Determine the concentration at equilibrium if you start with 2.3 grams of Hydrogen and 200grams of Iodine in a 2.3 liter container. If you now add an extra .25M of HI after equilibrium, calculate Qc. Recalculate now what the concentrations should be at equilibrium.
H2(g)+I2(g)<-->2HI(g)
Kc=56
To determine the concentration at equilibrium, we first need to calculate the initial moles of each species.
Given:
Mass of Hydrogen (H2) = 2.3 grams
Molar mass of H2 = 2 g/mol
Using the formula: Moles = Mass / Molar mass
Moles of H2 = 2.3 grams / 2 g/mol = 1.15 mol
Similarly:
Mass of Iodine (I2) = 200 grams
Molar mass of I2 = 254 g/mol
Moles of I2 = 200 grams / 254 g/mol ≈ 0.7874 mol
Since no HI is present initially, the initial moles of HI is 0.
Now let's calculate the initial concentrations (M) for each species using the volume of the container.
Given:
Volume of the container = 2.3 liters
Initial concentration of H2 (initial moles / volume) = 1.15 mol / 2.3 L ≈ 0.5 M
Initial concentration of I2 (initial moles / volume) = 0.7874 mol / 2.3 L ≈ 0.342 M
Initial concentration of HI = 0 M
Now, we will calculate the value of Qc using the equilibrium concentrations.
Qc = [HI]^2 / [H2][I2]
Let's assume at equilibrium, the concentrations of H2, I2, and HI are x M, x M, and 2x M respectively.
Substituting the values into the Qc expression:
Qc = (2x)^2 / (x)(x) = 4x^2 / x^2 = 4
Since Kc = 56 (given), we compare Qc and Kc to determine the direction in which the reaction will proceed.
If Qc < Kc, the reaction will proceed forward (towards the product side) to reach equilibrium.
If Qc > Kc, the reaction will proceed backward (towards the reactant side) to reach equilibrium.
If Qc = Kc, the reaction is already at equilibrium.
In this case, Qc = 4 and Kc = 56. Since Qc < Kc, the reaction will proceed forward (towards the product side) to reach equilibrium.
To calculate the concentrations at equilibrium, we will first set up an ICE (Initial, Change, Equilibrium) table:
H2(g) + I2(g) ⇌ 2HI(g)
-----------------------------------
Initial | 0.5 M | 0.342 M | 0 M
Change | -x | -x | +2x
Equilibrium | 0.5-x | 0.342-x | 2x
Since the reaction proceeds forward, the change in concentration for H2, I2, and HI will be -x, -x, and +2x respectively.
At equilibrium, the concentrations are given by the initial concentrations minus the changes, as shown in the table.
Now, we can set up an expression for Qc using the concentrations at equilibrium:
Qc = [HI]^2 / [H2][I2]
Qc = (2x)^2 / (0.5 - x)(0.342 - x)
Since Qc = 4, we can substitute this into the expression and solve for x:
4 = (2x)^2 / (0.5 - x)(0.342 - x)
Cross multiply:
4(0.5 - x)(0.342 - x) = (2x)^2
Simplify:
4(0.171 - 0.842x + x^2) = 4x^2
Expand and rearrange:
0.684 - 3.368x + 4x^2 = 4x^2
Combine like terms:
0.684 - 3.368x = 0
Solve for x:
3.368x = 0.684
x ≈ 0.203 M
Substitute this value of x back into the equilibrium expressions to find the concentrations at equilibrium:
[H2] = 0.5 - x ≈ 0.5 - 0.203 ≈ 0.297 M
[I2] = 0.342 - x ≈ 0.342 - 0.203 ≈ 0.139 M
[HI] = 2x ≈ 2(0.203) ≈ 0.406 M
Therefore, the concentrations at equilibrium are approximately:
[H2] ≈ 0.297 M
[I2] ≈ 0.139 M
[HI] ≈ 0.406 M