Respuesta :
Answer:
76.03
Explanation:
We have the equation
T= Delta H/ Delta S
For this we need to find the Delta H of vaporization for the boiling point. We can obtain this by taking delta H(products) - delta H(reactants). In this case we will take (-235.1 kj/mol) - (-277.1 kj/mol) = 42.6 kj/mol.
We will then need the delta S of vaporization, which can be obtained by taking delta S(products) - delta S(reactants). Here we will take (282.6 j/k*mol) - (160.6 j/k*mol) = 122 j/k*mol. We still need to convert this to kj/mol so we divide 122 by 1000 to get 0.122 kj/mol as delta S of vaporization.
Then we plug in these values into the equation mentioned above as follows.
T= 42.6/0.122
T= 349.18 K
All we need to do now is to convert this into degrees Celsius by taking 349.18 - 273.15 to end up with an answer of 76.03 degrees Celsius.
The boiling pressure of ethanol is [tex]76.03^oC[/tex].
According to Gibbs's free energy:
[tex]\triangle G=\triangle H-T\triangle S[/tex]...............(1)
The system is at equilibrium thus, [tex]\triangle G=0[/tex]
[tex]C_2H_5OH(l)[/tex] ⇄ [tex]C_2H_5OH(g)[/tex]
Enthalpy of vaporization:
It is the amount of energy that must be added to a liquid substance to transform a quantity of that substance into a gas. The enthalpy of vaporization is a function of the pressure at which that transformation takes place.
[tex]\triangle H_{vap}=\triangle H_{products}-\triangle H_{reactants}\\\\\triangle H_{vap}=(-235.1 kJ/mol) - (-277.1 kJ/mol) \\\\\triangle H_{vap}= 42.6 kJ/mol[/tex]
Calculation for [tex]\triangle S_{vap}[/tex]:
[tex]\triangle S_{vap}=\triangle S_{products} - \triangle S_{reactants}\\\\\triangle S_{vap}=(282.6 J/K*mol) - (160.6 J/K*mol) \\\\\triangle S_{vap}= 122 J/K*mol =0.122kJ/mol[/tex]
From equation 1:
[tex]\triangle H_{vap}=T\triangle S_{vap}\\\\T=\frac{\triangle H_{vap}}{\triangle S_{vap}}[/tex]
Now substituting the values in the above formula we will get:
[tex]T=\frac{\triangle H_{vap}}{\triangle S_{vap}}\\\\T=\frac{42.6}{0.122}\\\\ T= 349.18K[/tex]
Conversion of [tex]^oC \text{ into } K[/tex]:
[tex]T=349.18-273.15\\\\T=76.03^oC[/tex]
Thus, boiling pressure of ethanol is [tex]76.03^oC[/tex].
Find more information about Enthalpy of vaporization here:
brainly.com/question/5602087
