What is the approximate value of the equilibrium constant K P for the change C 2 H 5 OC 2 H 5 ( l ) ⇌ C 2 H 5 OC 2 H 5 ( g ) at 25 ° C. (The equilibrium vapor pressure for this substance is 570 torr at 25 ° C.) 760 torr = 1 atm R = 0.0821 L * atm / (mol * K)

Kp for this equilibrium ≅ 570

Explanation:

This is a situation where we have an heterogeneous mixture.

We get gas from liquid state:

C₂H₅OC₂H₅ (l) ⇌ C₂H₅OC₂H₅ (g)

In order to determine Kc → [C₂H₅OC₂H₅ (g)], in equilibrium

Liquid or solid do not count for Kc, neither Kp

Kp →  Partial pressure of C₂H₅OC₂H₅ (g)

Kc or Kp at equilibrium do not depend on the amount of liquid, there is; although it cannot be null since otherwise there would be no reaction.

Kp for this equilibrium ≅ 570

(Value for partial pressure in the only gas).

In order to determine Kc we can replace data at this formula

Kp = Kc (R . T)^ ⁽ⁿ ⁻ ⁿ°⁾

Where n - n° is the difference between the moles of produced gas. We only have 1 mol.

 Kp = Kc (298K . 0.082)¹

570 Torr / (298K . 0.082)  → Kc

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.