Question

The vapor pressure of water at 25 •C is 23.8 mmHg. Write K for the vaporization of water in the unit of atm. What is Kc for the vaporization process?
answer is: Kc= 1.28 x 10^-3

please tell me how to solve this problem

Answer 1

The degree of dissociation gives the amount of the substance that dissociated into the products. The value of Kc for water is 1.28 x 10⁻³.

What is equilibrium constant, Kc?

The equilibrium constant, Kc is the equilibrium concentration of the products over the reactants that are raised by the powers of their stoichiometry coefficient.

The reaction is shown as:

H₂​O(l) ⇋ H₂​O(g)

Kp = pressure of water = 0.03131579 atm

$\rm Kp = Kc(RT)^{\Delta n}$

Here Δ n = 1

Solving further,

Kp = Kc(RT)

0.03131579 = Kc(0.0821 × 298.15)

Kc =1.28 x 10⁻³

Therefore, Kc is 1.28 x 10⁻³.

Learn more about the equilibrium constant here:

brainly.com/question/14988161

#SPJ1