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Lerok [7]
3 years ago
15

Why experiment boiling point different from theoritical value

Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
6 0

Answer:

See Explanation

Explanation:

Boiling point, by definition, is that temperature at which the vapor pressure of the liquid equals the atmospheric pressure. When an experimental boiling point is determined in a practical laboratory environment one frequently finds that the observed boiling point is not the same as that published in the literature defined as the ‘standard boiling point’ value.  

Two primary factors are responsible for variations in experimentally measured boiling points. These two factors include variations in atmospheric pressure; i.e., not standard under experimental conditions and contamination/impurities dissolved in small amounts (e.g., dissolved salts) in the liquid sample.  

Standards for boiling points have been established for a long time. The physical boiling point of water was defined and published by The IUPAC System of weights and measures in 1908 as the temperature that water boils under an atmospheric pressure of 1.000 Bar (~ 1.00 Atmosphere) and given the value of 100⁰C (=373K). From Thermodynamic property table values for water a ‘Thermodynamic Boiling Point’ can also be defined. Calculation based upon the expression ΔH⁰ = T∙ΔS⁰ gives the boiling point to be 97.4⁰C (371K).  

It may be interesting to note that 100% pure, uncontaminated water will not boil, but evaporate from the liquid/atmosphere interface until all liquid phase transitions into gas phase (steam). The presence of micro contaminates in water are nucleation sites on which the liquid to gas transition takes place very rapidly and forms a bubble which, of course, rises to the surface of the liquid and discharges its water vapor into the atmosphere. Under conditions of boiling, theoretically for pure water samples, all molecules at 373K and 1.0Atm pressure have the ability to escape into the vapor phase. However, only the surface molecules in pure water samples will transition into the gas phase. Only when an impurity suspended in the water sample comes in contact with the water at 373K will the liquid explode into gas phase as ‘boiling bubbles.’

Calculation of the Thermodynamic Boiling Pt of Water from ΔH⁰ = T∙ΔS⁰.

The phase transition equation is represented by H₂O(l) <=> H₂O(g). Associated with the compounds of the reaction are Thermodynamic Property Data; typically found in the appendix section of most college chemistry text books.  

For H₂O(l) <=> H₂O(g) the following values for ΔH⁰ and S⁰ for water are as follows:

For H₂O(l) <=> H₂O(g)

ΔH⁰(l) = -285.8 Kj.mole  and ΔH⁰(g)  = -244.8 Kj.mole => ΔHRxn = [-244.8 – (-285.8)]Kj/mol = 44 Kj/mol

 S⁰(l) = 69.95 j/mol∙K    and S⁰(g) = 188.7 j/mol∙K    => ΔSRxn = (188.7 – 69.95)j/mol∙K = 118.8 j/mol∙K  

For the calculation, ΔS⁰ must be in j/mol∙K and must be dividing by 1,000 => ΔS⁰ = 0.1188 Kj/mol∙K.  

Applying to ΔH⁰ = T∙ΔS⁰.=> T = ΔH⁰/ΔS⁰ = 44 Kj/mol/0.1188Kj/mol∙K = 371K = (371 – 273)⁰C = 97.4⁰C.

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
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Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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