Question

A gas mixture in a 1.00L reaction vessel at 100ºC contains 4.5 mol of Br(g) and 33.1 mol F2(g). Compte the mole fraction of each gas in this mixture.

When this mixture is heated to 150ºC, Br2(g) and F2(g) react to form BrF5(g) according to the following balanced chemical reaction: Br2(g)+ 5F2(g)---------- 2 BrF5(g). At a certain point, 2.2 mol of BrF5(g) is present in this reaction vessel. Determine the mole fractions of Br2(g), F2(g) and BrF5(g) at this point in the reaction.

Assuming the reaction goes to complection (one or both of the reactants is limiting),

What will be the total pressure inside this vessel? What are the partial pressures of all the components present?

Calculate the root mean squared velocity of the components present after the reaction is complete?

Answer 1

A gas mixture in a 1.00L reaction vessel at 100ºC contains 4.5 mol of Br(g) and 33.1 mol F2(g). Compte the mole fraction of each gas in this mixture.

Mole fraction of Br₂(g) , X = number of moles of Br2 / Total number of moles

                                         = 4.5 mol / ( 4.5+ 33.1 ) mol

                                         = 0.12

Mole fraction of F₂(g) , X' = number of moles of F2 / total number of moles

                                         = 33.1 mol / ( 33.1+4.5) mol

                                         = 0.88

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after heating at 150 ^oC ,

                                 Br₂(g) + 5F₂(g) 2 BrF₅(g)

initial moles            4.5           33.1              0

Change                    -a             -5a             +2a

at time t                  4.5-a    33.1-5a         2.2

Given 2a = 2.2

            a = 1.1 mol

So the number of moles of Br₂ is = 4.5 - a = 4.5 - 1.1 = 3.4 mol

the number of moles of F₂ is = 33.1 - 5a = 33.1 - ( 5x1.1) = 27.6 mol

Mole fraction of Br₂(g) , X = number of moles of Br₂ / Total number of moles

                                         = 3.4 mol / ( 3.4+27.6+2.2 ) mol

                                         = 0.102

Mole fraction of F₂(g) , X' = number of moles of F₂ / total number of moles

                                         = 27.6 mol / ( 3.4+27.6+2.2) mol

                                         = 0.83

Mole fraction of BrF₅ is X'' = number of moles of BrF₅ / total number of moles

                                           = 2.2 mol / ( 3.4+27.6+2.2 ) mol

                                           = 0.07

Partial pressure of Br₂ is , p = nRT / V

Where n = number of moles of Br₂ = 3.4 mol

           R = gas constant = 0.0821 L atm / mol/K

           T = temperature = 150 + 273 = 423 K

           V = volume of the reaction vessel = 1.0 L

Plug the values we get p _Br2 = (3.4 x 0.0821 x423 ) / 1.0 = 118.1 atm

Simillarly p _F2 = ( 27.6 x 0.0821 x 423 ) / 1.0 = 958.5 atm

              p _BrF5 = ( 2.2 x 0.0821 x 423 ) / 1.0 = 76.4 atm

So the total pressure , P = p _Br2 + p _F2 + p _BrF5

                                       = 118.1 + 958.5 + 76.4

                                       = 1153 atm

rms velocity of Br₂ , u _Br2 =

Where R = gas constant = 8.314 x10 ⁷ ergs/mol-K

           T = temperature = 150+273 = 423 K

           M = molar mass of Br2 = 2 x 80 = 160 g/mol

Plug the values we get u _Br2 =

                                     u _Br2 =

Similarly rms velocity of F₂ is u _F2 =      M = molar mass of F₂ = 2 x 19 = 38 g/mol

                                     u _F2 =

rms velocity of BrF₅ is u_BrF5 =      M = molar mass of BrF₅ = 80 +(5 x 19) = 175 g/mol

                                     u_BrF5 =