Hydrogen Sulphide (H 2 S) occurs in the atmosphere naturally at a concentration
ID: 481439 • Letter: H
Question
Hydrogen Sulphide (H2S) occurs in the atmosphere naturally at a concentration of 0.05 ppb. Determine whether or not this concentration represents equilibrium with respect to H2O, SO2 and O2 (all in the gaseous phase).
A) Convert the concentration of H2S to its partial pressure, assuming an atmospheric pressure of 1atm.
B) Write the appropriate chemical reaction that governs the concentration of H2S in the atmosphere if equilibrium were maintained between SO2, H2O, H2S, and O2. Important: Place O2 on the right hand side of the raction. The equilibrium constant for this reaction is 10-86.74.
C) Calculate the partial pressure of H2S as if the raction is controlled by chemical equilibrium.
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Explanation / Answer
A) Convert the concentration of H2S to its partial pressure, assuming an atmospheric pressure of 1atm.
The partial pressure of H2S = Total pressure X mole fraction
ppb = parts per billion = 1 / 10^9 parts of air
There are 0.05 molecules of H2S per 10^9 molecules of air
So mole fraction = 0.05 X 10^-9
Hence partial pressure = 0.05 X 10^-9 atm
B) Write the appropriate chemical reaction that governs the concentration of H2S in the atmosphere if equilibrium were maintained between SO2, H2O, H2S, and O2. Important: Place O2 on the right hand side of the raction. The equilibrium constant for this reaction is 10-86.74.
the equation will be:
SO2 + H2O --> H2S + 1.5O2
C) Calculate the partial pressure of H2S as if the raction is controlled by chemical equilibrium.
Keq = pH2S X (pO2)1.5 / pSO2 X pH2O
Let the initial pressure of each SO2 and H2O = 1
so change = -x
Keq = 10-86.74 = (x)(x)^1.5 / (1-x)(1-x)
x <<1
so can be ignored in denominator
x = (10^-34.7)
x = 1.99 X 10^-35
which is much less than the given 0.05 X 10^-9 atm
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