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ivann1987 [24]
2 years ago
10

Using Raoult’s law, estimate the boiling pressure and mole fractions in the vapor phase that is in equilibrium with a liquid hav

ing 0.1238 mole fraction of ethanol, in a mixture ethanol/water at 85.3o C.
Chemistry
1 answer:
fenix001 [56]2 years ago
3 0

Answer:

a) boiling pressure:

P = 0.759 atm

b) mole fractions in the vapor phase:

Xethanol(v) = 0.2192

Xwater(v) = 0.7808

Explanation:

Raoult's law:

  • Pa = (P*a)×(Xa,(l))
  • Xa(l) + Xb(l) = 1

∴ a: water

∴ b: ethanol

∴ Xb(l) = 0.1238

⇒ Xa(l) = 1 - 0.1238 = 0.8762

  • Pt = Pa + Pb

vapor pressure of the pure components at T = 85.3°C:

∴ P*a(85.3°C) ≅ 400 torr (0.53 atm).....from literature

∴ P*b(85.3°C) ≅ 800 torr (1.053 atm).....from literature

total pressure of the mix:

⇒ Pt = ((0.53 atm)(0.8762)) + ((1.053 atm)(0.1238))

⇒ Pt = 0.5947 atm

boiling point:

  • Xb(l) = Xa(l) = 0.5
  • P = (P*b - Pt)(0.5) + P*a

⇒ P = ((1.053 atm - 0.5947 atm)(0.5)) + 0.53 atm

⇒ P = 0.759 atm

assuming that the gas system is ideal:

  • (Xb(v))(Pt) = (P*b)(Xb(l))

fraction in the vapor fase, ethanol:

⇒ Xb(v) = (P*b)(Xb(l)) / Pt

⇒ Xb(v) = ((1.053 atm)(0.1238)) / (0.5947 atm)

⇒ Xb(v) = 0.2192

∴ Xa(v) = 1 - Xb(v)

⇒ Xa(v) = 1 - 0.2192 = 0.7808

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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