Question

At 90°C following equilibrium is established:
H2(g) S(s) H2S(g) Kp=6.8*10^-2
If 0.2mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel what will be the partial pressure of H2S at equilibrium? ​

Answer 1

Answer: 0.38 atm

Explanation:

Step 1:

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

Step 2:

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

Step 3:

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