If a gaseous mixture is made by combining 3.78 g ar and 3.27 g kr in an evacuated 2.50 l container at 25.0 ∘ c, what are the partial pressures of each gas, p ar and p kr, and what is the total pressure, p total, exerted by the gaseous mixture?

The partial pressure of Ar = 0.925 atm

The partial pressure of Kr = 0.381 atm

The total pressure = 1.306 atm

Explanation:

Step 1: Data given

Mass of Ar = 3.78 grams

Mass of Kr = 3.27 grams

Volume = 2.50 L

Temperature = 25.0 °C

Molar mass of Ar = 39.95 g/mol

Molar mass of Kr = 83.80 g/mol

Step 2: Calculate moles of Ar

Moles Ar = mass Ar / molar mass Ar

Moles Ar = 3.78 grams / 39.95 g/mol

Moles Ar = 0.0946 moles

Step 3: Calculate moles of Kr

Moles Kr = 3.27 grams / 83.80 g/mol

Moles Kr = 0.039 moles

Step 4: Calculate partial press of each gas

p*V = n*R*T

⇒ p =the partial pressure of the gas

⇒ V = the volume in the container = 2.50 L

⇒ n = the number of moles of the gas

⇒ R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25°C = 298 K

The partial pressure of Ar = (n*R*T)/V

p(Ar) = (0.0946*0.08206*298)/ 2.5

p(Ar)= 0.925 atm

p(Kr) = (0.039*0.08206*298)/ 2.5

p(Kr) = 0.381 atm

Step 5: Calculate the total pressure

The total pressure = p(Ar) +p(Kr)

Total pressure = 0.925 atm + 0.381 atm = 1.306 atm

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

i dont know

explanation: