For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.570 mol IBr in a 2.50-L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2?
M = mols/L = 0.570/2.50L = about 0.228M
........I2 + Br2 ==> 2IBr
I.......0....0......0.228
C.......x....x.....0.228-2x
E.......x....x.....0.228-2x
Substitute the E line into Kc expression and solve.
do u have to use the quadratic to find x?
what do i do after i get
.051984+4x^2 - .912/(x^2)
.051984+4x^2 - .912x=280(x^2)
then subtract the 4x^2? so
.051984-.912x=276x^2
now what?
To here is ok.
.051984+4x^2 - .912x=280(x^2)
Just rearrange to
0.05918 + 4x^2 - 0.912x = 280x^2
Then 280x^2-4x^2 + 0.912x - 0.05198 = 0 and
276x^2+0.912x-0.05198 = 0 and use the quadratic formula. I have x = 0.0122 or close to that.
To determine the equilibrium concentrations of IBr, I2, and Br2, we can use an ICE table and apply the equilibrium expression. Here's how you can calculate the equilibrium concentrations:
1. Start by writing the balanced chemical equation for the reaction:
I2(g) + Br2(g) ⇌ 2 IBr(g)
2. Since the initial concentration of IBr is given as 0.570 mol in a 2.50-L flask, we can calculate the initial concentration of IBr:
Initial concentration of IBr = 0.570 mol / 2.50 L
3. Use an ICE table to track the changes in concentration:
I2(g) + Br2(g) ⇌ 2 IBr(g)
Initial: 0 0 0.570 mol / 2.50 L
Change: -x -x +2x
Equilibrium: 0 - x 0 - x 0.570 mol / 2.50 L + 2x
4. The expression for the equilibrium constant (Kc) is given as:
Kc = [IBr]^2 / ([I2] * [Br2])
5. Substitute the equilibrium concentrations into the equilibrium expression:
Kc = (0.570 mol / 2.50 L + 2x)^2 / ((0 mol-x) * (0 mol-x))
6. Simplify the equation by expanding the numerator:
Kc = (0.3228 + 0.912x + 4x^2) / (x^2)
7. Rearrange the equation to bring all terms to one side:
280x^2 - 4.56x + 0.3228 = 0
8. Solve the quadratic equation using the quadratic formula or factoring methods. In this case, we can use the quadratic formula:
x = [-(-4.56) ± √((-4.56)^2 - 4(280)(0.3228))] / (2 * 280)
Solving this equation will give us two values for 'x', but since 'x' represents a decrease in concentration, we will consider the smaller positive value.
9. Once you have determined the value of x, substitute it back into the expressions in the ICE table to find the equilibrium concentrations:
Equilibrium concentration of IBr: 0.570 mol / 2.50 L + 2x
Equilibrium concentration of I2: 0 mol - x
Equilibrium concentration of Br2: 0 mol - x
This calculation will give you the equilibrium concentrations of IBr, I2, and Br2 in mol/L units.