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sergeinik [125]
3 years ago
12

The normal boiling point of iodomethane, CH3I, is 42.43 8C, and its vapor pressure at 0.00 8C is 140. Torr. Calculate (a) the st

an- dard enthalpy of vaporization of iodomethane; (b) the standard entropy of vaporization of iodomethane; (c) the vapor pressure of iodomethane at 25.0 8C.
Chemistry
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

a=28600J; b=90.6 J/K; c=402 torr

Explanation:

(a) considering the data given

 Vapour pressure P1 =0  at Temperature T1 = 42.43˚C,

Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)

Using the Clausius-Clapeyron Equation

ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

In 760/140 = ΔH/8.314 J/mol/K  × (1/315.58K -- 1/273.15K)

ΔH vap= +28.6 kJ/mol or 28600J

(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.

Since ΔG at boiling point is zero,

ΔS =(ΔH°vap/Τb)

 ΔS = 28600 J/315.58 K

= 90.6 J/K

(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

ln P298 K/1 atm =  28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)

P298 K = 0.529 atm

                = 402 torr

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A cylinder was charged with 1.25 atm of oxygen gas, 6.73 atm of argon, and 0.895 atm of xenon. What is the mole fraction of each
katrin2010 [14]

Considering the Dalton's partial pressure, the mole fraction of each gas is:

  • x_{oxygen}= 0.14
  • x_{argon}= 0.76
  • x_{xenon}= 0.10

<h3>Dalton's partial pressure</h3>

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

P_{T} =P_{1} +P_{2} +...+P_{n}

where n is the amount of gases in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

P_{A} =x_{A} P_{T}

In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.

Mole fraction of each gas

In this case, you know that:

  • P_{oxygen }= 1.25 atm
  • P_{argon}= 6.73 atm
  • P_{xenon}= 0.895 atm
  • P_{T} =P_{oxygen} +P_{argon}+P_{xenon}= 1.25 atm + 6.73 atm + 0.895 atm= 8.875 atm

Then:

  • P_{oxygen} =x_{oxygen} P_{T}
  • P_{argon} =x_{argon} P_{T}
  • P_{xenon} =x_{xenon} P_{T}

Substituting the corresponding values:

  • 1.25 atm= x_{oxygen} 8.875 atm
  • 6.73 atm= x_{argon} 8.875 atm
  • 0.895 atm= x_{xenon} 8.875 atm

Solving:

  • x_{oxygen}= 1.25 atm÷ 8.875 atm= 0.14
  • x_{argon}= 6.73 atm÷ 8.875 atm= 0.76
  • x_{xenon}= 0.895 atm÷ 8.875 atm=0.10

In summary, the mole fraction of each gas is:

  • x_{oxygen}= 0.14
  • x_{argon}= 0.76
  • x_{xenon}= 0.10

Learn more about Dalton's partial pressure:

brainly.com/question/14239096

brainly.com/question/25181467

brainly.com/question/14119417

#SPJ1

3 0
2 years ago
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