Areaction 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

The value of the equilibrium constant:

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:

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

Mole fraction for HI:

Mole fraction for H₂:

Mole fraction for I₂:

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:

Therefore, the value of the equilibrium constant:

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3