Question

General Chemistry !! Workshop 9: Thermodynamics/Free Energy Applications 1. Using thermodynamic data, estimate the vapor pressure of water at 298 K Hint: Start by writing the equilibrium constant expression for this process. (2 pts) 2. A 7.0 g piece of potassium metal is tossed in 500 ml. of water held in an insulated container. The potassium reacts with some of the water, producing potassium hydroxide and hydrogen gas. (2 pt) a) Given the following information, determine the temperature of the resulting solution of potassium hydroxide? The starting temperature of both the potassium metal and the water was 20.0°C. The heat capacity of water is 4.18 J/g-deg. You may assume that resulting solution of potassium hydroxide has the same heat capacity as that of pure water. The molecular weight of Kis 39.1 g/mol Chemical equation:

Answer 1

Following is the - complete Answer -&- Explanation: for the first question (i.e. Question - 1 ), of the given: Question Set.....in...typed format....

$\Rightarrow$Answer:

Vapor pressure of water, at 298.0 K ( Kelvin ):  P_vap =  25.0 torr ( approx. )

$\Rightarrow$Explanation:

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

• Given:

1. We know: vapor pressure of water: at 37.0 ^oC OR, 310.0 K ( kelvin) :  P₁ = 47.1 torr
2. Therefore: T₁ = 310.0 K ( Kelvin )
3. Desired temperature: T₂ = 298.0 K ( Kelvin )
4. Vapor pressure at T₂  ; is P₂ =  Unknown ...
5. We know: Enthalpy of vaporization: of water:  $\Delta$H_vap = +40.8 kJ/mol
6. Ideal Gas constant: R = 0.008314 kJ / (mol.K )

• Step - 1:

We know the following formula: i.e. based on Clausius Clapeyron Equation:

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

Plugging in values in Equation - 1: we would get:the following:

$\Rightarrow$ ln ( 47.1 / P₂ ) = ( 40.8 kJ/mol / 0.008314 kJ/ mol. K ) x  [ ( 1/ 298.0 ) - ( 1/310.0 ) ]

$\Rightarrow$ln ( 47.1 / P₂ ) = 0.63746  

$\Rightarrow$( 47.1 / P₂ ) = exp ( 0.63746 ) = 1.89  ( approx. )

$\Rightarrow$P₂ = ( 47.1 / 1.89 ) = 24.92 torr...  $\approx$25.0 torr

$\Rightarrow$   P₂ = 25.0 torr ( approx. )

• Answer:

​​​​​​​$\Rightarrow$ Vapor pressure of water: at 298.0 K :  is P_vap=   P₂= 25.0 torr ( approx. )

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