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At 90° C , the following equilibrium is established :
If 0.20 mole of hydrogen and 1.0 mole of sulphur are heated to 90°C in a 1.0 dm3 flask, what will be the partial pressure of H2S gas at equilibrium?
If 0.20 mole of hydrogen and 1.0 mole of sulphur are heated to 90°C in a 1.0 dm3 flask, what will be the partial pressure of H2S gas at equilibrium?
- a)0.36 atm
- b)0.38 atm
- c)0.28 atm
- d)0.26 atm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
At 90° C , the following equilibrium is established :If 0.20 mole...
Initial moles of H₂ = 0.2
Initial moles of S = 1
Kp = 6.8 * 10⁻²
Given equation:
H2(g) + S(s) ⇋ H2S(g)
Initial moles: 0.2 1
At equilibrium: (0.2-α) (1-α) α
Here, in the above equation we can see that hydrogen is the limiting reagent.
∴ Kp = α/(0.2 – α)
⇒ 6.8 * 10⁻² = α/(0.2 – α)
⇒ 1.36*10⁻² – (6.8*10⁻²)α = α
⇒ α + 0.068α = 1.36*10⁻²
⇒ α = 1.36*10⁻² / 1.068 = 1.273 * 10⁻² ← moles of H₂S
So, at equilibrium moles of H₂ = 0.2 – α = 0.2 – 1.273 * 10⁻² = 0.1873
Now, using the Ideal Gas equation,
PV = nRT ….. (i)
Where P = total pressure of the vessel
n = total no. of moles = (0.2-α) + (1-α) + α = 1.2 – α = 1.2 – 1.273*10⁻² = 1.1873
V = volume of vessel = 1 litre
R = Ideal gas constant = 0.082 L atm K⁻¹mol⁻¹
T = total temperature = 90℃ = 90+273 = 363 K
Substituting all the values in eq. (i), we get
P * 1 = 1.1873 * 0.082 * 363
⇒ P = 35.34 atm
Thus,
The partial pressure of H₂S at equilibrium
= (mole fraction of H₂S) * (total pressure)
= [1.273*10⁻² / 1.1873] * 35.34
= 0.3789 atm
≈ 0.38 atm
Initial moles of S = 1
Kp = 6.8 * 10⁻²
Given equation:
H2(g) + S(s) ⇋ H2S(g)
Initial moles: 0.2 1
At equilibrium: (0.2-α) (1-α) α
Here, in the above equation we can see that hydrogen is the limiting reagent.
∴ Kp = α/(0.2 – α)
⇒ 6.8 * 10⁻² = α/(0.2 – α)
⇒ 1.36*10⁻² – (6.8*10⁻²)α = α
⇒ α + 0.068α = 1.36*10⁻²
⇒ α = 1.36*10⁻² / 1.068 = 1.273 * 10⁻² ← moles of H₂S
So, at equilibrium moles of H₂ = 0.2 – α = 0.2 – 1.273 * 10⁻² = 0.1873
Now, using the Ideal Gas equation,
PV = nRT ….. (i)
Where P = total pressure of the vessel
n = total no. of moles = (0.2-α) + (1-α) + α = 1.2 – α = 1.2 – 1.273*10⁻² = 1.1873
V = volume of vessel = 1 litre
R = Ideal gas constant = 0.082 L atm K⁻¹mol⁻¹
T = total temperature = 90℃ = 90+273 = 363 K
Substituting all the values in eq. (i), we get
P * 1 = 1.1873 * 0.082 * 363
⇒ P = 35.34 atm
Thus,
The partial pressure of H₂S at equilibrium
= (mole fraction of H₂S) * (total pressure)
= [1.273*10⁻² / 1.1873] * 35.34
= 0.3789 atm
≈ 0.38 atm
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At 90° C , the following equilibrium is established :If 0.20 mole...
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At 90° C , the following equilibrium is established :If 0.20 mole of hydrogen and 1.0 mole of sulphur are heated to 90°C in a 1.0 dm3 flask, what will be the partial pressure of H2S gas at equilibrium?a)0.36atmb)0.38 atmc)0.28 atmd)0.26atmCorrect answer is option 'B'. Can you explain this answer?
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