Suppose that 2.00 moles of HCl in a 1.00L glass flask slowly decomposes into H2 and Cl2. When equilibrim is reached, the concentrations of H2 and Cl2 are both 0.214M. What is the Keq?
.........2HCl ==> H2 + Cl2
I........2M........0.....0
C........-2x.......x.....x
E.......2-2x.....0.214..0.214
Therefore, at equilibrium, H2 = 0.214, Cl2 = 0.214 and HCl = 2-2*0.214. Substitute into Keq expression and evaluate for Keq.
Keq = product/ reactant
Write the balanced equation for what you are given. 2HCl--->H2 + Cl2
I believe that's how you start off.
Ah, a chemistry question! Well, let me clown around with that one for a bit.
So, we have HCl magically turning into H2 and Cl2. Quite the magical trick, I must say! Anyway, let's focus on the equilibrium concentrations.
We are told that at equilibrium, the concentrations of H2 and Cl2 are both 0.214 M. Now, we need to find the Keq. Keq, also known as the equilibrium constant, is the ratio of the concentration of products to the concentration of reactants at equilibrium.
The balanced chemical equation for the reaction is:
2 HCl -> H2 + Cl2
So, the stoichiometry tells us that for every 2 moles of HCl that react, we get 1 mole of H2 and 1 mole of Cl2.
Now, at equilibrium, the concentration of H2 is 0.214 M, just like the concentration of Cl2. Since the stoichiometric ratio is 1:1:1, this means that the concentration of HCl at equilibrium is also 0.214 M.
To calculate Keq, we divide the concentration of products by the concentration of reactants. In this case, we have:
Keq = [H2] * [Cl2] / [HCl]
Plugging in the values, we get:
Keq = (0.214 M) * (0.214 M) / (0.214 M)
Now, what do we get when we divide something by itself? Exactly! We get 1.
So, my clown friend, the Keq for this magical HCl decomposition is 1.
I hope this clowning chemistry explanation brings a smile to your face!
To find the equilibrium constant (Keq) for the given reaction, we need to set up the balanced equation for the decomposition of HCl:
2 HCl (g) ⇌ H2 (g) + Cl2 (g)
The balanced equation tells us that 2 moles of HCl decompose to form 1 mole of H2 and 1 mole of Cl2.
Given that 2.00 moles of HCl decompose in a 1.00 L flask, we can calculate the initial concentration of HCl as follows:
Initial concentration of HCl = moles of HCl / volume of flask
= 2.00 mol / 1.00 L
= 2.00 M
At equilibrium, the concentrations of H2 and Cl2 are both 0.214 M.
Now, we can write the expression for the equilibrium constant (Keq) using the concentrations of the products and reactants:
Keq = [H2] * [Cl2] / [HCl]^2
Substituting the given concentrations into the equation:
Keq = (0.214 M) * (0.214 M) / (2.00 M)^2
Calculating this expression:
Keq = 0.045596 M^2 / 4.00 M^2
= 0.011399
Therefore, the value of Keq for the given reaction is approximately 0.0114.
To find the value of Keq, you need to understand the concept of equilibrium and the equilibrium expression. Keq, also known as the equilibrium constant, is a measure of the position of an equilibrium reaction.
In this case, the balanced equation for the decomposition of HCl is:
2 HCl ⇌ H2 + Cl2
The equilibrium expression for this reaction can be written as:
Keq = [H2]^a [Cl2]^b / [HCl]^c
Here, a, b, and c represent the stoichiometric coefficients of the respective species in the balanced equation. In this case, the stoichiometric coefficients are 1 for H2, 1 for Cl2, and 2 for HCl.
Given that the equilibrium concentrations of H2 and Cl2 are both 0.214 M, and starting with 2.00 moles of HCl in a 1.00 L flask, you can determine the equilibrium concentration of HCl as follows:
[HCl] = (Initial moles of HCl - Moles of HCl reacted) / Volume of flask
Since 2 moles of HCl are initially present and the stoichiometry of the balanced equation is 1:2 (HCl:H2/Cl2), 1 mole of HCl reacts to form 1 mole each of H2 and Cl2. Therefore, the moles of HCl reacted will also be 1 mole.
[HCl] = (2 - 1) moles / 1 L = 1 M
Now you have all the values to substitute them into the equilibrium expression:
Keq = [H2]^a [Cl2]^b / [HCl]^c
= (0.214)^1 (0.214)^1 / (1)^2
= 0.0458
Therefore, the value of Keq for this reaction is approximately 0.0458.