.0200 moles
Equilibrium	0.19 moles	0.545 moles	0.0200 moles

The change is calculated as follows

the amount of NO2 at equilibrium is taken as 4x

so x = 0.0200 / 4 = 5 * 10-3

So amount of N2O is initial concentration - 2x = 0.2 - ( 2 * 5 * 10-3) = 0.19 moles

So amount of O2 is initial concentration - 3x = 0.2 - ( 3 * 5 * 10-3) = 0.545 moles

Kc = [NO2]4 / [N2O]2[O2]3

= [0.02]4 / [0.19]2[0.545]3 = 2.73 * 10-5

The Kc value at 250C is found to be 2.73 * 10-5