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Fynjy0 [20]
3 years ago
6

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

L21 5 1.1. Estimate the value of the parameters at 508C
Chemistry
1 answer:
Gala2k [10]3 years ago
8 0

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C   and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound              Liquid molar volume    (cm³/mol)

Ethanol (1)                    58.68

1 - propanol (2)            75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = \frac{V_2}{V_1} \  exp \ (\frac{-a_{12}/R}{T} )          ------------ equation (1)

where:

a_{12}/R = Wilson parameter = ???

V_2 = liquid molar volume of component 2 = 75.14 cm³/mol

V_1 = liquid molar volume of component 1  = 58.68 cm³/mol

T = temperature  = 25° C  = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

In (0.547) =  \ (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R=   0.60 * 298.15 \ K

-a_{12}/R=   - 178.89 \ K

a_{12}/R=    178.89 \ K

To calculate the temperature dependent parameters of the Wilson equation  ∧₂₁

∧₂₁ = \frac{V_1}{V_2} \  exp \ (\frac{-a_{12}/R}{T} )          ---------- equation (2)

1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )

1.1 = 0.7809 \ exp \  (\frac{-a_{12}/R}{298.15 \ K} )

\frac{1.1}{0.7809}=    exp \  (\frac{-a_{12}/R}{298.15 \ K} )

1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )

-a_{12}/R =     0.3426 * 298.15 \ K

-a_{12}/R =102.15 \ K

a_{12}/R = -102.15 \ K

From equation (1) ; let replace  178.98 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for a_{12}/R

V_2 = 75.14 cm³/mol

V_1 = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )

∧₂₁ =  0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

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