Question

2. (a) Using data commonly found on the Internet about the properties of water, calculate the normal boiling point of water using this equation: lnK = - (deltaH/R)(1/T) + C

(b) What are some possible reasons that your answer may differ from the accepted value of

100 °C?

Answer 1

Following is the - complete; Answer -&- Explanation: for the given: Question: in....typed format...

$\Rightarrow$Answer:

1. part - (a):  Normal boiling point of water: T_bp = 377.6 K = 104.6 ^oC   
2. part - (b):  As explained below: under Step - 3, in the Explanation...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer.

• Given:

1. Given:  clausius-clapeyron equation:     ln K = - ( $\Delta$ H_vap / R  ) + C
2. We know: vapor pressure of pure water : P₁ = 23.8 mmHg = 23.8 Torr, at T₁  = 298 K ( Kelvin )
3. We know: enthalpy of vaporization of pure water:   $\Delta$H_vap = 40.7 kJ /mol ( approx. )
4. We know: at normal boiling point of Water: vapor pressure of water: P₂ = 760.0 Torr  ( i.e. 1.0 atm )
5. Let's assume: normal boiling point of water: T₂ = ( unknown )

• ​​​​​​​Step - 1:

Using the Calusius-Clapeyron Equation: we can derive the following Equation:

$\Rightarrow$ln ( P₁ / P₂ ) = -   $\Delta$H_vap / R x (  1/ T₁ - 1/T₂  ) -------------------------------Equation - 1

$\Rightarrow$ Where:

1. P₁ =   23.8 Torr,
2. P₂ = 760.0 Torr ( i.e. 1.0 atm. )
3.   $\Delta$H_vap =   40.7 kJ /mol ( approx. ) = 40,700 J / mol
4. Ideal Gas constant: R = 62.364 L. Torr/ ( mol. K ) = 8.314 J / (mol. K )
5. T₁ = 298.0 K  ( Kelvin )
6. Normal boiling point of Water:   T₂ = ( unknown )

• ​​​​​​​Step - 2:

​​​​​​​Now, plugging in values: in : Equation - 1, we would get, the following:

$\Rightarrow$(  1/ T₁ - 1/T₂  ) = 0.0007074  

$\Rightarrow$   (  1/ 298   - 1/T₂  ) = 0.0007074  

$\Rightarrow$   1/T₂ =   0.002648

$\Rightarrow$ T₂ = 377.6 K = ( 377 - 273 ) =  104.6 ^oC

$\Rightarrow$ Therefore: Calculated normal boiling point of water : T_bp = 104.6 ^oC   

• Step - 3:

​​​​​​​$\Rightarrow$ We know: accepted value of boiling point of water = 100.0 ^oC   

$\Rightarrow$Here, the calculated value of the boiling point of water, is slightly more than the accepted value, because: the values used above: depends on the value of enthalpy of vaporization, and the value of vapor pressure at 298 K , which may have some errors. Due to the above said errors, the calculated value of the normal boiling point of Water, may vary from : the accepted value: of normal boiling point of Water...

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