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)
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.
To determine the relative rate of diffusion of the two gases, we can use Graham's law of diffusion. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's denote the rate of diffusion of sulphur (IV) oxide as R1 and the rate of diffusion of hydrogen sulphide as R2. We can set up the following equation:
(R1 / R2) = √(M2 / M1)
Where M1 and M2 represent the molar masses of the respective gases.
The molar mass of sulphur (IV) oxide (SO2) is calculated as follows:
M1 = (32.07 g/mol) + (16.00 g/mol) + (16.00 g/mol) = 64.07 g/mol
The molar mass of hydrogen sulphide (H2S) is calculated as follows:
M2 = (1.01 g/mol) + (1.01 g/mol) + (32.07 g/mol) = 34.09 g/mol
Now we can substitute these values into the equation:
(R1 / R2) = √(34.09 g/mol / 64.07 g/mol)
(R1 / R2) = √(0.531)
(R1 / R2) ≈ 0.73
So, the relative rate of diffusion of the two gases is approximately 0.73.
To determine at what point free sulphur, the product of their chemical interaction, will first appear, we need to consider the stoichiometry of the reaction between sulphur (IV) oxide and hydrogen sulphide. The balanced equation for this reaction is as follows:
SO2 + 2H2S → 2H2O + 3S
In this reaction, each molecule of sulphur (IV) oxide reacts with two molecules of hydrogen sulphide to form two molecules of water and three molecules of sulphur.
Given that they are sealed in a 100 cm tube and diffusing towards each other, the gases will mix equally at the midpoint of the tube, which is at 50 cm.
At this point, the gases will react and produce free sulphur. However, based on the stoichiometry of the reaction, it is clear that the amount of sulphur formed will depend on the ratio of the reactant gases in the mixture. Without knowing the initial amounts of sulphur (IV) oxide and hydrogen sulphide, we cannot determine the exact point at which free sulphur will first appear.
Therefore, the correct answer is that free sulphur, the product of their chemical interaction, will first appear when the gases have mixed at the midpoint of the tube (50 cm), but the amount of sulphur formed will depend on the initial ratio of the reactants.