Question

N2O4 is a volatile liquid that, in the vapor phase, undergoes the reaction: N204 (g) --> 2NO2 (g). 1.0 g of N2O4 is injected in liquid form into 1.0 L container holding 1.0 bar of (non-reactive) air at 345 K. After a period of time, the liquid has entirely evaporated and the total pressure of the systen (still at 345 K) is 1.56 bar. Find the amount (in moles) of N2O4 (g) in the system at that time. Report your answer with proper units. Show steps in your work.

Answer 1

moles of N2O4 = 1 / 92 = 0.0109

P V = n RT

P = 0.0109 x 0.0821 x 345 / 1 = 0.309 atm

initial pressure = 0.9869 + 0.309 = 1.30 atm

N204 (g) -------------------> 2NO2 (g)

1.30                                       0

1.3-x                                       2x

total pressure = 1.3 - x + 2x = 1.54

x = 0.24

equilibrium pressure of N2O4 = 1.3 - x = 1.3 - 0.24 = 1.06 atm

P V = n R T

1.06 x 1 = n x 0.0821 x 345

n = 0.0374

the amount (in moles) of N2O4 (g) in the system at that time = 0.0374 moles