At a different temperature (this means that Keq will be different than part a)), 6.0 mol of IF5 and 8.0 mol of I4F2 are placed in a 10.0 L container. At equilibrium, 6.0mol of I4F2 are left. Calculate the Keq for the new temperature.

Keq for the new temperature is 26.8

Explanation:

Let's propose the equilibrium:

2IF₅ +  I₄F₂  ⇄  3I₂  +  6F₂

Now we propose the situations:

                2IF₅     +      I₄F₂    ⇄    3I₂   +     6F₂

Initial      6 mol           8 mol           -              -

Initially we added 6 mol and 8 mol of our reactants  

React.        x                 x/2           3/2x         3x

By stoichiometry x amount has reacted, so a half of x react to the I₄F₂ and we finally produced 3/2x and 3x in the product side

Eq.          (6 - x)        (8 - x/2)        3/2x          3x

Notice we have the concentration left for the I₄F₂, so we can find the x value, the amount that has reacted:

8 - x/2 = 6

x = 4, so the concentrations in the equilibrium are:

2 moles of IF₅, 6 moles I₄F₂, 6 moles of I₂ and 12 moles of F₂

As we need molar concentration to determine Keq, we must divide the moles by the volume of the container:

2/10 = [IF₅] → 0.2 M

6/10 = [I₄F₂] → 0.6 M

6/10 = [I₂] → 0.6 M

12/10 = [F₂] → 1.2 M

Let's make, expression for Keq:

Keq = ([I₂]³ . [F₂]⁶) / [IF₅]² . [I₄F₂]

Keq = 0.6³ . 1.2⁶ / 0.2² . 0.6 → 26.8

it's c. elements are placed in rows based on how may valencies electrons have.

the product of the pressure and volume is also constant