How would you rank the following gases, NO, Ar, #N_2#, #N_2O_5#, in order of increasing speed of effusion through a tiny opening and increasing time of effusion?

Question

Avery Adams · Accepted Answer

#Z_("eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff",N_2O_5)#

Clearly, the rate is the reciprocal of the time it takes, since the rate is in #"s"^(-1)# and the time is in #"s"#. Therefore:

#t_("eff",N_2) < t_("eff",NO) < t_("eff",Ar) < t_("eff",N_2O_5)#

The rate of effusion, which I will denote as #Z_"eff"#, is defined in Graham's Law of Effusion as:

#Z_("eff",1)/(Z_("eff",2)) = sqrt(M_(m,2)/(M_(m,1))#

where #M_(m,i)# is the molar mass of compound #i# and #Z_("eff",i)# is the rate of effusion of compound #i#.

Since #Z_("eff",1) prop 1/sqrt(M_(m,1))# the higher the molar mass, the slower the rate of effusion, which makes sense since the larger the gas, the less often it can get through a small hole.

#M_(m,NO) = 14.007 + 15.999 = "30.006 g/mol"# #M_(m,Ar) = "39.948 g/mol"# #M_(m,N_2) = 2xx14.007 = "28.014 g/mol"# #M_(m,N_2O_5) = 2xx14.007 + 5xx15.999 = "108.009 g/mol"#

Thus, the molar masses are ranked as follows:

#M_(m,N_2) < M_(m,NO) < M_(m,Ar) < M_(m,N_2O_5)#

which implies that:

#color(blue)(Z_("eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff",N_2O_5))#

Mateo Aguilar · Answer

undefined The gases would be ranked in order of increasing speed of effusion and increasing time of effusion as follows:

1. Ar (Argon)
2. NO (Nitric oxide)
3. N2 (Nitrogen)
4. N2O5 (Dinitrogen pentoxide)