Question

Two gases,sulphur (IV) oxide and hydrogen sulphide,are admitted at opposite end of a 100cm tube,the tube is sealed and the gases allowed to diffuse towards each other.What is the relative rate of diffusion of the two gases? Calculate at what point free sulphur, the product of their chemical interaction,will first appear, explaining your reasoning.(H=1, S=32 O=16)

Answer 1

r = rate; mm = molar mass
(rH2S/rSO2) = sqrt(mmSO2/mmH2)
(rH2S/rSO2) = sqrt(64/34) = 1.37 so
rH2S = 1.37*rSO2
Draw a 100 cm tube.
|..................................||.........|
H2S end......100-x......S....x...|SO2 end
100-x = distance from H2S end
x = distance from SO2 end.
If x is distance from SO2 end where S forms, then
d=r*t = 1.37x = distance traveled by H2S
x + 1.37x = 100 or x = 42.2 cm from SO2 end
100-x = 100-42.2 = 57.8 cm from H2S end.