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a certain amount of nobr(g) is sealed in a flask, and the temperature is raised to 350 k. the following equilibrium is established: nobr(g) uv no(g) 1 12 br2(g) the total pressure in the flask when equilibrium is reached at this temperature is 0.675 atm, and the vapor density is

Respuesta :

The partial pressures of each species of the equilibrium are

  • NOBr = 0.39 atm
  • NO = 0.19 atm

Vapour density = 2.219 g / L

Temperature = 350 K

Total pressure = 0.675 atm

Since volume of flask is constant and the mass is conserved,

Initial mass of NOBr = 2.219 g

1 mol NOBr = 109.92 g

Number of moles of NOBr = 2.219 / 109.92 = 0.02019 mol

Total number of moles at equilibrium is,

[tex]n_{total}[/tex] = ( 0.02019 - x ) + x + 0.5 x

[tex]n_{total}[/tex] = 0.02019 + 0.5 x

Acording to ideal gas equation.

[tex]n_{total}[/tex]  = P V / R T

[tex]n_{total}[/tex] = ( 0.675 * 1 ) / ( 0.08206 * 350 )

[tex]n_{total}[/tex] = 0.00662

Partial pressure of NOBr,

P = ( 0.02019 - x ) * 0.675 / ( 0.02019 + 0.5 x )

P = 0.39 atm

Partial pressure of NO,

P = x * 0.675 / ( 0.02019 + 0.5 x )

P = 0.19 atm

Partial pressure of Br₂,

P = 0.5 x * 0.675 / ( 0.02019 + 0.5 x )

P = 0.095 atm

Therefore, the partial pressures of

  • NOBr = 0.39 atm
  • NO = 0.19 atm
  • Br₂ = 0.095 atm

The given question is incomplete. The complete question is:

A certain amount of NOBr(g) is sealed in a flask, and the temperature is raised to 350 K. The following equilibrium is established:

NOBr(g) ↔ NO(g) + ½ Br2(g)

NOBr(g)↔NO(g)+½Br2(g)

The total pressure in the flask when equilibrium is reached at this temperature is 0.675 atm, and the vapor density is 2.219 g L^-1. Calculate the partial pressure of each species.

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