At 25°C, K = 0.090 for the following reaction.
H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g)
Calculate the concentrations of all species at equilibrium for the following case.
1.3 mol pure HOCl is placed in a 1.6 L flask
[HOCl]________
[Cl2O]________
[H2O]________
Thank you so much!
To calculate the concentrations of all species at equilibrium, we need to set up an ICE (Initial, Change, Equilibrium) table. Here's how you can do it step by step:
Step 1: Write the balanced equation for the reaction
H2O(g) + Cl2O(g) ⇌ 2 HOCl(g)
Step 2: Set up the ICE table
Species Initial (mol/L) Change (mol/L) Equilibrium (mol/L)
HOCl ? -2x 1.3 - 2x
Cl2O ? -x
H2O ? -x
Step 3: Apply the equilibrium expression and expression for K
The equilibrium expression for this reaction is:
K = ([HOCl]^2) / ([Cl2O] * [H2O])
Given that K = 0.090 at 25°C, we can set up the equation:
0.090 = (1.3 - 2x)^2 / (x * x)
Step 4: Solve for x
Solve the equation for x by rearranging and solving the quadratic equation:
0.090 * x^2 = (1.3 - 2x)^2
0.090 * x^2 = 1.69 - 5.2x + 4x^2
0.010x^2 + 5.2x - 1.6881 = 0
You can solve this quadratic equation using a calculator or any other preferred method. Once you find the value of x, you can substitute it back into the ICE table to determine the equilibrium concentrations.
Step 5: Calculate the concentrations at equilibrium
[HOCl] = 1.3 - 2x (substitute the value of x)
[Cl2O] = -x (substitute the value of x)
[H2O] = -x (substitute the value of x)
Calculating the concentrations based on the determined value of x will give you the final values for each species at equilibrium.
Remember to convert the final concentration values to mol/L if necessary.
I hope this explanation helps you understand how to solve this equilibrium problem! Let me know if you have any further questions.