Small amounts of NH3 and HCl(g) are released simultaneously at the opposite ends of a 2 metres long tube. At what point the formation of NH4Cl would start? (Either end of the tube can be used to tell the answer.)

Question

Serenity Allen · Accepted Answer

The formation of #"NH"_4"Cl"# would start at 0.8 m from the #"HCl"# end of the tube.

According to Graham's law of diffusion, a gas's rate of diffusion is inversely related to its molecular mass squared. The law can be drafted as #color(blue)(bar(ul(|color(white)(a/a)r_1/r_2 =sqrt(M_2/M_1)color(white)(a/a)|)))" "# where #r_1# and #r_2# are the rates of effusion of two gases and #M_1# and #M_2# are their molecular masses. If Gas 1 is #"NH"_3# and gas 2 is #"HCl"#, the formula becomes #r_1/r_2 = sqrt((36.46 color(red)(cancel(color(black)("u"))))/(17.03 color(red)(cancel(color(black)("u"))))) = sqrt(2.141) = 1.463# Thus, the #"NH"_3# travels 1.463 times at fast as #"HCl"#. A white ring of #"NH"_4"Cl"# will form when the two gases meet. Let #x =# the distance travelled by the #"HCl"#. Then #1.463x =# the distance travelled by the #"NH"_3#. #x + 1.463x = "2 m"# #2.463x = "2 m"# #x = "2 m"/2.463 = "0.8 m"# So, the white ring will be formed at a distance of 0.8 m from the #"HCl"# end. .

Serenity Allen · Answer

undefined The formation of NH4Cl would start where NH3 and HCl gases meet. This occurs at the midpoint of the tube, which is 1 meter from either end.