Question: The following equilibrium was studied by analyzing the equilibrium mixture for the amount of HCL produced. LaCl3(s)+ H2O(g) goes to LaOCl(s)+2HCl(g). A vessel whose volume was 1.25L was filled with 0.0125 mol of lanthanum (III)chloride and 0.0250 mol H2O. After the mixture came to equilibrium in the closed vessel at 619 degree C the gaseous mixture was removed and dissolved in more water sufficent silver (I)ion was added to precipitate the chloride ion completely as silver chloride. If 3.59g AgCl was obtained, what is the value of Kc at 619 degree C?
I worked out the problem could you please show me how you would work it so i can check my answer
I wouldn't do it that way. First, the expression for Kc does not involve AgCl, LaCl3, or LaOCl because all three are solids and a concn of a solid is 1 by definition. (You can't get the concn to increase by adding another speck of a solid to a solid).
LaCl3(s) + H2O(g) ==> LaOCl(s) + 2HCl(g)
I would convert grams AgCl to moles as you did, which is 3.59/143 = approximately 0.025 which is moles HCl. Then moles H2O will be 1/2 that or 0.0125. Since you started with 0.0250, there will be 0.250-0.0125 = 0.0125 moles H2O at equilibrium.
Then (H2O) = 0.0125 moles/1.25L = ?? M
(HCl) = 0.0250/1.25 = ?? M.
Kc = (HCl)^2/(H2O) = ??
Note that the solids don't show up in Keq. You need to go through and change the numbers a little since I used approximations for the molar mass AgCl. I obtained 0.04 for Kc. Check my work.
We would much prefer you to show your work and let us check your answer. We can help better that way.
I first changed the 3.59g of AgCl into moles and than divided by liters given to get .0200M of Agcl. I also changed the mol of H20 and LaCl3 to Molarity as well. I got .01M for LaCl3 and.02 for H20. Than I plugged into equation Kc=(.0200)/(.010)(.02). To give me the Kc value as 200
thanks! Once you mentioned about solids i totally remembered they don't play apart when your solving somthing.
That's why we like to see your work. Seeing your work made me put those remarks in about the solids.
To solve this problem, we need to use the given information and the equation for the equilibrium reaction to determine the value of the equilibrium constant, Kc.
First, let's write the balanced equation for the reaction:
LaCl3(s) + H2O(g) ⇌ LaOCl(s) + 2HCl(g)
The equation shows that for every 1 mole of lanthanum (III) chloride (LaCl3) and water (H2O) that react, we get 1 mole of lanthanum oxychloride (LaOCl) and 2 moles of hydrogen chloride (HCl).
Next, let's calculate the initial and equilibrium concentrations of HCl. We are given the initial moles of LaCl3 and H2O, and we can use stoichiometry to determine the moles of HCl produced:
Initial moles of H2O = 0.0250 mol
Initial moles of HCl = 0 mol (since no HCl is initially present)
Since the reaction stoichiometry shows that 1 mole of H2O produces 2 moles of HCl, the equilibrium moles of HCl will be twice the moles of H2O reacted:
Equilibrium moles of HCl = 2 * 0.0250 mol = 0.0500 mol
Now, let's calculate the equilibrium concentration of HCl:
Equilibrium concentration of HCl = (moles of HCl) / (volume of vessel)
Equilibrium concentration of HCl = 0.0500 mol / 1.25 L = 0.0400 M
To find the value of Kc, we need to compare the equilibrium concentrations of the products (LaOCl and HCl) to the initial concentrations of the reactants (LaCl3 and H2O). Since the concentration of a solid (LaOCl) does not change during the reaction, it is not included in the equilibrium expression. The equilibrium expression for this reaction can be written as:
Kc = [HCl]^2 / [LaCl3] * [H2O]
where [HCl], [LaCl3], and [H2O] are the equilibrium concentrations of HCl, LaCl3, and H2O, respectively.
Plugging in the values we calculated, we get:
Kc = (0.0400 M)^2 / (0.0125 M)(0.0250 M)
Kc = 0.0016 / 0.0003125
Kc ≈ 5.12
Therefore, the value of Kc at 619 degree C is approximately 5.12.