Question

The reaction N2O(g) + O2(g) NO2(g) +NO(g) has ΔHº = -42.9 kJ and ΔSº = -26.1 J/K. Suppose 0.400 mol of N2O and 0.400 of O2 are placed in a 8.00 L container at 400 ºC and this equilibrium is established. What percentage of the N2O has reacted? (Note: Assume that ΔH and ΔS are relatively insensitive to temperature, so ΔHº298 and ΔSº298 are about the same as ΔHº673 and ΔSº673 respectively.) Enter your answer with 2 significant digits.

Answer 1

First calculate the dG as follows:

dG = dH - TdS

dG = -42.9 kJ *1000 J /1.0KJ -673 K *-26.1 J/K

dG = -42900 +17565.3 J

dG =-25334.7 J

Now equilibrium constant :

dG = -2.3RT log K_eq

- 25334.7= -2.3 * 8.34 J /mol K *673 log K_eq

log K_eq=25334.7/12869.24

log K_eq=1.97= 93.325

The reaction N2O(g) + O2(g) = NO2(g) +NO

I                  0.05            0.05            0        0

C                 -x                -x                +x     +x

E       0.05-x                   0.05-x         x        x

K eq = [NO2] [NO] / [N2O] [ O2]

93.325= x*x / (0.05-x)2

taking under root

9.66 = x / 0.05-x

0.483 -9.66 x= x

0.483 = 10.66 x

x= 0.045 M

at equilibrium N2O = 0.05-x = 0.05-0.045 = 5*10^-3

Now calculate the percentage of the N2O has reacted as follows:

5*10^-3 / 0.05*100= 10%