Question

A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine: 2HI (g) H2(g) + I2(g) When the system comes to equilibrium at 425 °C, PHI = 0.708 atm, and PH2 = PI2 The value of Kp at this temperature is ________.

Answer 1

Answer:

The value of the equilibrium constant: $K_{p} = 0.25$

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:

$K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}$

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

$\chi_{a} = \frac{number \, of \,moles \,of \,given \,gas (n_{a})}{total \,number \,of \,moles (n)}$

Mole fraction for HI: $\chi_{1} = \frac {n_{1}}{n} = \frac {2}{4} = 0.5$

Mole fraction for H₂: $\chi_{2} = \frac {n_{2}}{n} = \frac {1}{4} = 0.25$

Mole fraction for I₂: $\chi_{3} = \frac {n_{3}}{n} = \frac {1}{4} = 0.25$

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:

$K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}$

$K_{p} = \frac{0.354 atm\times 0.354 atm}{(0.708 atm)^{2}} = 0.25$

Therefore, the value of the equilibrium constant: $K_{p} = 0.25$