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

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.