Equilibrium Problems Kc = .090: Calculating Concentration

[marioland]

Kc = .090

a). Calculate the concn. of all species at equilibrium if 1.00g of water and 4.78g of ClO are mixed in a 1.0L flask.

b). Calculate the concn. of all species at equilibrium if 1.0 mol of pure HOCl is placed in a 2.0L flask. Hint: consider what is now the reactant.





H2O + Cl2O --> 2HOCl

I| 18mol 88 mol 0
C| -x -x +x
E| 18-x 88-x x

Kc= .090 = [HOCl]2 / [H2O].[Cl2O] = x2/(18-x)(88-x)


Solve for x and find all species.Is this the correct way ? Am I confusing you... *hmmm*

 

[lippyka]

a). 0.09 = x2/(18-x)(88-x)Solve for x to find the equilibrium concentrations of all species: x = 0.748 mol/L [H2O] = 17.252 mol/L [Cl2O] = 87.252 mol/L [HOCl] = 0.748 mol/L b). 1 mol/L = [HOCl] = x Kc = .090 = (x)2 / (1)(1+x) Solve for x to find the equilibrium concentrations of all species: x = 0.894 mol/L [H2O] = 0.106 mol/L [Cl2O] = 0.894 mol/L [HOCl] = 0.894 mol/L

 

[M98Ranger]

a) To calculate the concentration at equilibrium, we first need to determine the initial moles of each species present. We know that we have 1.00g of water, which has a molar mass of 18g/mol, so we have 1.00/18 = 0.0556 mol of water. Similarly, 4.78g of ClO has a molar mass of 52.46g/mol, so we have 4.78/52.46 = 0.091 mol of ClO.

Using the equilibrium expression, Kc = [HOCl]2 / [H2O].[Cl2O], we can set up an ICE table:

I| 0.0556 mol 0.091 mol 0
C| -x -x +x
E| 0.0556-x 0.091-x x

Substituting these values into the equilibrium expression, we get:

0.090 = (x)2 / (0.0556-x)(0.091-x)

Solving for x, we get x = 0.033 mol.

Therefore, at equilibrium, the concentrations of each species will be:

[HOCl] = 0.033 mol/L
[H2O] = 0.0226 mol/L
[Cl2O] = 0.058 mol/L

b) In this case, we are starting with 1.0 mol of pure HOCl in a 2.0L flask. This means that initially, the concentration of HOCl is 1.0 mol/L. Using the same equilibrium expression and ICE table as above, we get:

0.090 = (x)2 / (1.0-x)(1.0-x)

Solving for x, we get x = 0.083 mol.

Therefore, at equilibrium, the concentrations of each species will be:

[HOCl] = 0.083 mol/L
[H2O] = 1.0-0.083 = 0.917 mol/L
[Cl2O] = 1.0-0.083 = 0.917 mol/L

It is important to note that in this case, the concentration of HOCl does not change significantly since it is the only reactant and is in excess. The concentration of the other species will be much lower compared to