A 5.00L reaction vessel is filled with 1.00mol of H2, 1.00mol of I2, and 2.50mol of HI. If equilibrium constant for the reaction is 129 at 500k, what are the equilibrium concentrations of all species?
H2(g) + I2(g) <-> 2HI(g) Kc = 129
(H2) = 1.00/5.00 = 0.2M
(I2) = 1.00/5.00 = 0.2M
(HI) = 2.50/5.00 = 0.5M
.........H2 + I2 ==> 2HI
I........0.2...0.2...0.5
C........-x....-x....+2x
E.......0.2-x..0.2-x..0.5+2x
Kc = (0.5+2x)^2/(0.2-x)(0.2-x)
Note: In the C line, how do we know it is -x. -x, and +2x instead of +x +x, and -2x;i.e., how do we which way the reaction will occur. You run a Qrxn = (HI)^2/(H2)(I2)= (0.5)^2/(0.2)^2 = about 6. Compare with K = 129 and that means the numerator is too small and denominator is too large so the reaction must go to the right.
What does x equal out too?
To find the equilibrium concentrations of all species, we need to use the concept of the equilibrium constant and the stoichiometry of the reaction.
The equilibrium constant (Kc) expression for the given reaction is:
Kc = [HI]^2 / ([H2] * [I2])
We are given the values of the equilibrium constant (Kc = 129) and the initial moles of each species. We can use these values to determine the equilibrium concentrations.
Step 1: Calculate the initial concentration of each species.
The initial concentration of a species can be calculated using the formula: initial concentration = moles / volume.
Initial concentration of H2: 1.00 mol / 5.00 L = 0.20 M
Initial concentration of I2: 1.00 mol / 5.00 L = 0.20 M
Initial concentration of HI: 2.50 mol / 5.00 L = 0.50 M
Step 2: Set up an expression for the equilibrium concentrations.
Let's assume that the equilibrium concentrations of H2, I2, and HI are x, x, and 2x respectively.
Therefore, the equilibrium concentrations are:
[H2] = x M
[I2] = x M
[HI] = 2x M
Step 3: Write the equilibrium constant expression using the equilibrium concentrations:
Kc = ([HI]^2) / ([H2] * [I2])
Substituting the equilibrium concentrations from Step 2:
129 = (2x)^2 / (x * x)
Step 4: Solve the equation to find the value of x.
129 = 4x^2 / x^2
129 = 4
Since 129/4 is greater than 1, this equation does not have a real solution. It means the reaction does not reach equilibrium under the given conditions.
Therefore, there are no equilibrium concentrations for this reaction at 500 K.
h2, i2 = 11.4
hi = 22.8