For the reaction H2(g) + I2(g) ¡ê 2 HI(g), you have the initial concentrations [H2] = 0.15 and [I2] = 0.05. Keq for the reaction at this temperature is 4.5 x 10-6. Make a reaction table. Include rows for initial concentration, change in concentration, and equilibrium concentration. Write down the equation for Keq. Solve for x. What are the equilibrium concentrations for H2, I2, and HI?

Question

For the reaction H2(g) + I2(g) ¡ê 2 HI(g), you have the initial concentrations [H2] = 0.15 and [I2] = 0.05. Keq for the reaction at this temperature is 4.5 x 10-6. Make a reaction table. Include rows for initial concentration, change in concentration, and equilibrium concentration. Write down the equation for Keq. Solve for x. What are the equilibrium concentrations for H2, I2, and HI?

Answer 1

That's 0.15 WHAT and 0.05 WHAT? You add whatever it is. It is either mols, molar, or atm pressure. I will assume it is M since you say "concentrations." but you can change that as needed.

..........H2 + I2 ==> 2HI
I......0.15M...0.05M...0
C........-x.....-x.....2x
E......0.15-x...0.05-x..2x

Keq = 4.5E-6 = (HI)^2/(H2)(I2)
Substitute from the ICE chart and solve for x and the concns of each. Post your work if you get stuck.

Answer 2

heeeeellppppppp

Answer 3

To make a reaction table for this equilibrium reaction, we need to consider the initial concentrations, changes in concentrations, and the equilibrium concentrations for the reactants and products.

The balanced equation for this reaction is:

H2(g) + I2(g) ⇌ 2 HI(g)

Let's start by filling in the initial concentrations:

[H2] = 0.15 M
[I2] = 0.05 M
[HI] = 0 M (since it is a product and not present initially)

Next, we need to determine the change in concentration for each species. Since the stoichiometry of the reaction is 1:1:2 for H2, I2, and HI respectively, the change in concentration for H2 and I2 would be -x (as they react) and the change for HI would be +2x (as it is produced in a 2:1 ratio).

For the equilibrium concentrations, we need to subtract/add the changes from the initial concentrations. Therefore:

[H2] = 0.15 - x
[I2] = 0.05 - x
[HI] = 2x

The equation for the equilibrium constant (Keq) in this reaction is given by:

Keq = [HI]² / ([H2] * [I2])

Substituting the equilibrium concentrations:

Keq = (2x)² / ((0.15 - x) * (0.05 - x))

To solve for x, we can use the given Keq value:

4.5 x 10^-6 = (2x)² / ((0.15 - x) * (0.05 - x))

Using algebraic methods, we can solve for x. After solving for x, we can substitute its value back into the equilibrium concentrations equations to find the equilibrium concentrations for H2, I2, and HI.