0.1 mole of #N_2O_4(g)# was sealed in a tube under one-atmosphere conditions at #25^o#C. Calculate the number of moles of #NO_2(g)# present, if the equilibrium #N_2O_4(g) -> 2NO_2(g)# #(K_p = 0.16)# is reached after some time?

Calculate the volume of the container given initial conditions of the #"N"_2 "O"_4# gas.

#"V" = (n * R * T)/ (P)#
#color(white)("V") = (0.1 color(white)(l) "mol" * 0.0205 color(white)(l) L * atm * "mol"^(-1) * "K"^(-1) * (273.15 + 25) color(white)(l) "K")/(1.00 color(white)(l) "atm")#
#color(white)("V") = 0.611 color(white)(l) "L"#

Let the increase in the partial pressure of #"NO"_2# be #x color(white)(l) "atm"# and construct a partial pressure RICE table:

• R #color(white)(I)"N"_2 "O"_4 (g) rightleftharpoons 2 color(white)(l) "NO"_2 (g)#
• I #color(white)(R) 1 color(white)(l) "atm" color(white)( (g) rightleftharpoons 2 color(white)(l) ) 0 color(white)(l) "atm"#
• C#color(white)(I) -x/2 color(white)(._4 (g) rightleftharpoons l) +x#
• E#color(white)(I) 1 color(white)(l) "atm" - x/2 color(white)(-2ll)x#

#(x^2)/(1-x//2) = K_p = 0.16#

#x ~~ 0.36 color(white)(l) "atm"#

Again, by the ideal gas equation:

#n("NO"_2) = (P * V) / (R * T)#
#color(white)(n("NO"_2)) = (0.36 color(white)(l) "atm" * 0.611 color(white)(l)"L")/(0.0205 color(white)(l) L * atm * "mol"^(-1) * "K"^(-1) * (273.15 + 25) color(white)(l) "K")#
#color(white)(n("NO"_2)) = 0.036 color(white)(l) "mol"#