At 2,000 K, the equilibrium constant for the reaction below is Kc = 3.92.
Br2(g) ⇌ 2 Br(g)
What is the percent dissociation of Br2 at 2,000 K, if the initial concentration of Br2 is 2.25 mol L−1 ?
To find the percent dissociation of Br2 at 2,000 K, we need to determine the concentrations of Br2 and Br(g) at equilibrium using the given equilibrium constant (Kc) and initial concentration of Br2.
The equilibrium constant expression for the reaction is: Kc = [Br(g)]^2 / [Br2(g)]
Let's assign the concentration of Br2 at equilibrium as x mol/L. Since 2 moles of Br(g) are formed for every 1 mole of Br2 that reacts, the concentration of Br(g) at equilibrium will be 2x mol/L.
Using these concentrations, we can rewrite the equilibrium constant expression as: Kc = (2x)^2 / x
Substituting the given value for Kc (3.92), we have: 3.92 = (2x)^2 / x
Next, let's solve the equation for x. We can multiply both sides of the equation by x to remove the denominator and rearrange the equation as a quadratic equation:
3.92x = 4x^2
Rearranging, we get: 4x^2 - 3.92x = 0
Factoring, we find: x(4x - 3.92) = 0
This yields two possible solutions: x = 0 and x = 0.98
Since the concentration of a reactant cannot be zero at equilibrium, we discard x = 0 as a valid solution. Therefore, the concentration of Br2 at equilibrium is 0.98 mol/L.
To find the percent dissociation, we can use the formula:
Percent dissociation = (change in concentration / initial concentration) x 100
In this case, the change in concentration is given by the difference between the initial concentration and the concentration at equilibrium: (2.25 - 0.98) mol/L = 1.27 mol/L
Therefore, the percent dissociation is:
Percent dissociation = (1.27 / 2.25) x 100 = 56.44%
So, the percent dissociation of Br2 at 2,000 K, with an initial concentration of 2.25 mol/L, is approximately 56.44%.
To find the percent dissociation of Br2 at 2,000 K, we need to calculate the concentration of Br2 at equilibrium and then determine the percentage that has dissociated.
First, let's set up an ICE table to track the changes in concentration:
Br2(g) ⇌ 2 Br(g)
Initial: 2.25 mol L^(-1) 0 mol L^(-1)
Change: -x mol L^(-1) +2x mol L^(-1)
Equilibrium: 2.25 - x mol L^(-1) 2x mol L^(-1)
Since the equilibrium constant (Kc) is given as 3.92, we can use it to set up the equation for the equilibrium expression:
Kc = [Br]^2 / [Br2]
Substituting the equilibrium concentrations from the ICE table, we can write:
3.92 = (2x)^2 / (2.25 - x)
Now, let's solve this equation for x:
3.92 (2.25 - x) = (2x)^2
8.82 - 3.92x = 4x^2
4x^2 + 3.92x - 8.82 = 0
Solving this quadratic equation, we find two possible values for x: x ≈ 0.814 or x ≈ -2.164. Since the concentration cannot be negative, we discard the negative solution and consider x ≈ 0.814 as the concentration of Br2 at equilibrium.
Now, let's calculate the percent dissociation:
Percent Dissociation = (Change in concentration / Initial concentration) x 100
Percent Dissociation = (0.814 mol L^(-1) / 2.25 mol L^(-1)) x 100
Percent Dissociation ≈ 36.2%
Therefore, the percent dissociation of Br2 at 2,000 K, with an initial concentration of Br2 as 2.25 mol L^(-1), is approximately 36.2%.
...................Br2(g) ⇌ 2 Br(g)
I.....................2.25........0
C.......................-x.........2x
E.................2.25-x.........2x
Kc = (Br)^2/(Br)
Plug the E line into Kc expression and solve for (Br).
Then % dissoc = [(Br)/2.25]*100