Determine the equilibrium concentration of H2O if the starting concentrations of both reactants are 0.003923 M, with [HOCl]o = 0.000 M.

Question

The value of K_c for the reaction between water vapor and dichlorine monoxide is 0.0900 at 25°C.

H2​O(g)+Cl2​O(g)2HOCl(g)

a. Determine the equilibrium concentration of H₂O if the starting concentrations of both reactants are 0.003923  M, with [HOCl]_o = 0.000 M.

b. Determine the equilibrium concentration of Cl₂O.

c. Determine the equilibrium concentration of HOCl.



Answer 1

H2​O(g) + Cl2​O(g)2HOCl(g)

0.003923…0.003923…….0…….I

-x……………….-x………..+2x….C

0.003923-x…0.003923….2x……E

Kc = 0.090 = [HOCl]² / [H2O][Cl2O]

0.090 = (2x)² / (0.003923-x)² and assume x is small, we simplify to

0.090 = (2x)² / (0.003923)²

0.090 = 4x² / 1.55x10^-5

4x² = 1.39x10^-6

x = 5.88x10^-4 M

Unfortunately, this value is NOT insignificant relative to 0.003923 and so it cannot be ignored and the calculation cannot be simplified in the above manner.

You will have to solve for x using the quadratic formula or by using some other simplifying assumption.

0.090 = (2x)² / (0.003923)²

0.30 = 2x / 0.003923

x = 0.000588 M

[H20] = 0.003923 - 0.000588 = 0.003335 M

[Cl2O] = 0.003335 M

[HClO] = 2 x 0.003335 = 0.00667 M