chem eqm

At 25°C and a total pressure of 1 atm, an equilibrium mixture of N2O4 and NO2 contains 68.69% of N2O4 by mol.
(a) Calculate the value of Kp
(b) Calculate the degree of dissociation of N2O4 .
(c) If the Kp is 0.6707 atm at 45°C, deduce whether the dissociation of N2O4 is exothermic, endothermic, or thermoneutral ?


(a)
N2O4(g) ⇌ 2NO2(g)

PN2O4 = XN2O4•PT = 68.69% * 1 = 0.6869 atm
PNO2 = XNO2•PT = (1 - 68.69%) * 1 = 0.3131 atm

Kp = (PNO2)²/PN2O4 = (0.3131)²/0.6968 = 0.143 atm


(b)
Let Po atm be the initial pressure of N2O4.

Po = 0.6869 + (0.3131/2) = 0.84345 atm

Degree of dissociation of N2O4 = (0.84345 - 0.6869)/0.84345 = 0.1856


(c)
Kp = 0.1856 at 25°C, and KP = 0.6707 at 45°C
(Kp at 25°C) < (KP at 45°C)

At higher temperature, the eqm position shift to the right (favours the forward reaction).

Since it favours the endothermic reaction in a higher temperature,
it can deduced that the forward reaction is endothermic.

2011-01-01 22:56:51 補充：
你的疑問是哪裡要除以 total pressure ?

2011-01-03 22:15:39 補充：
什麼？

2011-01-05 01:10:45 補充：
沒有需要除以 total pressure。