Question

0.400 mol of HI (g) was placed in a 1.00 L flask that had no NO2 (g) initially. After equilibration occurred, the concentration of HI (g) was found to be 0.328 M. What is the value of Kc for this reaction?

Answer 1

The equilibrium constant (Kc) for the reaction is 0.01202.

Step-by-step explanation:

The reaction is:

2HI (g) <=> H
`_2` (g) + I
`_2` (g)

The equilibrium constant expression for this reaction is:

Kc = [H
`_2`][I
`_2`] / [HI
`]^2`

where:

[H
`_2`] is the concentration of H
`_2` (g) at equilibrium

[I
`_2`] is the concentration of I
`_2` (g) at equilibrium

[HI] is the concentration of HI (g) at equilibrium

We can calculate the equilibrium concentrations of HI, H2, and I2 using the following ICE table:

HI H
`_2` I
`_2`

Initial 0.400 M 0.00 M 0.00 M

Change -0.072 M +0.036 M +0.036 M

Equilibrium 0.328 M 0.036 M 0.036 M

Now we can plug the equilibrium concentrations into the equilibrium constant expression to solve for Kc:

Kc = ([0.036
`]^2`) / (0.328
`)^2`

Kc = 0.01202.

Therefore, the equilibrium constant (Kc) for the reaction is 0.01202.