Question

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)

Answer 1

Kp for this equilibrium ≅ 570

Step-by-step 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