Question

The vapor pressure of water at 25 degree C is o.0313 atm. Calculate the values of K_p and K_c at 25 degree C for the equilibrium: H_2O (l) H_2O(g). Vinegar contains acetic acid, a weak acid that is partially dissociated in aqueous solution: CH_3COOH(aq) H^+(aq) + CH_3CO^-_2 (aq) a) Write the equilibrium constant expression for K_c. b) What is the value of K_c if the extent of dissociation of 1.0 M CH_3COOH is 0.42%? Determination of an Equilibrium Constant

Answer 1

3)

H2O (l)   <—> H2O(g)

Kp = p(H2O(g))

   = 0.0313

T = 25.0 oC =(25.0 + 273) K= 298.0 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = 1

Kp= Kc (RT)^Δn

0.0313 = Kc *(0.0821*298.0)^(1)

Kc = 1.279*10^-3

Answer:

Kp = 0.0313

Kc = 1.279*10^-3

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