Respuesta :
Answer:
The value of the equilibrium constant: [tex]K_{p} = 0.25[/tex]
Explanation:
Given reaction: 2HI (g) ⇌ H₂(g) + I₂(g)
Number of moles of- HI: n₁ = 2 mole; H₂: n₂ = 1 mole; I₂: n₃ = 1 mole
Total number of moles: n = n₁ + n₂ + n₃ = 2 + 1 + 1 = 4 moles
The equilibrium constant for the given reaction is given as:
[tex]K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}[/tex]
Given: Temperature: T = 425 °C = 425 + 273 = 698 K
The partial pressure: pHI = 0.708 atm,
and, pH₂ = pI₂
∵ partial pressure of a given gas: pₐ = Χₐ . P
Here, P is the total pressure
Χₐ is the mole fraction of given gas and is given by the equation
[tex]\chi_{a} = \frac{number \, of \,moles \,of \,given \,gas (n_{a})}{total \,number \,of \,moles (n)} [/tex]
Mole fraction for HI: [tex]\chi_{1} = \frac {n_{1}}{n} = \frac {2}{4} = 0.5[/tex]
Mole fraction for H₂: [tex]\chi_{2} = \frac {n_{2}}{n} = \frac {1}{4} = 0.25[/tex]
Mole fraction for I₂: [tex]\chi_{3} = \frac {n_{3}}{n} = \frac {1}{4} = 0.25[/tex]
Thus, Χ₂ = Χ₃ = 0.25
The partial pressure of HI is given by;
pHI = Χ₁ P
0.708 atm = 0.5 × P
⇒ P = 1.416 atm
As the partial pressures: pH₂ = pI₂
∴ pH₂ = pI₂ = Χ₂ P = Χ₃ P = 0.25 × 1.416 atm = 0.354 atm
Therefore, the value of Kp can be calculated as:
[tex]K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}[/tex]
[tex]K_{p} = \frac{0.354 atm\times 0.354 atm}{(0.708 atm)^{2}} = 0.25[/tex]
Therefore, the value of the equilibrium constant: [tex]K_{p} = 0.25[/tex]
