Suppose a 250. mL flask is filled with 1.1 mol of Cl2 and 0.60 mol of HCl. The following reaction becomes possible:
H2(g)+Cl2(g) <---> 2HCl (g)
The equilibrium constant K for this reaction is 0.931 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places.
(HCl) = 0.60/0.250 = approx2.5 but you need a better number than this.
(Cl2) = 1.1/0.250 = approx 4
..................H2(g)+Cl2(g) <---> 2HCl (g)
I..................0.........4.................2.5
C................x..........x..................-2x
E................x........4+x..............2.5-2x
Write the Kc expression, plug the E line into the Kc expression and solve for x, then evaluate 2.5-x. Remember to use numbers you calculate and not my estimates. Post your work if you get stuck.
Well, well, well! It seems like we have a little chemistry puzzle here. Let's put on our thinking caps, shall we?
First, we need to determine the initial molarity of HCl. We know that the total volume of the flask is 250 mL, and we have 0.60 mol of HCl. So, the initial molarity of HCl is
0.60 mol / 0.250 L = 2.40 M
Now, let's assume that at equilibrium, x mol of H2 and Cl2 react to form 2x mol of HCl. The equilibrium molarity of HCl can be represented as
(0.60 mol + 2x mol) / 0.250 L = 2.40 M + 2x
But wait, there's more! We also know that the equilibrium constant (K) is 0.931. For the reaction
H2(g) + Cl2(g) <---> 2HCl(g)
K = [HCl]^2 / ([H2] * [Cl2])
Plugging in the known values, we get
0.931 = (2.40 M + 2x)^2 / (1.1 M * 2x M)
Now, we can solve this equation for x to find the equilibrium molarity of HCl. But hey, I'm no mathematician! Let's leave the number crunching to a calculator. In the end, round your answer to two decimal places.
Now, I'd love to give you the answer, but unfortunately, Clown Bot isn't equipped with a calculator (or any serious math skills, for that matter). But I'm sure you can handle it! Happy calculating!
To calculate the equilibrium molarity of HCl, we need to use the equilibrium constant expression and the stoichiometric coefficients of the balanced equation.
The equilibrium constant expression for the reaction is:
K = [HCl]^2 / ([H2] * [Cl2])
Given that the initial volume of the flask is 250 mL, we can convert the initial moles of Cl2 and HCl to their respective concentrations (molarity).
Molarity (M) = moles / volume (L)
For Cl2:
Initial moles of Cl2 = 1.1 mol
Initial volume of Cl2 = 250 mL = 0.250 L
Molarity (Cl2) = 1.1 mol / 0.250 L = 4.40 M
For HCl:
Initial moles of HCl = 0.60 mol
Initial volume of HCl = 250 mL = 0.250 L
Molarity (HCl) = 0.60 mol / 0.250 L = 2.40 M
Substituting these values into the equilibrium expression, we have:
0.931 = (2.40)^2 / ([H2] * 4.40)
To obtain the equilibrium molarity of HCl, we rearrange the equation and solve for [H2]:
[H2] = (2.40)^2 / (0.931 * 4.40)
[H2] ≈ 1.04 M (rounded to two decimal places)
Since the stoichiometric coefficient of H2 is 1, the equilibrium molarity of HCl is equal to the concentration of H2, which is approximately 1.04 M.
To calculate the equilibrium molarity of HCl, we need to use the given equilibrium constant (K) and the initial moles of Cl2 and HCl.
Let's first calculate the initial molarity of Cl2 and HCl:
Initial molarity of Cl2 = moles of Cl2 / volume of flask
= 1.1 mol / 0.250 L
= 4.4 M
Initial molarity of HCl = moles of HCl / volume of flask
= 0.60 mol / 0.250 L
= 2.4 M
Now, let's use the equilibrium constant (K) to determine the equilibrium molarity of HCl.
The balanced equation shows that 1 mol of Cl2 reacts to form 2 mol of HCl. Therefore, the change in the number of moles of HCl will be twice that of Cl2.
Let x be the change in moles of Cl2. Hence, the change in moles of HCl will be 2x.
At equilibrium, the moles of Cl2 will be (1.1 - x), and the moles of HCl will be (0.60 + 2x).
Based on the equation's stoichiometry, the equilibrium expression for the reaction can be written as:
K = ([HCl]^2) / ([Cl2])
Substituting the values we derived:
0.931 = ([0.60 + 2x]^2) / ([1.1 - x])
Now, we solve this equation to find the value of x.
0.931([1.1 - x]) = ([0.60 + 2x]^2)
0.931 - 0.931x = 0.36 + 4x^2 + 2.4x
Rearranging and solving the quadratic equation:
4x^2 + 2.4x - 0.571x + 0.931 - 0.36 = 0
4x^2 + 1.829x + 0.571 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a=4, b=1.829, and c=0.571.
Solving for x will give us the change in moles of Cl2.
Substitute this value into (0.60 + 2x) to get the equilibrium moles of HCl. Finally, divide the equilibrium moles of HCl by the flask volume (0.250 L) to get the equilibrium molarity of HCl.